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ECE2262 Electric Circuit

Chapter 7: FIRST AND SECOND-ORDERRL AND RC CIRCUITS

Response to First-Order RL and RC Circuits

Response to Second-Order RL and RC Circuits

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7.1. Introduction

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■ In dc steady state, a capacitor looks like an open circuit and an inductor looks like a short circuit.

■ The voltage across a capacitor must be a continuous function of time

■ The current flowing through an inductor must be a continuous function of time

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7.2. Response to First-Order RL and RC Circuits to Constant Sources

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■ Natural Response : t = 0 ! The currents and voltages that arise when stored energy in an inductor or capacitor is suddenly released to a resistive network. This happens when the inductor or capacitor is abruptly disconnected from its DC source.

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■ Step Response : t = 0 +

The currents and voltages that arise when energy is being acquired by an inductor or capacitor due to the sudden application of a DC voltage or current source (initial values are zero !)

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Total Response = Natural Response + Step Response

Natural Response = Zero Input Response ! v 0 !( )

Step Response = Zero-State Response (Forced Response) ! v 0 !( ) = 0

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■ The First Order Differential Equation - Constant Source

dx t( )dt

+ ax t( ) = A

x t( ) = xp t( )particularsolution

!+ xh t( )naturalresponse

!

dxp t( )dt

+ ax p t( ) = A , dxh t( )dt

+ axh t( ) = 0

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• Let x p t( ) = K1 ! aK1 = A ! K1 =Aa

! dx t( )dt

+ ax t( ) = A Then the complete solution is

x t( ) = Aa+ xh t( )

where xh t( ) must (?) be the solution of the homogenous equation

• dxh t( )dt

+ axh t( ) = 0

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• Choose (?) the solution in the form: xh t( ) = K2e!t

! K2!e!t + aK2e

!t = 0 ! dxh t( )dt

+ axh t( ) = 0

! ! + a = 0 ! ! = "a

! xh t( ) = K2e!at .

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• Total Response - the general solution of the 1’st order differential equation

dx t( )dt

+ ax t( ) = A

x t( ) = Aa+ K2e

! at

! If x 0 !( ) is given then K2 can be uniquely determined

! x !( ) = Aa

assuming that a > 0

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A. Natural Response of RL and RC Circuits

The currents and voltages that arise when stored energy in an inductor or capacitor is suddenly released to a resistive network. This happens when the inductor or capacitor is abruptly disconnected from its DC source.

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A1. Natural Response of RL Circuits

For t < 0 the inductor appears as a short circuit and therefore all the source current Is appears in the inductive branch, iL 0 !( ) = Is , iR 0 !( ) = 0 We wish to find i t( ) and v t( ) after the switch has been opened, i.e., when the source has been disconnected and the inductor begins releasing energy.

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The circuit for t ! 0

KVL: Ldi t( )dt

+ Ri t( ) = 0 ! di t( )dt

+ RLi t( ) = 0

! i t( ) = K2e!RLt ! i t( ) = i 0 !( )e!

RLt

! i 0 +( ) = i 0 !( )

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■ Natural Response of an RL circuit

i t( ) = I0e!t /"

I0e!t /"

" =LR

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■ The voltage across resistor

v t( ) = i t( )R = RI0e! t /"

v t( ) = RI0e!t /"

v 0 !( ) = 0 , v 0 +( ) = RI0

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■ The energy delivered to the resistor during any interval of time after the switch has been opened

w t( ) = p z( )dz0

t

! = Ri2 z( )dz0

t

!

= RI02 e!2z/" dz

0

t

#

= I02L2

1! e!2t /"{ }

As t!" the energy dissipated in the resistor approaches the initial energy stored in the inductor.

