TRANSIENT RESPONSE TO DC AND AC NETWORKS

ELECTRIC CIRCUITS

1.6 TRANSIENT RESPONSE TO DC AND AC NETWORKS

The transients in the network are due to the energy stor- ing elements of the opposite kind (L and C). If the network consists of only resistors, no transients will occur at the time of switching. A resistor can accommodate any amount of current through and any amount of voltage across it. Since the energy in system cannot change instantaneously, so the energy stored in the inductor and capacitors cannot change instantaneously, that is, the inductor current and capacitor voltage cannot change instantaneously. Hence, this behaviour leads to the transients in the network (Fig. 1.54).

N −∞ 0⁻ ∞

0⁻ = −0.000...1 0⁺ = +0.000...1

0⁺ 0 V

+−

t = 0

Figure 1.54

|

Transients in a network.

Now, the behaviour of the inductor and capacitor at t = 0⁺ and t=∞ is as follows:

The inductor current at t = 0⁻ and t = 0⁺ i t

V t dt

t L( )= L( )

−∞

∫

⁼ ¹ ⁺¹

0 0

V t dt L

V t dt

L( ) L( )

−

∫ ∫

∞

=i +

V t dt

t L(0 ) L( )

−

∫

−

We want i only at t = 0⁺, so

i i

V t dt

L(0 ) L(0 ) L( ) 1

0 0 +

= ⁻ +

∫

−

i i

L(0⁺)= L(0⁻)+0

i i

L(0⁺)= L(0⁻)

So, the inductor current just after the switching is equal to the current before switching, so the inductor current cannot change instantaneously.

For all excitations except impulse,

E E

L(0⁺)= L(0⁻) If VL(t) = d (t), then i t i

L( )= L( )+ ( )t dt

0 1

0 0

−

∫

d i t i

L( )= L(0 )+

− 1

So, the inductor current can change instantaneously for impulse voltage across it,

E L i

L(0 ) L( ) 1

0 ²

− −

= Joules

E L i

L(0 ) L( ) 1

0 ²

+ +

= Joules

The capacitor voltage at t = 0⁻ and t = 0⁺ V t

i t dt

t C( )= ( )

−∞

∫

= +

1 1

0 0

i t dt C

i t dt

C( ) C( )

−

∫ ∫

∞

=V +

i t dt

t C(0 ) C( )

−

∫

−

We want voltage only at t = 0⁺

V V

i t dt

C(0 ) C(0 ) C( ) 1

0 0 +

= ⁻ +

∫

−

VC (0⁺) = V

C (0⁻) EL (0⁺) = E

C (0⁻)

The voltage across the capacitor cannot change instantaneously for all excitations except impulse for impulse current.

V V

C(0 ) C(0 )

+ 1

= ⁻ +

1. Inductive impedance at t = 0⁺, s→ ∞ ZL = sL

as s → ∞, Z

L = ∞. Here, L is open circuit behaviour.

2. Capacitive impedance at t = 0⁺, s→ ∞ Z

C = 1

as s→ ∞, Z

C = =

∞ 0. Here, C is short circuit behaviour.

3. Inductive impedance at t=∞,s→ ∞ ZL = sL

as s → 0, Z

L = 0. Here, L is short circuit behaviour.

4. Capacitive impedance at t=∞,s→0 Z

C = 1 as s → 0, Z

C =∞. Here, C is open circuit behaviour.

In circuits, voltage and current conditions are continu- ally changing. We therefore need to include the effects of time and transient events on circuit conditions. The transients are of two kinds based on the type of voltage:

1. DC Transients 2. AC Transients

1.6.1 Initial and Steady State

Whenever the impedance source is connected to the network for a long time, the network is said to be in steady state. In steady state, energy stored in the network is maximum and constant, that is, the energy stored in the inductor and capacitor is maximum and constant.

Hence, 1 2

Li2 =

maximum and constant and i

L = maximum and constant.

