RC and RL Circuits with Piecewise Constant Sources
M. B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)
β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources
C
A
B
v
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
C
A
B
i
v
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + v β VTh = RThCdv
dt+ v β dv
dt+
v
RThC=
VTh
RThC.
* Homogeneous solution:
dv
dt+
1
Οv = 0 , where Ο = RTh C is the βtime constant.β
(V
Coul/secΓ Coul
V
)β dv
v= β dt
Οβ log v = β t
Ο+ K0 β v (h) = exp [(βt/Ο) + K0] = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making i = 0, and we get v (p) = VTh as a particular solution (whichhappens to be simply a constant).
* v = v (h) + v (p) = K exp(βt/Ο) + VTh .
* In general, v(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on v .
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
β‘ VTh
* If all sources are DC (constant), we havev(t) = A exp(βt/Ο) + B , Ο = RThC .
* i(t) = Cdv
dt= C Γ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, i β 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
β‘ VTh
* If all sources are DC (constant), we havev(t) = A exp(βt/Ο) + B , Ο = RThC .
* i(t) = Cdv
dt= C Γ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, i β 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
β‘ VTh
* If all sources are DC (constant), we havev(t) = A exp(βt/Ο) + B , Ο = RThC .
* i(t) = Cdv
dt= C Γ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, i β 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits with DC sources (continued)
C C
A
B
A
B
i
vv
Circuit(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
RTh
β‘ VTh
* If all sources are DC (constant), we havev(t) = A exp(βt/Ο) + B , Ο = RThC .
* i(t) = Cdv
dt= C Γ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, i β 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
Plot of f (t)= eβt/Ο
t/Ο eβt/Ο 1β eβt/Ο
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787Γ10β2 0.9502
4.0 1.8315Γ10β2 0.9817
5.0 6.7379Γ10β3 0.9933
* For t/Ο = 5, eβt/Ο ' 0, 1β eβt/Ο ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/Ο
exp(βx)
1β exp(βx)
M. B. Patil, IIT Bombay
Plot of f (t)= eβt/Ο
t/Ο eβt/Ο 1β eβt/Ο
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787Γ10β2 0.9502
4.0 1.8315Γ10β2 0.9817
5.0 6.7379Γ10β3 0.9933
* For t/Ο = 5, eβt/Ο ' 0, 1β eβt/Ο ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/Ο
exp(βx)
1β exp(βx)
M. B. Patil, IIT Bombay
Plot of f (t)= eβt/Ο
t/Ο eβt/Ο 1β eβt/Ο
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787Γ10β2 0.9502
4.0 1.8315Γ10β2 0.9817
5.0 6.7379Γ10β3 0.9933
* For t/Ο = 5, eβt/Ο ' 0, 1β eβt/Ο ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/Ο
exp(βx)
1β exp(βx)
M. B. Patil, IIT Bombay
Plot of f (t)= eβt/Ο
t/Ο eβt/Ο 1β eβt/Ο
0.0 1.0 0.0
1.0 0.3679 0.6321
2.0 0.1353 0.8647
3.0 4.9787Γ10β2 0.9502
4.0 1.8315Γ10β2 0.9817
5.0 6.7379Γ10β3 0.9933
* For t/Ο = 5, eβt/Ο ' 0, 1β eβt/Ο ' 1.
* We can say that the transient lasts for about 5 time constants.
0
1
0 1 2 3 4 5 6
x= t/Ο
exp(βx)
1β exp(βx)
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
Plot of f (t)=Aeβt/Ο + B
* At t = 0, f =A + B.
* As t ββ, f β B.
* The graph of f (t) lies between (A + B) and B.
Note: If A > 0, A + B > B. If A < 0, A + B < B.
* At t = 0,df
dt= Aeβt/Ο
(β 1
Ο
)= β A
Ο.
If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.
* As t ββ,df
dtβ 0, i.e., f becomes constant (equal to B).
0 5 Ο t
A < 0
A+ B
B
0 5 Ο t
A > 0
A+ B
B
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L
i
v
A
B
L
RTh
β‘ VTh
* If all sources are DC (constant), VTh = constant .
