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NAME: EUGUNE CHOW MUN WAI
MATRICS CARD NO: A15MJ0033
Question 1
1. By using nodal analysis, analyse the circuit of Fig. 1(a) and Fig. 1(b) for v1, v2 and v3.
Figure 1(a)
Solution :
At supernode v1-v2,
I1=I2+I3+I4+I5
2=V1-V3 + V1 + V2-V3 + V2
0.5 0.125 0.25
2=2V1-2V3+V1+8V2-8V3+4V2
2+3V1+12V2-10V3 : equation 1
By using KVL at loop :
-V1+2V0+V0=0
V1=3V0 , V0=V2
V1=3V2-: equation 2
V3=13V: equation 3
Sub equation 2 and 3 into equation 1 :
2=3(3V2)+12V2-10(13)
2=9V2+12V2-130
132=21V2
V2=6.286V
When V2=6.286V , sub into equation 2 , so V1 = 18.858V
Answer : V1 = 18.858V , V2=6.286V , V3=13V
Figure 1(b)
Solution :
Figure A
Figure B
At supernode V1,V2,V3 (shown at Figure A):
We cannot use KCL because it able cancel each other .But from the circuit , we able know that V2 =
12V
So to find V1 and V3 , we use KVL (shown at Figure B):
Loop 1 : -V1-10+12=0
V1=2V
Loop 2 : -12+20+V3=0
V3=-8V
Answer : V1=2V , V2=12V , V3=-8V
Question 2
11kΩ and 10kΩ resistors are parallel. Therefore,
R11||10 =
11 x 1011+10 = 5.238kΩ as shown in the figure below.
By Using supermesh:
Applying mesh to Loop 1 and Loop 2:
-25 + 2i +4i2 + 5.238i = 0
-25 + 7.238i + 4i2 = 0 ………… (1)
Applying KCL at the node:
i = 20 + i2
Substitute i = 20 + i2 into Equation (1):
-25 + 7.238(20 + i2) + 4i2 = 0
119.76 + 11.238i2 = 0
i2= -10.66mA
v = 4i2
= 4(-10.66) = -42.64V
Substitute i2 = -10.66 into Equation (1):
-25 + 7.238i + 4(-10.66) = 0
i = 9.345mA
Question 3
3) (a)
Combining the 3A ammeter and the 2 ohm resistor by using ohm’s law,
V=IR
V=3× 2=6V
The voltage and the resistor becomes series after source transformation. The diagram is
shown below:
Combining the two resistors in series,
R=2+4=6 Ω
The diagram is shown below:
By using the loop, combining all the voltage, and the resistors are combining as parallel:
V T=10−1=9 V
Req=[ 13+ 1
6]−1
Req=[ 12]−1
Req=2Ω
By using the source transformation, the current can be calculated:
I=VR
V=3× 2=6V
I=93=3 A
The two voltage and current diagram is shown after source transformation,
(b)
By using source transformation, the voltage of 4 ohm resistor and 1 ampere current is found:
V=IR
V=1 × 4=4 V
The diagram is shown as below,
Combining the voltage between 1V and 4V, and the 2 ohm and 4 ohm resistors:
V=1+4=5V
R=[ 12+ 1
4]−1
R=[ 34]−1
R=43
Ω
The diagram is shown as below after combining,
Combining to find the total resistance and the voltage as well as the current:
V=5+4=9V
R=1.333+3=4.333 Ω
I=VR
I= 94.333
=2.077 A
After the source transformation, the final diagram is shown as below,
4.333Ωa
b
9 Vrms 60 Hz 0°
4.333Ω
a
b
2.077 A 1kHz 0°
Question 4
Solution:
Using current division:
Current flowing through 5Ω resistor is, I = 6
5+6(2)
= 1211
A
Vx = IR
= (12/11)(5)
= 5.455V
The 5Ω and 6Ω resistors will be ignored because of short circuit.
Therefore, Vx = 0
The 5Ω and 6Ω resistors will be ignored because of short circuit.
Therefore, Vx = 0
Vx = 5.455 + 0 + 0
= 5.455V.