1 2

CV = maximum and constant and V

C = maximum and constant.

when i

L = maximum and constant

V L

di dt

L V

= ⇒ L =0

The inductor short circuits in steady state. In steady state condition, inductor acts as a constant current source. Since the current is maximum and constant, V

= maximum and constant

i C

dV dt

C= ^C ⇒iC=0

The capacitor is open in steady state. In steady state, the capacitor acts as constant voltage source, since the voltage across capacitor is maximum and constant.

1.6 TRaNSIENT RESpONSE TO DC aND aC NETWORKS 47

The equivalent circuits in transient for current free inductor and voltage free capacitor are shown in Figs. 1.55 and 1.56, respectivley.

V_L(t) i_L(t)

I₀

I₀ L

− V_L(t)

i_L(t) +

−

Figure 1.55

|

Current free inductor.

V_C(t) i_C(t)

V₀ V₀

C +

− V_C(t)

i_C(t) +

−

− +

− +−

Figure 1.56

|

Voltage free capacitor.

1.6.2 DC Transients

Capacitors and inductors, when wired in series with resistors, react over a period of time to sudden changes in DC voltage (transient), as they absorb and release energy. This response of capacitors and inductors to DC transient voltage can be determined for the following cases.

Case 1: Source-free circuits

These are circuits without any independent source.

There are three types of source-free circuits:

1. Source-free RL circuit: Here L consists of initial condition I

2. Source-free RC circuit: Here C consists of initial condition V

3. Source-free RLC circuit: Here L consists of initial condition I

0 and C consists of initial condition V

0. Case 2: Circuits with sources

With sources, initial condition is (t = 0) and final condition is (t=∞), so

iL(0⁺), V

C(0⁺), i

L(∞),V

C(∞)

Since the inductor and capacitor effects are eliminated from the network at t = 0⁺ and at t=∞, the nature of the circuit at these two instances is resistive.

Case 3: Laplace transform

The Laplace transform approach is used for solving problems in the presence of sources for t > 0. It can be used for RLC circuits but not required for RL and RC circuits. The Laplace transform approach involves

R R

←LT→ W L← →sL W C

← → 1 W

The transformations are depicted as follows in Fig. 1.57 with characteristic equations.

V_R(s) I_R(s)

R R

− V_R(t)

i_R(t) +

−

V t Ri t

R( )= R( ) (a) V s I s R

R( )= ( )

V_L(s) I_L(s)

I₀ s L

I₀

sL +

− V_L(t)

i_L(t) +

−

sL LI₀ V_L(s)

I_L(s) +

−

+−

(b) V t L

di t dt

( ) L

( )

V L

di dt

= L =L sI( ( )s i ( ⁺)]

L − L0 I s

sL V s

I s

L( )= L( )+

1 0

V_C

V₀ i_C

V₀ C

C +

− V_C −

i_C +

−

1 sC V_C(s)

I_C(s) +

−

+−

+− −V₀

(c)

i C

dV t dt

C( )

= I s C sV s V

C( )= [ C( )− C(0⁺)]

Figure 1.57

|

Laplace transformations and characteristic equations.

Problem 1.78: Determine the time constant of the circuit shown in the figure given below.

2Ω

2Ω 1Ω

1V + 1H

−

Solution: We have t =

L Req











s

Now, R

eq = 2||2 + 1 = 2 Ω So, t =

1 2 s

Ans. (0.5) For resistive circuit shown in Fig. 1.59, at t = 0⁺, we have

i(0⁺) = ke⁻⁰, i

L(0⁺) = I

0, k = I

So,

i t( )=I e ⁽^R^/^{L t}⁾

− for t ≥ 0 i t( )=I e ^t^/

− t

where t = L R

s At t = 5t, i(t) = 0.0067 I

0< 0.1% of I

0≈ 0 This can be represented as shown in Fig. 1.60.

t = t 0.368I₀ 0.00657I₀

t = 5t

Figure 1.60

|

Plot of time vs. current in source free RL circuit.

1.6.2.1 Source Free RL, RC and RLC Circuits

Source Free RL Circuit

In a source-free RL circuit, the source is suddenly disconnected and we select the inductor current as the response since inductor current cannot change instantaneously.