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■ Time Constant ! = LR

! = 1

! = 0.3

i t( ) = I0e! t /"

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Graphical Interpretation of the Time Constant

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■ Calculation of the natural response of an RL circuit 1. Find the initial current I0 through the inductor

2. Find the time constant of the circuit: ! = LR

3. Use i t( ) = I0e! t /" to generate i t( ) from I0 and !

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Example In the following circuit the switch has been closed for a long time before it is opened at t = 0

1. At t = 0 ! the voltage across inductor is zero (short circuit) !

iL 0 !( ) = iL 0 +( ) = 20A

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2. t = 0 +

iL

Req = 2 + 40 ||10( ) = 10! The time constant ! = L

Req = 210

= 0.2sec

iL t( ) = I0e! t /" = 20e!5t , t ! 0 , iL 0 !( ) = 20A

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3. The current in the 40! resistor

iL

i0 t( ) = 1010 + 40

! "iL t( )( ) = !4e!5t , t ! 0 + , i0 0 !( ) = 0

v0 t( ) = 40i0 t( ) = !160e!5t , t ! 0 +

Note: Time Constant: L / RTh = 2 /10 = 0.2s - the same as before

iL RTh = 40 ||10( )+ 2 = 10!

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4. Power and Energy dissipated in the 10 ! resistor

iL

• p10! t( ) = v02 t( )10

= 2560e!10t W, t ! 0 +

• w10! = p10! z( )dz0

"

# = 256 J

• The initial energy stored in the 2 H inductor:

wL 0( ) = 12Li2 0( ) = 1

22 ! 202 = 400 J.

256400

!100% = 64 % - the percentage of energy dissipated in the 10 ! resistor

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Example In the following circuit the initial currents in inductors L1 and L2 have been established by sources not shown.

i i4

1. Find i1,i2 ,i3 for t ! 0 ! need v t( )

Equivalent Circuit: L : L1 || L2 =5 ! 2025

= 4H

Req : 10 ||15( ) + 4{ } || 40 = 8!

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I0

The time constant ! = LReq

= 4 / 8 = 0.5 sec.

i t( ) = I0e! t /" = 12e!2t , t ! 0 , i 0 !( ) = 12

v t( ) = 8i t( ) = 96e!2t , t ! 0 + , v 0 !( ) = 0

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i i4

i1 + i2 + i = 0 , i t( ) = 12e!2t , t ! 0

CD: i3 =15

15 +10i4 =

35i4

i4 =40

40 + 15 ||10 + 4( ) i = 45i ! i3 =

1225i

i3 =1225

!12e"2t = 5.76e!2t t ! 0 +

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i i4

i1 + i2 + i = 0 , i t( ) = 12e!2t , t ! 0 , v t( ) = 96e!2t , t ! 0 +

i1 t( ) = 1L1

v z( )0

t

! dz + !8( ) = 1.6 ! 9.6e!2t , t ! 0 , i1 0 !( ) = !8

i2 t( ) = 1L2

v z( )0

t

! dz + !4( ) = !1.6 ! 2.4e!2t , t ! 0 , i2 0 !( ) = !4

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2. The initial energy stored in the inductors

i i4

w = 125 ! 82 + 1

220 ! 42 = 320 J

3. As t!" then i1!1.6 , i2 !"1.6 Hence, a long time after the switch has been opened, the energy trapped in the two inductors is

w = 125 !1.62 + 1

220 ! "1.6( )2 = 32 J

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4. The total energy delivered to the resistive network

I0

i t( ) = 12e!2t

w = i2 z( )! 80

"

# dz = 1152 e!4 z0

"

# dz = 288 J ! = 320 - 32

This is the difference between the initially stored energy (320 J) and the energy trapped in the parallel inductors ( 32 J)

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i t( )

i1 t( ) i2 t( )

i 0 !( ) = 12 , i1 0 !( ) = !8 , i2 0 !( ) = !4

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A2. Natural Response of RC Circuits

• t = 0 ! ! the capacitor behaves as an open circuit ! v 0 !( ) =Vg

• t = 0 +

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■ Natural Response

Cdv t( )dt

+v t( )R

= 0 with v 0 !( ) = v 0 +( ) =Vg =V0

v t( ) =V0e! t /" , t ! 0

Time Constant: ! = RC

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v t( ) =V0e! t /" " = RC

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■ Calculation of the natural response of an RC circuit 1. Find the initial voltage V0 across the capacitor

2. Find the time constant of the circuit: ! = RC

3. Use v t( ) =V0e! t /" to generate v t( ) from V0 and !

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Example

1. t = 0 ! (position x) ! vC 0 !( ) = 100V , t = 0 + (position y) ! vC 0 +( ) = 100V and 32 + 240 || 60 = 80k! Time Constant = ReqC = 80 !103( ) 0.5 !10"6( ) = 40 ms.

vC t( ) =V0e! t /" = 100e!25t V, t ! 0

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2. v0 t( )

240 || 60 = 48k!