* KVL: VTh = RTh i + Ldi
dt.
* Homogeneous solution:
di
dt+
1
Οi = 0 , where Ο = L/RTh
β i (h) = K exp(βt/Ο) .
* Particular solution is a specific function that satisfies the differential equation. We know that all timederivatives will vanish as t ββ , making v = 0, and we get i (p) = VTh/RTh as a particular solution(which happens to be simply a constant).
* i = i (h) + i (p) = K exp(βt/Ο) + VTh/RTh .
* In general, i(t) = A exp(βt/Ο) + B , where A and B can be obtained from known conditions on i .
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
β‘ VTh
* If all sources are DC (constant), we havei(t) = A exp(βt/Ο) + B , Ο = L/RTh .
* v(t) = Ldi
dt= LΓ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, v β 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
β‘ VTh
* If all sources are DC (constant), we havei(t) = A exp(βt/Ο) + B , Ο = L/RTh .
* v(t) = Ldi
dt= LΓ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, v β 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
β‘ VTh
* If all sources are DC (constant), we havei(t) = A exp(βt/Ο) + B , Ο = L/RTh .
* v(t) = Ldi
dt= LΓ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, v β 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RL circuits with DC sources (continued)
i
v
A
B
v
CircuitA
B
(resistors,voltage sources,current sources,CCVS, CCCS,VCVS, VCCS)
i
L L
RTh
β‘ VTh
* If all sources are DC (constant), we havei(t) = A exp(βt/Ο) + B , Ο = L/RTh .
* v(t) = Ldi
dt= LΓ A exp(βt/Ο)
(β 1
Ο
)β‘ Aβ² exp(βt/Ο) .
* As t ββ, v β 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.
* Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can beexpressed asx(t) = K1 exp(βt/Ο) + K2 ,where K1 and K2 can be obtained from suitable conditions on x(t).
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: Can Vc change βsuddenly?β
i
0 V t
5 V
Vs
Vs
C= 1Β΅FVc
R= 1 k
Vc(0)= 0V
* Vs changes from 0V (at t = 0β), to 5V (at t = 0+). As a result of this change, Vc will rise. How fastcan Vc change?
* For example, what would happen if Vc changes by 1V in 1Β΅s at a constant rate of 1V /1Β΅s = 106 V /s?
* i = CdVc
dt= 1Β΅F Γ 106 V
s= 1A .
* With i = 1A, the voltage drop across R would be 1000V ! Not allowed by KVL.
* We conclude that Vc (0+) =Vc (0β)β A capacitor does not allow abrupt changes in Vc if there is a finiteresistance in the circuit.
* Similarly, an inductor does not allow abrupt changes in iL.
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
R
Vs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
v(0β) = Vs(0β) = 0 V
v(0+) β v(0β) = 0 V
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
As t β β , i β 0 β v(β) = Vs(β) = V0
Imposing (1) and (2) on Eq. (A), we get
i.e., B = V0 ,A = βV0
t = 0+: 0 = A+ B ,
t β β: V0 = B .
v(t) = V0 [1β exp(βt/Ο)]
0 V
i
t
Cv
RVs
Vs
V0
(A)Let v(t) = A exp(βt/Ο) + B, t > 0
(1)
(2)
Conditions on v(t):
Note that we need the condition at 0+ (and not at 0β)
because Eq. (A) applies only for t > 0.
v(0β) = Vs(0β) = V0
v(0+) β v(0β) = V0
As t β β , i β 0 β v(β) = Vs(β) = 0 V
Imposing (1) and (2) on Eq. (A), we get
t = 0+: V0 = A+ B ,
i.e., A = V0 ,B = 0
t β β: 0 = B .
v(t) = V0 exp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1β exp(βt/Ο)]
=CV0
Οexp(βt/Ο) =
V0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t = 0+: v = 0 , Vs = V0 β i(0+) = V0/R .
t β β: i(t) = 0 .