Figure 1.58 (a) shows source free RL cicuit and the cor- responding equivalent circuit is shown in Fig. 1.58(b).

R V_R

I₀ V_L L i

− +

−

t ≥ 0

I₀ L

(a) (b)

Figure 1.58

|

Source free RL (a) circuit and (b) equivalent circuit.

Based on current direction, separating the current with a current source, we get

−V

L− V

R = 0

−L di dt

= = 0 L

di dt

iR + = 0 di

dt R L

+ =



 

 0 Let

d dt

= , so Di R L i + = 0

or D

R L

+ =



 

 0 Characteristic equation is D

R L + = 0

D R L

=− Solution of first order equation,

i t( )=ke⁻⁽^R^/^{L t}⁾ (for t ≥ 0)

R I₀

i(t)

Figure 1.59

Problem 1.79: Determine the time constant in the circuit given in the following figure.

10Ω

20Ω 2Ω

5A 2H 2H

Solution: Current passing through 1 H inductor and 2 Ω resistor is the same.

1.6 TRaNSIENT RESpONSE TO DC aND aC NETWORKS 49

t = = =

L Req

5 50

1 10

i t e ^t

L( ) = 2 7. ⁻¹⁰A (iii) At t = 0⁺

+ 40Ω

10Ω

2.4A V(0⁺) V_L(0⁺)

−

V(0⁺) = 40 (−2.4) = −96 V

−V

40 + V

10 + V

L (0⁺) = 0

− 96 + 10 × 2.4 + V

L(0⁺) = 0 VL(0⁺) = 72 V 10Ω

20Ω 2Ω

5A 2H

10Ω

20Ω 2Ω

5A 2H

t = = =

L R

eq eq

2 12

1 6 s

Ans. (1/6)

Problem 1.80: For the circuit shown in the following figure, the switch is closed for long time and it is opened at t = 0. Determine i

L(0⁺), V(0⁺), i

L(t) for t ≥ 0.

t = 0 +

40Ω

10Ω

24V

5H i_L

−

Solution: (i) At t = 0⁻

+ 40Ω

10Ω

24V

(0⁻) i_L

−

i i

L(0 ) . A L( )

24 10

2 4 0

− = = = ⁺

i40 0

24 40

W( ⁻)= =0 6. A VL(0⁻) = 0 V (ii) At t ≥ 0

+ 40Ω

10Ω

5H V_L 2.4A

−

i t e ^t

L( )= 2 4. ⁻^/^tA

Source Free RC Circuit

A source-free RC circuit is a circuit in which DC source is suddenly disconnected and the energy which is already stored in the capacitor is dissipated through the resistor (Fig. 1.65).

(0V) C v i_R

v₀ i_C

−

Figure 1.61

|

Source free RC circuit.

Here,

iR + i

C = 0 v

R C

dv dt + ⋅ =0 dv

dt RC v

+ =

⋅ 0 D

RC v

+ =



 



⇒ v t( )=ke⁻^t^/^RC, t≥0 At t = 0, the circuit is as shown in Fig. 1.62

v₀

C v

+−

t ≥ 0 Figure 1.62

V t ke t

V ke

(

( ) )

)

( )

0 0

0 +

 = for

( v

− RC ≥

⇒

⇒k=V

V( )t =V e ^t^/ ( t )

0 ⁻ ^t for ≥0 where, t = RC seconds

This equation shows that voltage response of circuit is an exponential decay of its initial value, that is, initial voltage V

0 ( Fig. 1.63). When there is no external source of excitation present in the circuit, then the response obtained is called natural response of the circuit. Here, the time constant is t =RC.

0.368V₀ V₀

0.00657V₀

0 T 5T

Figure 1.63

|

Plot of voltage vs. time for source free RC cicuit.