VD: v0 t( ) = 4848 + 32

vC t( ) = 60e!25t V , t ! 0+ , v0 0 !( ) = 0

3. i0 t( ) = v0 t( )60k

= e!25tmA , t ! 0+

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4. The power and energy dissipated in the 60 k! resistor

p60 t( ) = i02 t( )! 60k = 60e!50tmW , t ! 0+

w60 = p60 z( )0

!

" dz = 1.2 mJ

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B. Step Response for RC and RL Circuits

The currents and voltages that arise when energy is being acquired by an inductor or capacitor due to the sudden application of a DC voltage or current source

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B1. Step Response of RC Circuits

t = 0

Vs

Rt = t0t = 0

• We wish to determine v t( ) for t > 0 , given the initial value v 0 !( ) and the values of the DC voltage source Vs and the resistance R .

• v 0 !( ) - the capacitor voltage just before the switch is closed. • v 0 !( ) = v 0 +( ) due to the continuity of the capacitor voltage

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For t > 0 by KVL

Vs

R

Vs = Ri t( ) + v t( ) and i t( ) = C dv t( )dt

Vs = RCdv t( )dt

+ v t( )

dv t( )dt

+ 1RC

v t( ) = 1RC

Vs

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dv t( )dt

+ 1!v t( ) = 1

!Vs , ! = RC - Time Constant

The first order differential equation yielding the response v t( ) with respect to the forcing input Vs and the initial value v 0 !( ) (natural response)

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RC - circuit (step) response to constant source Vs :

v t( ) = v 0 !( ) !Vs"# $%e

!t /& +Vs

for t ! 0

v 0( ) = v 0 !( )

v t( )!Vs as t!" - steady state condition Under DC conditions the capacitor becomes an open circuit, taking the full value of the DC voltage source Vs

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The solution v t( ) = v 0 !( )!Vs"# $%e

! t /& +Vs can be rewritten

v t( ) =Vs 1! e! t /"#$ %&zero-state response

! "# $#+ v 0 !( )e! t /"

zero-input response

! "# $#

! zero-input response = natural response

! zero-state response = forced response

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If the switching time is not t = 0 but t = t0 then the step response is

v t( ) = v t0 !( ) !Vs"# $%e

! t!t0( )/& +Vs

for t ! t0

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Example (a) v 0 !( ) = 4V , Vs = 2V , R = 2! , C = 0.5F !" = 1s

v t( ) = v 0 !( )!Vs"# $%e! t /& +Vs ! v t( ) = 2e! t + 2 , t ! 0

(b) v 1!( ) = 4V , Vs = 6V , ! = 0.5s ! v t( ) = !2e!2 t!1( ) + 6 , t !1

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Example The switch is moved from position a to b at t = 0 . The switch in position a has been for a long time.

Vs

i1i2

v t( )

ab2!

1! 2A1F

+"

2!

2!

1. t < 0 - the capacitor is in the DC state and behaves like an open circuit, i.e., the source 2 A flows trough the 1! resistor. Hence, v 0 !( ) = 2 "1= 2V

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2. t > 0

Vs

i1i2

v t( )

ab2!

1! 2A1F

+"

2!

2!

Thevenin equivalence:

• Voc = Voc =22 + 2

!Vs = 12Vs - steady state value

• RTh = 2||2 =1 !

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Vs v t( )

ab

1! 2A1F

+"

2!1!

Vs2

Time Constant : ! = RTh "C = 1 s. Hence, for t > 0

v t( ) = 2 ! Vs2

"#$

%&'e! t + Vs

2 = 2e! t + Vs

21! e! t"# $%

v t( ) = v 0 !( ) !Vs"# $%e

!t /& +Vs

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3. i1 t( ) and i2 t( ) for t > 0

Vs

i1i2

v t( )

ab2!

1! 2A1F

+"

2!