Aβ² =V0
R, Bβ² = 0 β i(t) =
V0
Rexp(βt/Ο)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(βt/Ο)]
= βCV0
Οexp(βt/Ο) = βV0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t β β: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 β i(0+) = βV0/R .
Aβ² = βV0
R, Bβ² = 0 β i(t) = βV0
Rexp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1β exp(βt/Ο)]
=CV0
Οexp(βt/Ο) =
V0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t = 0+: v = 0 , Vs = V0 β i(0+) = V0/R .
t β β: i(t) = 0 .
Aβ² =V0
R, Bβ² = 0 β i(t) =
V0
Rexp(βt/Ο)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(βt/Ο)]
= βCV0
Οexp(βt/Ο) = βV0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t β β: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 β i(0+) = βV0/R .
Aβ² = βV0
R, Bβ² = 0 β i(t) = βV0
Rexp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1β exp(βt/Ο)]
=CV0
Οexp(βt/Ο) =
V0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t = 0+: v = 0 , Vs = V0 β i(0+) = V0/R .
t β β: i(t) = 0 .
Aβ² =V0
R, Bβ² = 0 β i(t) =
V0
Rexp(βt/Ο)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(βt/Ο)]
= βCV0
Οexp(βt/Ο) = βV0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t β β: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 β i(0+) = βV0/R .
Aβ² = βV0
R, Bβ² = 0 β i(t) = βV0
Rexp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1β exp(βt/Ο)]
=CV0
Οexp(βt/Ο) =
V0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t = 0+: v = 0 , Vs = V0 β i(0+) = V0/R .
t β β: i(t) = 0 .
Aβ² =V0
R, Bβ² = 0 β i(t) =
V0
Rexp(βt/Ο)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(βt/Ο)]
= βCV0
Οexp(βt/Ο) = βV0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t β β: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 β i(0+) = βV0/R .
Aβ² = βV0
R, Bβ² = 0 β i(t) = βV0
Rexp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1β exp(βt/Ο)]
=CV0
Οexp(βt/Ο) =
V0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t = 0+: v = 0 , Vs = V0 β i(0+) = V0/R .
t β β: i(t) = 0 .
Aβ² =V0
R, Bβ² = 0 β i(t) =
V0
Rexp(βt/Ο)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(βt/Ο)]
= βCV0
Οexp(βt/Ο) = βV0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t β β: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 β i(0+) = βV0/R .
Aβ² = βV0
R, Bβ² = 0 β i(t) = βV0
Rexp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [1β exp(βt/Ο)]
=CV0
Οexp(βt/Ο) =
V0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t = 0+: v = 0 , Vs = V0 β i(0+) = V0/R .
t β β: i(t) = 0 .
Aβ² =V0
R, Bβ² = 0 β i(t) =
V0
Rexp(βt/Ο)
0 V
i
t
Cv
RVs
Vs
V0
Compute i(t), t > 0 .
(A) i(t) = Cd
dtV0 [exp(βt/Ο)]
= βCV0
Οexp(βt/Ο) = βV0
Rexp(βt/Ο)
Using these conditions, we obtain
(B) Let i(t) = Aβ² exp(βt/Ο) + Bβ² , t > 0 .
t β β: i(t) = 0 .
t = 0+: v = V0 , Vs = 0 β i(0+) = βV0/R .