1. When the network consists of several independent sources, multiple resistors and multiple inductors (separable) or single inductor, then inductor current for t ≥ 0 is

i t i i ie e ^t

L L L

( )= (∞)+[ ( )0 − (∞)] ⁻/^t

t = = =

L R

eq eq

V t L di t

( ) L

( ) ,

= (for t ≥ 0)

2. When the network consists of several independent sources, multiple resistors, and multiple capacitors (separable) or single capacitor, then the capacitor voltage is

V t V V V e ^t

C( )= C(∞)+[ C( )0 − L(∞)]⁻^/^t (for t ≥ 0) t = RC = R

eq C = R

eq C

i t C dV t

( ) C

( )

Solution: First we calculate R

eq across capacitor Req =

+ +

= +

= =

5 8 12 5 8 12

5 20 5 20

5 20 25

×( ) × ×

( )

W t = R

eqC = 4 × 0.1 = 0.4 s v=v( )0e⁻^t^/^t =10e⁻^t^/^{0 4}^. V

⇒V e ^t

= 10 ⁻^{2 5} V

Using voltage division rule, we get

v V e ^t e ^t

x = +

= =

12 12 8

0 6 10 ^{2 5} 6 ^{2 5}

× . ×( ⁻ ^. ) ⁻^.

i v

= x

= =

6 12

0 5

2 5

e 2 5

−. −

Problem 1.82: The switch in the circuit has been closed for a long time, and is opened at t = 0. Find v(t) for t ≥ 0. Calculate the initial energy stored in the capacitor.

t = 0

9Ω

1Ω 3Ω

20mF

24V + v

− −

Solution: For t < 0, since capacitor is open circuit to DC, we calculate voltage across 9 Ω resistor which is the initial voltage of capacitor. By using voltage division rule,

v t

C( )= ( ) V

= 9

9 3 24

9 12

24 18

× × (for t < 0)

⇒ v

C (0) = V

0 = 18 V Now, for t > 0

9Ω 1Ω

20mF 15V

−

Req = 1 + 9 = 10 Ω t = R

eqC = 10 × 20 × 10⁻³ = 0.2 s V(t) = V

0 e^−t/t = 18e^−t/0.2⇒ V(t) = 18e^−5t V The initial energy stored in the capacitor is

W CV

C( )0 .

1 2

20 10 18 18 3 24

2 3

= = × × ⁻ × × = J

Ans. (3.24) Problem 1.81: In the figure shown, let V

C (0) = 10 V.

Find v,V

C, v

x and i

x for t > 0.

4Ω 5Ω

8Ω

0.1F V_x

i₀₁ V_C

− +

−

1.6 TRANSIENT RESPONSE TO DC AND AC NETWORKS 51

Problem 1.83: For the circuit shown in the figure below, the switch is closed for a long time and it is opened at t = 0. Determine V

C (0⁺), i(0⁺) and V

(2 ms).

2kΩ i

500Ω 500Ω

50V

t = 0

1µF +

−

Solution: At t = 0⁻

2kΩ

500Ω 500Ω

50V

−

VC (0⁻) = 50 2 5 10

2 10

. × × × = 40 V = V

C (0⁺) At t = 0⁺

500Ω

40V 2kΩ i(0⁺)

+ + +

−

i( ) . 0

40 2 5 10

+ = =

× mA

At t ≥ 0

500Ω

1µF V_C

2kΩ i

−

V t e ^t

/ .

( ) = 40 ^{2 5} ¹⁰ ¹ ¹⁰

3 6

− × × × ⁻

= 40 ^{2 5} ¹⁰

e⁻^t ^×

/ . −

V VC(0⁻) = V

C(0⁺) = 40 V t = 2.0 ms

VC (2 ms) = 40e⁻^2/2.5 = 17.97 V

Ans. (40, 16, 17.97) Source Free RLC Circuits (Parallel and Series)

Consider the circuit given in Fig. 1.64 for source free parallel RLC circuit.

R C

i L

Figure 1.64

|

Source free parallel RLC circuit.