2!

i1 t( ) = Vs ! v t( )2

and i2 t( ) = v t( )2

i1 t( ) = !e! t + Vs41+ e! t"# $% , i1 0 +( ) = Vs

2!1

i2 t( ) = e! t + Vs41! e! t"# $% , i2 0 +( ) = 1

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4. i1 t( ) and i2 t( ) for t < 0

Vs

i1i2

v t( )

ab2!

1! 2A1F

+"

2!

2!

i1 t( ) = i2 t( ) = Vs4

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5. Plot v t( ) , i1 t( ) , i2 t( ) if Vs = 8V

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B2. Step Response of RL Circuits

Is R

• We wish to determine i t( ) for t > 0 , given the initial value i 0 !( ) and the values of the DC current source Is and the resistance R .

• i 0 !( ) - the initial current just before the switch is closed. • i 0 !( ) = i 0 +( ) due to the continuity of the inductor current

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For t > 0 by KCL

Is R

Is =v t( )R

+ i t( ) and v t( ) = L di t( )dt

di t( )dt

+ 1!i t( ) = 1

!Is , ! = L / R - Time Constant

The first order differential equation yielding the response i t( ) with respect to the forcing input Is and the initial value i 0 !( ) (natural response)

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RL - circuit (step) response to constant source Vs :

i t( ) = i 0 !( )! Is"# $%e

!t /& + Is

for t ! 0

i 0( ) = i 0 !( )

i t( )! Is as t!" - steady state condition Under DC conditions the conductor becomes a short circuit, taking the full value of the DC current source Is

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The solution i t( ) = i 0 !( )! Is"# $%e

! t /& + Is can be rewritten

i t( ) = Is 1! e! t /"#$ %&zero-state response

! "# $#+ i 0 !( )e! t /"

zero-input response

! "# $#

! zero-input response = natural response

! zero-state response = forced response

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If the switching time is not t = 0 but t = t0 then the step response is

i t( ) = i t0 !( )! Is"# $%e

! t!t0( )/& + Is

for t ! t0

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B3. A General Solution for Step and Natural Responses In the RC and RL circuits the unknown variable (signal) can be determined from the general formula

x t( ) = x f + x t0( ) ! x f"# $%e

! t!t0( )/&

• x f - the final value (steady state) - the value of x !( )

• x t0( ) - initial value at either t0 ! or t0 + • t0 -switching time, commonly t0 = 0 • ! - time constant: ! = RC , ! = L / R

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1. Identify the variable of interest for the circuit. For RC circuits, it is most convenient to choose the capacitive voltage; for RL circuits, it is best to choose the inductive current.

2. Determine the initial value of the variable, which is its value at t0 .Note that if you choose capacitive voltage or inductive current as your variable of interest, it is not necessary to distinguish between t0 ! and t0 + . This is because they both are continuous variables. If you choose another variable, you need to remember that its initial value is defined at t0 + .

3. Calculate the final value of the variable, which is its value as t!"

4. Calculate the time constant for the circuit.

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Example The switch in position a has been for a long time.

1. t = 0 ! ! the capacitor looks like an open circuit !

vC 0 !( ) = 6020 + 60

" !40( ) = !30 V

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2. t = 0 +

After the switch has been in position b for a long time the the capacitor looks like an open circuit ! vC !( ) = 90V

3. Time constant: ! = RC = 400k ! 0.5µF = 200 !10"3 = 0.2 s.

4. vC t( ) = 90 + !30 ! 90[ ]e! t /0.2 = 90 !120e!5t , t ! 0 .

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5. i t( ) , t ! 0 +

• i 0 +( ) = 90 ! vc 0 +( )400k

= 90 ! !30( )400k

= 300 µA

• i !( ) = 0 , • time constant: ! = RC = 0.2 s - the same

i t( ) = 0 + 300 ! 0[ ]e!5t = 300e!5tµA , t ! 0 +

Note: i t( ) = C dvC t( )dt

gives the same result

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Note: vC t( ) = 0 if 90 !120e!5t = 0 ! t = 15log 4

3!"#

$%& = 57.54 ms.