Aβ² = βV0
R, Bβ² = 0 β i(t) = βV0
Rexp(βt/Ο)
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1Β΅Fv
R=1 kVs
Vs5 V
v(t) = V0 [1β exp(βt/Ο)]
i(t) =V0
Rexp(βt/Ο)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
β2 0 2 4 6 8
t0 V
i
C1Β΅F
v
R=1 kVs
Vs5 V
v(t) = V0 exp(βt/Ο)
i(t) = βV0
Rexp(βt/Ο)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
β5
time (msec)
β2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1Β΅Fv
R=1 kVs
Vs5 V
v(t) = V0 [1β exp(βt/Ο)]
i(t) =V0
Rexp(βt/Ο)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
β2 0 2 4 6 8
t0 V
i
C1Β΅F
v
R=1 kVs
Vs5 V
v(t) = V0 exp(βt/Ο)
i(t) = βV0
Rexp(βt/Ο)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
β5
time (msec)
β2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1Β΅Fv
R=1 kVs
Vs5 V
v(t) = V0 [1β exp(βt/Ο)]
i(t) =V0
Rexp(βt/Ο)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
β2 0 2 4 6 8
t0 V
i
C1Β΅F
v
R=1 kVs
Vs5 V
v(t) = V0 exp(βt/Ο)
i(t) = βV0
Rexp(βt/Ο)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
β5
time (msec)
β2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1Β΅Fv
R=1 kVs
Vs5 V
v(t) = V0 [1β exp(βt/Ο)]
i(t) =V0
Rexp(βt/Ο)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
β2 0 2 4 6 8
t0 V
i
C1Β΅F
v
R=1 kVs
Vs5 V
v(t) = V0 exp(βt/Ο)
i(t) = βV0
Rexp(βt/Ο)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
β5
time (msec)
β2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1Β΅Fv
R=1 kVs
Vs5 V
v(t) = V0 [1β exp(βt/Ο)]
i(t) =V0
Rexp(βt/Ο)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
β2 0 2 4 6 8
t0 V
i
C1Β΅F
v
R=1 kVs
Vs5 V
v(t) = V0 exp(βt/Ο)
i(t) = βV0
Rexp(βt/Ο)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
β5
time (msec)
β2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
t0 V
i
C= 1Β΅Fv
R=1 kVs
Vs5 V
v(t) = V0 [1β exp(βt/Ο)]
i(t) =V0
Rexp(βt/Ο)
0
5
v (
Vo
lts) v
Vs
5
0
i (m
A)
time (msec)
β2 0 2 4 6 8
t0 V
i
C1Β΅F
v
R=1 kVs
Vs5 V
v(t) = V0 exp(βt/Ο)
i(t) = βV0
Rexp(βt/Ο)
v (
Vo
lts)
5
0
v
Vs
i (m
A)
0
β5
time (msec)
β2 0 2 4 6 8
M. B. Patil, IIT Bombay
RC circuits: charging and discharging transients
0 V 0 V
0
5
v (
Volts)
v (
Volts)
5
0
i i
t t
time (msec) time (msec)β1 0 1 2 3 4 5 6 β1 0 1 2 3 4 5 6
C= 1Β΅F C= 1Β΅Fv v
R RVs Vs
Vs Vs5 V 5 V
R = 1 kΞ©
R = 100 Ξ©
R = 1 kΞ©
R = 100 Ξ©
v(t) = V0 exp(βt/Ο)v(t) = V0 [1β exp(βt/Ο)]
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(βt/Ο) + B1 , t < t1 ,x(t) = A2 exp(βt/Ο) + B2 , t1 < t < t2 ,x(t) = A3 exp(βt/Ο) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a βlongβ time(long compared to Ο), all derivatives are zero.
β iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (tβ0 ) , and iL(t+
0 ) = iL(tβ0 ) .
* Compute A1, B1, Β· Β· Β· using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(βt/Ο) + B1 , t < t1 ,x(t) = A2 exp(βt/Ο) + B2 , t1 < t < t2 ,x(t) = A3 exp(βt/Ο) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a βlongβ time(long compared to Ο), all derivatives are zero.
β iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (tβ0 ) , and iL(t+
0 ) = iL(tβ0 ) .
* Compute A1, B1, Β· Β· Β· using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(βt/Ο) + B1 , t < t1 ,x(t) = A2 exp(βt/Ο) + B2 , t1 < t < t2 ,x(t) = A3 exp(βt/Ο) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a βlongβ time(long compared to Ο), all derivatives are zero.
β iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (tβ0 ) , and iL(t+
0 ) = iL(tβ0 ) .