The nodal equation for the circuit is v

R L

v dt i t C dv dt

t t

+ + =

0 0

− ( )

∫

Initial conditions are as follows:

i(0⁺) = I

v(0⁺) = V

Differentiating both sides w.r.t. t, we get C

d v dt R

dv dt L

2 2

1 1

+ + =

Solution of above differential equation is as follows.

Assuming v = Ae^st, CAs e

R Ase

st st Aest

2 1 1

+ + =

Ae Cs R

s L

st 2 1 1

+ + =



 



For this equation to be satisfied for all time, one of its three factors must be zero. So,

Cs R

s L

2 1 1

+ + =

This is an auxiliary/characteristic equation:

RC RC LC

1 2

=− +  −

 



RC RC LC

1 2

=− −  −

 



v A e^{s t}

1 1

= 1 and v A e^{s t}

2 2

= 2

C d v

dt R dv

dt L v

2 1 2

1 1

+ + =

C dv dt R

dv dt L

2 v

2 2

1 1

+ + =

From the above two equations, we have C

d v v

dt R

d v v

dt L

v v

1 2

1 1

( ) ( )

( )

+ +

+ + =

v t( ) =A e^{s t}+A e^{s t}

1 2

2 0 2

=−a+ a −w

2 0 2

=− −a a −w where a =

1 2RC

, w0

= LC

Consider the source free RLC series circuit shown in Fig. 1.69(a) along with its dual circuit shown in Fig. 1.65(b)

R L

i C v_C

v_L +

−

R C

i_L i_L

L v

−

(a) (b)

Figure 1.65

|

(a) Series RLC circuit. (b) Dual of series RLC circuit.

The fundamental integral-differential equations for Fig. 1.65 are

For Fig. 1.65(a) L di dt

Ri C

i dt v t

t t

+ + =

0 0

− _C( )

∫

For Fig. 1.65(b) C di dt R

v L

v dt i t

t t

+ + =

1 1

0 0

− _L( )

∫

The respective second order equation obtained by differentiating these equation w.r.t. t is

L d i dt

R di dt C

2 2

1 0

+ + =

C d v dt R

dv dt L

2 2

1 1

+ + =

Using i=Ae^st, we have

LAs e RASe C

st st Aest

2 1

+ + =

or Ae Ls Rs

st 2 1

+ + =



 



Auxiliary equation is Ls Rs C

2 1

+ + =

R L

L LC

1 2

2 2

, =−

−

± 

 



s1 2

2 0 2 , =− ±a a −w Therefore, a =−R

L 2

and w₀ 1

= LC

Important Relations for Source Free RLC The important relations for source free RLC circuits are listed in Table 1.4.

Table 1.4

|

Summarised realtions for source free RLC

Type Condition Criteria a w₀ Response

Parallel series Overdamped a >w₀ 1 2RC

and R

L 2

1 LC

A e^{s t} A e^{s t}

1 2

+ s

1 2

2 0 2 , =−–+ – −w Parallel series Critically

damped

a =w

0a <w₀ 1 2RC

and R L 2

1 LC Parallel series Undamped

1 2RC

and R

L 2

1 LC

e^−a^t(A t A )

1 + 2

e^−a^t(B cosw t B sinw t)

d d

1 + 2 where

wd = w a

0 2− 2

4Ω I_S=5A

I_C 1F

1Ω a

V_L +

− Problem 1.84: S is in position `a’ for a long time. S

moved to position `b’ at t = 0. At t = 0⁺, the values of IC and V

L are

1.6 TRANSIENT RESPONSE TO DC AND AC NETWORKS 53

Applying KVL equations to the equivalent circuit, we have

Ri L di dt

V u t

+ =

0 ( )

Now, i(t) = 0 for t < 0. For positive time, u(t) is unity.

For t > 0

Ri L di dt

+ =

Ldi V Ri

0− =

Integrating both sides, we get

−L −

V Ri t k

ln( )

0 = +

value of `k’ can be found by using initial conditions of the circuit. We have i (0⁻) = 0 and i(0⁺) = 0.