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Example No initial charge on the capacitor. Determine i t( ) and v t( )

1. vC 0 !( ) = 0 ! vC 0 +( ) = 0 - no initial charge

the capacitor is an open circuit ! i 0 !( ) = 0

! x t( ) = x !( ) + x 0 +( )" x !( )#$ %&e" t /' !

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2. Since vC 0 +( ) = 0 ! i 0 +( ) = 2020 + 30

! 7.5m = 3 mA.

3. i !( ) = 0 (the capacitor is an open circuit)

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4. Time Constant: ! = RThC ! RTh = 20 + 30 = 50k!

RTh

! = 50 "103 " 0.1"10#6 = 5 ms

5. i t( ) = 0 + 3! 0[ ]e! t /5"10!3 = 3e!200t mA, t > 0 + , i 0 !( ) = 0

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6. v t( ) : v t( ) = vC t( ) + 30k ! i t( ) , i t( ) = 3e!200tmA

vC t( ) = vC !( ) + vC 0 +( )" vC !( )#$ %&e

" t /' • vC 0 +( ) = 0 • vC !( ) (the capacitor is an open circuit) = 7.5m ! 20k = 150V

• vC t( ) = 150 + 0 !150[ ]e!200t = 150 !150e!200t V

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• v t( ) = vC t( ) + 30k ! i t( ) = 150 ! 60e!200t V

• v 0 !( ) = 20k " 7.5m = 150V • v 0 +( ) = 20k ! i20 = 20 !3050

! 7.5m = 90V

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Example 7.3. Find i t( ) .The circuit has been in the steady-state before the switch is closed.

Note: for t ! 0 + we have i t( ) = vC t( ) / 6k

! vC t( ) = vC !( ) + vC 0 +( )" vC !( )#$ %&e" t /' !

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1. t = 0 !

vC 0 !( )! 36 + 2k " i 0 !( ) = 0 and i 0 !( ) = 36 !1212k

= 2mA

vC 0 !( ) = 36 ! 2k " 2m = 36 - 4 = 32 V

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2. t = 0 +

vC !( )+

"

vC !( ) = 66 + 2

" 36 = 27V

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3. Time Constant: ! = RThC = 32!103 !100 !10"6 = 0.15 s

RTh = 2 || 6 =32k!

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4. • vC t( ) = vC !( ) + vC 0 +( )" vC !( )#$ %&e" t /' ,

vC !( ) = 27V , vC 0 !( ) = vC 0 +( ) = 32V , ! = 0.15 s

vC t( ) = 27 + 5e! t /0.15 V , t ! 0

• i t( ) = vC t( ) / 6k , t ! 0 + and i 0 !( ) = 2mA i t( ) = 9

2+ 56e! t /0.15mA , t ! 0 +

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Example 7. 4. Find v t( ) .The circuit has been in the steady-state before the switch closure at t = 0 .

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1. t = 0 !

4!6A

iL 0 !( ) = 4 || 64 || 6 + 3

" 6A = 83A

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2. t = 0 + ! iL 0 !( ) = 83A = iL 0 +( )

KCL at v1 0 +( ) : 24 ! v1 0 +( )

4!v1 0 +( )6

! 83!v1 0 +( )12

= 0 ! v1 0 +( ) = 203!

!24 + v 0 +( )+ v1 0 +( ) = 0 ! v 0 +( ) = 24 ! v1 0 +( ) ! v 0 +( ) = 523V

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3. The steady-state: v !( ) = 24V

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4. The Thevenin equivalent resistance and the time constant

• RTh = 4 || 6{ } ||12 = 2!

• ! = L / RTh = 4 / 2 = 2s

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5. ! x t( ) = x !( ) + x 0 +( )" x !( )#$ %&e" t /' !

• v !( ) = 24V , • v 0 +( ) = 523V , • ! = 2s

v t( ) = 24 + 523! 24"

#$%&'e! t /2 = 24 ! 20

3e! t /2 V, t ! 0 +

Note: v 0 !( )

i4! 0 "( ) = 244 + 3 || 6

= 246

= 4A ! v 0 !( ) = 4"# 4A = 16V

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v t( ) = 24 ! 203e! t /2 , V, t ! 0 + , v 0 !( ) = 16V , v 0 +( ) = 52

3V = 17.333V