* Compute A1, B1, Β· Β· Β· using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(βt/Ο) + B1 , t < t1 ,x(t) = A2 exp(βt/Ο) + B2 , t1 < t < t2 ,x(t) = A3 exp(βt/Ο) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a βlongβ time(long compared to Ο), all derivatives are zero.
β iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (tβ0 ) , and iL(t+
0 ) = iL(tβ0 ) .
* Compute A1, B1, Β· Β· Β· using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(βt/Ο) + B1 , t < t1 ,x(t) = A2 exp(βt/Ο) + B2 , t1 < t < t2 ,x(t) = A3 exp(βt/Ο) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a βlongβ time(long compared to Ο), all derivatives are zero.
β iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (tβ0 ) , and iL(t+
0 ) = iL(tβ0 ) .
* Compute A1, B1, Β· Β· Β· using the conditions on x(t).
M. B. Patil, IIT Bombay
Analysis of RC/RL circuits with a piece-wise constant source
* Identify intervals in which the source voltages/currents are constant.For example,
(1) t < t1
(3) t > t2
(2) t1 < t < t2
t1 t2
Vs
0 t
* For any current or voltage x(t), write general expressions such as,x(t) = A1 exp(βt/Ο) + B1 , t < t1 ,x(t) = A2 exp(βt/Ο) + B2 , t1 < t < t2 ,x(t) = A3 exp(βt/Ο) + B3 , t > t2 .
* Work out suitable conditions on x(t) at specific time points using
(a) If the source voltage/current has not changed for a βlongβ time(long compared to Ο), all derivatives are zero.
β iC = CdVc
dt= 0 , and VL = L
diL
dt= 0 .
(b) When a source voltage (or current) changes, say, at t = t0 ,Vc (t) or iL(t) cannot change abruptly, i.e.,
Vc (t+0 ) = Vc (tβ0 ) , and iL(t+
0 ) = iL(tβ0 ) .
* Compute A1, B1, Β· Β· Β· using the conditions on x(t).
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ξ©
R2= 40Ξ©
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
Ο = L/RTh
= 0.1 s
= 0.8 H/8Ξ©
RTh = R1 β R2 = 8Ξ©
β i(tβ0 ) = 0 A β i(t+0 ) = 0 A .
At t = tβ0 , v = 0 V, Vs = 0 V .
10 V
t
v(β) = 0 V, i(β) = 10 V/10 Ξ© = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(β), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ξ©
R2= 40Ξ©
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
Ο = L/RTh
= 0.1 s
= 0.8 H/8Ξ©
RTh = R1 β R2 = 8Ξ©
β i(tβ0 ) = 0 A β i(t+0 ) = 0 A .
At t = tβ0 , v = 0 V, Vs = 0 V .
10 V
t
v(β) = 0 V, i(β) = 10 V/10 Ξ© = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(β), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ξ©
R2= 40Ξ©
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
Ο = L/RTh
= 0.1 s
= 0.8 H/8Ξ©
RTh = R1 β R2 = 8Ξ©
β i(tβ0 ) = 0 A β i(t+0 ) = 0 A .
At t = tβ0 , v = 0 V, Vs = 0 V .
10 V
t
v(β) = 0 V, i(β) = 10 V/10 Ξ© = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(β), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ξ©
R2= 40Ξ©
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
Ο = L/RTh
= 0.1 s
= 0.8 H/8Ξ©
RTh = R1 β R2 = 8Ξ©
β i(tβ0 ) = 0 A β i(t+0 ) = 0 A .
At t = tβ0 , v = 0 V, Vs = 0 V .
10 V
t
v(β) = 0 V, i(β) = 10 V/10 Ξ© = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(β), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ξ©
R2= 40Ξ©
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
Ο = L/RTh
= 0.1 s
= 0.8 H/8Ξ©
RTh = R1 β R2 = 8Ξ©
β i(tβ0 ) = 0 A β i(t+0 ) = 0 A .
At t = tβ0 , v = 0 V, Vs = 0 V .
10 V
t
v(β) = 0 V, i(β) = 10 V/10 Ξ© = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(β), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
t0 t1
R2
R1
Vs
Vs
R1= 10Ξ©
R2= 40Ξ©
L= 0.8H
t0=0
t1=0.1 s
Find i(t).