Setting i = 0 at t = 0, we get

−L R

V k

ln 0 =

−L − −

V Ri V t

[ln( ) ln ]

0 0 =

V Ri

e ^Rt ^L

0 0

− ₋

= ^/

i V

R V

R e ^Rt ^L

= ⁰− ⁰ ⁻ ^/ (where t > 0) i

V R

e ^Rt ^L u t

= ⁰ − ⁰ ⁻ ^/ ( )











So, i t V

e ^tR ^L ( )= ⁰(1− ⁻ ^/ )

For charging inductor (with DC source), current response rises exponentially, and voltage across inductor response decays exponentially.

Note: The general formula for calculating the current through inductor is

i t( )=[ (i0⁺)−i(∞)]e⁻^tR^/^L+i(∞) Solution: At t = 0^-

4Ω I_S=5A

I₂(0⁻) V_C(0⁻)

1Ω

C V

C Volts

×5 4

5 4

= ⇒ =

IL = A

= 5

4 4 1

× 4

4Ω 4V

4A I_C

V_C 1Ω +

− By superposition theorem

4Ω 1Ω 4V

I₁

I1 = 1 A

4Ω

4A 1Ω I₂

I2 = 4 A IC = I

1 + I

2 = 5 A (increasing) Therefore, —5 A (deceasing) VL = 0

Ans. (5, 0) 1.6.2.2 Source Driven RL, RC and RLC Circuits

Source Driven RL Circuit

When a DC source is applied to RL circuit, the voltage or current source can be modelled as a step function and the response is known as step response. Consider the driven RL circuit show in Fig. 1.66(a) and the equivalent circuit in Fig. 1.66(b).

V₀

i(t) R

L t = o

+− V₀ u(t)

i(t) R

+ L

−

(a) (b)

Figure 1.66

|

(a) Driven RL circuit with switch and (b) equivalent circuit.

Problem 1.85: In the network shown in the figure below, a steady state is reached with the switch K open. At t = 0, the switch K is closed. For the ele- ment values given, determine the values of V

A(0^—) and V

A(0⁺).

20Ω 10Ω

10Ω

10Ω 2H 5V

V_B V_A

i_L

Solution: The steady state is reached with switch K open.

20Ω 10Ω

10Ω

V_A

i_L i₁

i₂

iL =

5 30

×4= 2 3

A Therefore,

V i

A(0 ) 20 V

10 40

2 3

20 10

− = × = × × =

iL(0^—) = i

L(0⁺) At t = 0, K is closed

20Ω 10Ω

10Ω

10Ω 2/3A 5V

V_B V_A

V V V V

A−5 A A− B

10 10 20

+ + =

V V V

B B A i

−5 −

10 20

+ + =

2(VA− 5 +V

A) + V

A−V

B = 0 5VA−V

B = 10 2(VB− 5) +V

B− V

A = − ×2 3

3 10

40 3

10 3

V V

B− A − −

= =

15 5

50 3

V V

B− A −

= 5V V 10

A− B = 14

20 3 VB =− VB =− =

×20 − 3 14

10 21

5 10

10 21

200 21

40 21 10

3 40 21

VA = = (0 ) =

− ⇒ +



 



, Ans.

5 10

10 21

200 21

40 21 10

3 40 21

VA = = (0 ) =

− ⇒ +



 



V ,

Source Driven RC Circuit

When a DC source is applied to an RC circuit, the voltage or current source can be modelled as a step function and the response is known as step function.

Figure 1. 67 (a) shows the circuit for a driven RC circuit and the equivalent circuit is shown in Fig.

1.67(b).

V₀ V(t)

i(t) R

C t = 0

− − V (t)

i(t) R

+ C

− V₀ u(t) −

(a) (b)

Figure 1.67

|

(a) Driven RC circuit with switch (b) equivalent circuit.