Question 5
When the switch is on (t<0), the inductor acts as a straight wire while the capacitor acts as an open circuit. Thus, the circuit can be redrawn to circuit 1a.
Circuit 1a
From the circuit, it could be seen that:
R2 ||(R3 +R4)||(R5 +R6)
Req =( 115
+ 140+40
+ 110+10
)kΩ = 7.742kΏ
Rtotal = 10kΏ +7.742kΏ = 17.742kΏ
V(0-)= V(0) =
7 .74217 . 742
(30)=13. 09 V
i(0-)= i(0) =
13 . 09
17 . 742×103=0 . 738 mA
When t > 0, the switch is opened and the circuit becomes as follows in circuit 2a and can be redrawn into the diagram in circuit 2b:
Circuit 2a Circuit 2b
Circuit 2b can then be further simplified to become circuit 2c.
where Req = 25.413kΏ
α=25 . 413×103
2(2 )=6353 . 25
ωo=1
√2(120
)=√10
α > ωo Hence, the circuit is overdamped.
S1,2=−α±√α 2−ωo
2=−(6353 . 25)±√(6353 .25 )2−(√10)2
S1=−7 . 879×10−4
S2=−12706 .5
i( t )=A1 es1 t
+ A 2 es2 t
15kΏ
∴i( t )=A1e−7.879 t+ A 2 e−12706.5 t
When t = 0
i(0) = A1+A2 =0.738×10-3
A2 =0.738×10-3-A1
di(0 )dt
=−7 . 879×10−4 A1−12706 .5 A 2
1L
( RI o+V o )=−7 .879×10−4 A1−12706 .5 A 2
12
((25. 413×103 )(0 . 738×10−3)+13 . 09 )=−7 . 879×10−4 A1−12706 .5 A 2
15 . 922=−7 . 879×10−4 A1−12706 . 5(0 .738×10−3−A 1 )15 . 922=12706 . 499 A1−9 .377
A1=1 .991×10−3
A2=0 . 738×10−3−1 . 991×10−3=−1 . 253×10−3
∴i( t )=1 . 991×10−3e−7. 879 t−1 . 253×10−3e−12706 .5 t
V ( t )=I ( t ) R=(1.991×10−3e−7.879×10−4
−1 .253×10−3 e−12706 .5 t )(25 .413×103)
∴V ( t )=50 .6 e−7 .879×10−4
−31 . 842 e−12706 .5 t
Question 6
Analyze the calculate io(t) and vo(t) of the circuit shown in Figure 4 for t>0 .
For t < 0,
At t = 0- ,
vo(0) = 8
8+12(30 )
= 24V
For t > 0,
α = 1
(2 RC ) 1
√ LC
= 1
2(8)(14) =
1√ 1 (1/4 )
= ¼ = 2
Since α less than we have an under-damped response .
d = √ (ω )2−α2
= √4−( 116
)
= 1.9843
vo(t) = (A1cosdt + A2sindt)e-αt
A1= 24
io(t) = C(dvo/dt) = 0 when t = 0
dvo/dt = -α(A1cosd dt + A2sind dt)e-αt + (-d A1sind dt + d A2cosd dt)e-αt
at t = 0, dvo(0)/dt = 0 = -αA1 + d A2
Thus, A2 = (α/d)A1 = (1/4)(24)/1.9843 = 3.024
vo(t) = (24cos1.9843t + 3.024sin1.9843t)e-t/4 volts.
i0(t) = Cdv/dt
= 0.25[–24(1.9843)sin1.9843t + 3.024(1.9843)cos1.9843t –0.25(24cos1.9843t) 0.25(3.024sin1.9843t)]e-t/4
= [0.000131cos1.9843t – 12.095sin1.9843t]e-t/4A.
Question 7
7. Referring to the circuit shown in Figure 7, analyse the value of v(t) for all time.
Solution:
When t < 0, u(t) = 0, LC circuit is disconnected from the voltage source.
i(0) = 0A, v(0) = 0V
When t > 0, u(t) = 1, capacitor acrs like open circuit and inductor acts like short circuit.
v(t) = 50V
i(t) = 505
i(t) = 10A