(1) t < t0
(2) t0 < t < t1
(3) t > t1
There are three intervals of constant Vs:
R2
R1
Vs
RTh seen by L is the same in all intervals:
Ο = L/RTh
= 0.1 s
= 0.8 H/8Ξ©
RTh = R1 β R2 = 8Ξ©
β i(tβ0 ) = 0 A β i(t+0 ) = 0 A .
At t = tβ0 , v = 0 V, Vs = 0 V .
10 V
t
v(β) = 0 V, i(β) = 10 V/10 Ξ© = 1 A .
If Vs did not change at t = t1,
we would have
t1t0
Vs
i(t), t > 0 (See next slide).
Using i(t+0 ) and i(β), we can obtain
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
and we need to work out the
solution for t > t1 separately.
In reality, Vs changes at t = t1,
Consider t > t1.
For t0 < t < t1, i(t) = 1β exp(βt/Ο) Amp.
i(t+1 ) = i(tβ1 ) = 1β eβ1 = 0.632 A (Note: t1/Ο = 1).
i(β) = 0 A.
Let i(t) = A exp(βt/Ο) + B.
It is convenient to rewrite i(t) as
i(t) = Aβ² exp[β(tβ t1)/Ο ] + B.
Using i(t+1 ) and i(β), we get
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
(SEQUEL file: ee101_rl1.sqproj)
0 0.2 0.4 0.6 0.8
0
1
i (A
mp)
time (sec)
Combining the solutions for t0 < t < t1 and t > t1,
we get
M. B. Patil, IIT Bombay
RL circuit: example
i
v
t
10 V
0 0.2 0.4 0.6 0.8
i (A
mp)
1
0
time (sec)
i(t) = 0.632 exp[β(tβ t1)/Ο ] A.
t1t0
R2
R1
Vs
Vs
R1 = 10Ξ©
R2 = 40Ξ©
L = 0.8H
t0 = 0
t1 = 0.1 s
(SEQUEL file: ee101_rl1.sqproj)
0 0.2 0.4 0.6 0.8
0
1
i (A
mp)
time (sec)
Combining the solutions for t0 < t < t1 and t > t1,
we get
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5
β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: The switch has been closed for a long time and opens at t = 0.
t=0i
5 k 1 k
6 V
R2R1
ic
R3=5k
vc5 Β΅F
AND
i i
5 k1 k
5 k
5 k
5 k
6 V
ic
t < 0
ic
vc
t > 0
vc5Β΅F 5Β΅F
t = 0β: capacitor is an open circuit β i(0β) = 6 V/(5 k+ 1 k) = 1 mA.
vc(0β) = i(0β)R1 = 5V β vc(0
+) = 5V
β i(0+) = 5 V/(5 k+ 5 k) = 0.5 mA.
Let i(t) = Aexp(-t/Ο) + B for t > 0, with Ο = 10 kΓ 5Β΅F = 50 ms.
i(t) = 0.5 exp(-t/Ο) mA.
Using i(0+) and i(β) = 0 A, we get
0
1
time (sec)
i (mA)
0 0.5β0.5
0
0
5
time (sec)
time (sec)
0 0.5 0 0.5
ic (mA)vc (V)
(SEQUEL file: ee101_rc2.sqproj)
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
i C(m
A)
VS,V
C(Volts)Vc
5 k
iC
C1Β΅F
2
1
β1
0
10
0
0 2010 30 40 50
t (msec)
Find expressions for VC (t) and iC (t) insteady state (in terms of R, C , V0, T1,T2).
V0
0T1 T2
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
i C(m
A)
VS,V
C(Volts)Vc
5 k
iC
C1Β΅F
2
1
β1
0
10
0
0 2010 30 40 50
t (msec)
Find expressions for VC (t) and iC (t) insteady state (in terms of R, C , V0, T1,T2).