Since the voltage of a capacitor cannot change instantaneously, so

V(0⁻) = V(0⁺) = 0 The KVL equations are as follows:

V Ri t C

i t dt

= ( )+

∫

( ) ^{(for t}^>⁰⁾ 0 =R +

di dt

i C 0 = +

di dt

i RC i t( )=k e⁻^t^/^RC

At V(0⁻) = V(0⁺) = 0 V. Therefore, i

V R

= ⁰ = ⁰

or k

V R

= ⁰

and i t

V R

e ^t ^RC ( )= ⁰ ⁻^/

For charging capacitor (with DC source), current response decays exponentially. Voltage across the capacitor response rises exponentially.

The general formula for calculation of capacitor voltage is

V t V V e ^t ^RC V

C C C

( )=[ (0⁺)− (∞)] ⁻ + C(∞) Problem 1.86: For the network shown in the figure given below, the switch is in position 1 for a long time and is moved to 2 at t = 0. Determine i(0⁺).

1.6 TRANSIENT RESPONSE TO DC AND AC NETWORKS 55

2 1

C₁

C₂

V R L

i(t)

(a) 0 (b) −V R

(d) V

R 2 Solution: The equivalent circuit is

V_C2 V_C1

i_L

R +

− At t = 0⁻, −V +V ⁻ + =

C ( )

1 0 0 0

V V V

C ( ) C ( )

1 0⁻ = = 1 0⁺ iL(0⁻)= 0 A At t = 0⁺,

R (0⁺)

i₁ R

+−

Ri V Ri

i V

V R

1 1

0 0 0

0 2

( ) ( )

( ) . R ( )

+ +

+ + =

= −

Ans. (c)

Solution: At t = 0⁻,

1Ω 2Ω

1Ω 60V

i_L(0⁻) V_C(0⁻)

At t = 0⁻, steady state is reached

i i

L(0 ) A L( )

60 3

20 0

− = = = ⁺

VC (0⁻) = 20 V = V

C (0⁺) At t ≥ 0

1Ω 1Ω

2Ω

60V

20V 20A

C L

+−

At t = 0⁺

2Ω

0.5Ω N 60V

20V 20A

V_L(0⁺)

i_C(0⁺) + +

+ −

−

Applying nodal analysis at N, we get 20 60

2 5

20 0 0

− .

( ) + +i ⁺ =

−16 + 20 + i

C(0⁺) = 0 iC (0⁺) = −4 A VL(0⁺) = 20 V

Ans. (4, 20) Problem 1.87: For the circuit shown in the figure

given below, the switch is open for a long time.

Determine i

C(0⁺) and V

L(0⁺).

1Ω 2Ω

1Ω 60V

t = 0

i_C(t)

L C

Solution:

(i) At t = 0, the switch is closed. So,

V V

C ( ) C ( )

1 0⁻ = 1 0⁺ =0

V V

C ( ) C ( )

2 0 0 0

− = ⁺ =

(ii) At t → ∞, the circuit is

10V

5V 1Ω

1Ω +

−

5V +

−

− i_L1(∞)

i_L2(∞) V_C1(∞) +

−V_C2(∞)

At t→ ∞, the steady state is reached, so

i i

L ( ) L ( )

1∞ = = A= 2 ∞

10 2

5 VC ( )

1 ∞ = 0V and V

C ( )

2 ∞ = 5V

Ans. (0, 5) Source Driven RLC Circuit (Parallel and

Series)

Figure 1.68 shows the source driven RLC parallel circuit.

I₀ R L V(t)

i(t)

C t = 0

−

Figure 1.68

|

Source driven parallel RLC circuit.

Here,

V t L

i t C dV t

dt I ( )

( )

+ + =

Inductor voltage, V t L di t

dt ( )

( )

= d i t

dt RC

di t dt

i t LC

I LC

2 2

1 0

( ) ( ) ( )

+ + =

Figure 1.69 depicts the source driven RLC series circuit.

V₀ V (t)

i(t)

R L

t = 0

+ +

− −

Figure 1.69

|

Source driven RLC circuit in series.