V0
0T1 T2
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
0 < t < T1 Let V(1)C (t) = Aeβt/Ο + B
V(1)C (0) = V1, V
(1)C (β) = V0
β B = V0, A = V1 β V0.
V(1)C (t) = β(V0 β V1)eβt/Ο + V0 (1)
T1 < t < T2 Let V(2)C (t) = Aβ² eβt/Ο + Bβ²
V(2)C (T1) = V2, V
(2)C (β) = 0
β Bβ² = 0, Aβ² = V2 eT1/Ο .
V(2)C (t) = V2eβ(tβT1)/Ο (2)
Now use the conditions:
V(1)C (T1) = V2, V
(2)C (T1 + T2) = V1.
V2 = β(V0 β V1)eβT1/Ο + V0 (3)
V1 = V2eβ(T1+T2βT1)/Ο = V2eβT2/Ο (4)
Rewrite with a β‘ eβT1/Ο , b β‘ eβT2/Ο .
V2 = β(V0 β V1)a + V0 (5)
V1 = b V2 (6)
Solve to get
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab, with a = eβT1/Ο , b = eβT2/Ο .
V(1)C (t) = β(V0 β V1)eβt/Ο + V0, V
(2)C (t) = V2eβ(tβT1)/Ο .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab, with a = eβT1/Ο , b = eβT2/Ο .
V(1)C (t) = β(V0 β V1)eβt/Ο + V0, V
(2)C (t) = V2eβ(tβT1)/Ο .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab, with a = eβT1/Ο , b = eβT2/Ο .
V(1)C (t) = β(V0 β V1)eβt/Ο + V0, V
(2)C (t) = V2eβ(tβT1)/Ο .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab, with a = eβT1/Ο , b = eβT2/Ο .
V(1)C (t) = β(V0 β V1)eβt/Ο + V0, V
(2)C (t) = V2eβ(tβT1)/Ο .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab, with a = eβT1/Ο , b = eβT2/Ο .
V(1)C (t) = β(V0 β V1)eβt/Ο + V0, V
(2)C (t) = V2eβ(tβT1)/Ο .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iC
VC
VS
V0
V2
V1
0
T1 T2
T1 T2
V1 = b V01β a
1β ab, V2 = V0
1β a
1β ab, with a = eβT1/Ο , b = eβT2/Ο .
V(1)C (t) = β(V0 β V1)eβt/Ο + V0, V
(2)C (t) = V2eβ(tβT1)/Ο .
Current calculation:
Method 1:
iC (t) = CdVC
dt(home work)
Method 2:
Start from scratch!
STOP
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )
M. B. Patil, IIT Bombay
RC circuit: example
VS
R
C Vc
iC
0
I1
βI2
iCβ
VC
VS
V0
V2
V1
0T1 T2
T1 T2
β
0 < t < T1 Let i(1)C (t) = Aeβt/Ο + B
i(1)C (0) = I1, i
(1)C (β) = 0
β B = 0, A = I1.
i(1)C (t) = I1eβt/Ο (1)
T1 < t < T2 Let i(2)C (t) = Aβ² eβt/Ο + Bβ²
i(2)C (T1) = βI2, i
(2)C (β) = 0
β Bβ² = 0, Aβ² = βI2 eT1/Ο .
i(2)C (t) = βI2eβ(tβT1)/Ο (2)
Now use the conditions:
i(1)C (T1)β i
(2)C (T1) = β = V0/R,
i(1)C (0)β i
(2)C (T1 + T2) = β = V0/R.
I1eβT1/Ο β (βI2) = β (3)
I1 β (βI2eβ(T1+T2βT1)/Ο ) = β (4)
a I1 + I2 = β (5)
I1 + b I2 = β (6)
Solve to get
I1 = β1β b
1β ab, I2 = β
1β a
1β ab
(a = eβT1/Ο , b = eβT2/Ο )M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =Η«A
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
βI2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dtβ Q =
β«iC dt.
Q(t0 + T ) = Q(t0)β Q(t0 + T )β Q(t0) = 0
ββ« t0+T
t0
iC dt = 0.β« T
0iC dt = 0β
β« T1
0iC dt +
β« T1+T2
T1
iC dt = 0
ββ« T1+T2
T1
iC dt = ββ« T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =Η«A
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
βI2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dtβ Q =
β«iC dt.
Q(t0 + T ) = Q(t0)β Q(t0 + T )β Q(t0) = 0
ββ« t0+T
t0
iC dt = 0.β« T
0iC dt = 0β
β« T1
0iC dt +
β« T1+T2
T1
iC dt = 0
ββ« T1+T2
T1
iC dt = ββ« T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =Η«A
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
βI2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dtβ Q =
β«iC dt.
Q(t0 + T ) = Q(t0)β Q(t0 + T )β Q(t0) = 0
ββ« t0+T
t0
iC dt = 0.β« T
0iC dt = 0β
β« T1
0iC dt +
β« T1+T2
T1
iC dt = 0
ββ« T1+T2
T1
iC dt = ββ« T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =Η«A
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
βI2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dtβ Q =
β«iC dt.
Q(t0 + T ) = Q(t0)β Q(t0 + T )β Q(t0) = 0
ββ« t0+T
t0
iC dt = 0.β« T
0iC dt = 0β
β« T1
0iC dt +
β« T1+T2
T1
iC dt = 0
ββ« T1+T2
T1
iC dt = ββ« T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =Η«A
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
βI2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dtβ Q =
β«iC dt.
Q(t0 + T ) = Q(t0)β Q(t0 + T )β Q(t0) = 0
ββ« t0+T
t0
iC dt = 0.
β« T
0iC dt = 0β
β« T1
0iC dt +
β« T1+T2
T1
iC dt = 0
ββ« T1+T2
T1
iC dt = ββ« T1
0iC dt.
M. B. Patil, IIT Bombay
RC circuit: example
Q
Q
insulator
conductor
conductor
Unit: Farad (F)
VS
R
VC
C Vc
iC
C =Η«A
tt
iC
iC
iC =dQ
dt= C
dVC
dt
0
I1
βI2
iC
VC
VS
V0
V2
V1
0T1 T2
T1 T2Charge conservation:
Periodic steady state: All quantities are periodic, i.e.,x(t0 + T ) = x(t0)
Capacitor charge: Q(t0 + T ) = Q(t0)
iC =dQ
dtβ Q =
β«iC dt.
Q(t0 + T ) = Q(t0)β Q(t0 + T )β Q(t0) = 0
ββ« t0+T
t0
iC dt = 0.β« T
0iC dt = 0β
β« T1
0iC dt +
β« T1+T2
T1
iC dt = 0
ββ« T1+T2
T1
iC dt = ββ« T1
0iC dt.
M. B. Patil, IIT Bombay
β10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1Β΅F
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
β10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1Β΅F
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
β10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1Β΅F
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
β10
10
0
0
2
4
6
8
10
0 1 2 3 4 5 6
t (msec)
VS
R
Vc
1 k
iC
C1Β΅F
T1= 1.5msec, T2= 1.5msec.
V1= 1.8 V, V2= 8.2 V.
I1= 8.2mA, I2= 8.2mA.
VS
VC
iC (mA)
t (msec)
0 1 2 3 4 5 6
T1= 1msec, T2= 2msec.
V1= 0.9 V, V2= 6.7 V.
I1= 9.1mA, I2= 6.7mA.
t (msec)
0 1 2 3 4 5 6
T1= 2msec, T2= 1msec.
V1= 3.4 V, V2= 9.1 V.
I1= 6.7mA, I2 = 9.1mA.
SEQUEL file: ee101 rc1b.sqproj
M. B. Patil, IIT Bombay
t (msec)
0 1 2 3 4 5 6
0
2
4
6
8
10
β100
100
0
VS
R
T1= 1msec, T2= 2msec.
V1β 0V, V2= 10V.
I1= 100mA, I2= 100mA.
Vc
0.1 k
iC
C1Β΅F
iC (mA)
VS
VC
M. B. Patil, IIT Bombay