Here, L

di t dt

Ri t V t V ( )

( ) ( )

+ + =

0 (for t > 0) Capacitor current i t C

dV t dt ( )

( )

= d V t

dt R L

dV t

dt LC

V t V LC

2 2

1 0

( ) ( )

( )

+ + =

Problem 1.88: Determine the steady state voltages across the capacitors.

10V

1Ω

1F 1F 1H

1H t = 0

V_C2 +

− V_C1

− +−

Problem 1.89: In a series RLC circuit with DC excitation, the entire steady state voltage will be dropped across

t = 0

R S

L V

C i(t)

(a) R only (b) R and L only (c) R and C only (d) C only

Solution: We have t→ ∞, the circuit is in steady state:

R V

V_C(∞) 0V +

−

V V

L(∞)+0− =0

⇒ V V

C(∞)= volts

Ans. (d)

1.6 TRANSIENT RESPONSE TO DC AND AC NETWORKS 57

1.6.3 AC TRANSIENTS

In AC circuits with respect to selection of operating frequency, circuit elements and switching operation, it is possible to obtain transient free response. In DC circuits, it is not possible to obtain transient free response. The network with AC excitation is analysed in steady state using only by “phases”.

1.6.3.1 Phase Notations

The phase notation is defined for the cosinusoidal signal.

The sinusoidal signals are converted into cosinusoidal signal by subtracting 90° from the phase.

V( )t V cos( t )

= m w +f = RP⋅[ ⁺ ]

( )

V e^j^w^t ^f

= RP⋅[ ⋅ ]

Vme^j^−f e^j^w^t = RP⋅[V⋅e^j^w^t] where V is the phasor of the given V(t), V =V e^j

f is the exponential form used for multiplication and division, V =V

m∠fis polar form, and V =V +j

m(cosf sinf)is rectangular form used for addition and subtraction. Here RP denotes the real part of the function.

For current excitation, I t( )=I cos( t+ )

m w q

Current phasor is I=I e^j

m q=I

m∠q=I +j

m(cosq sinq) Impedance phasor is Z

V I

= Admittance phasor is Y

I V

= If the voltage excitation is V( )t =V sin( t )

m w +f =

Vmcos(wt+f−90°)

=RP⋅[ ⁺ ^°]

( )

V e^j^{w f−90}^t =RP⋅ ^° ⋅ [Vme^j⁽^f⁻⁹⁰⁾ e^j^w^t] Therefore, V V e^j

m = m ⁽^f^−90°⁾=

Vm∠f−90°

= +

Vm[cos(f−90°) jsin(f−90°)]

1.6.3.2 RL Circuit with AC Excitation

The RL circuit with AC excitation is shown in Fig. 1.70.

V(t) = V_m sin(ωt+f)

i(t) S R

L t = 0

Figure 1.70

|

RL circuit with AC voltage excitation.

The current response can be determine as i(t) = i

tr(t) + i

ss(t)

=ke +i t

R L

− t

ss( )

By using Laplace transformation application, we get I s

V s

H s

R sL ( )

( )

= =

+ 1 H j

R j L ( w)

= w +

1 H j

R L

( )

w = 1w

2 2

+ H j

R L

L R ( )

( ) w tan

= w +

2 2

∠ 1

 



− ⁻

i t ke

R L

L R

R L t

( )

sin tan

= + +

− ^m − −

2 2



 













w w f w



The current response can also be determined by using phasor method in steady state as shown in Fig. 1.71.

jwL R

V∠f -90°

+ +

−

Figure 1.71

|

Phasor method for determining current response.

V I R j L I

m∠f−90°− ⋅ − w ⋅ =0 I

R j L

= +

m∠f ° w

−90

= +

R L

L R

( ) . tan

∠ °



 



w w

−

2 2 1

= + V

R L

L R

2 2

90 ( )

tan

w ∠f w

 

 °

− ⁻ −

Therefore,

a b∠ a j^b

Problem 1.90: Write the phasor form of (i) V(t) = 5 cos(2t + 30°) (ii) V(t) = 5 sin(2t + 30° Solution: Phasor forms are given by

(i) V = 5e^j^30°= 5∠ °30 =5[cos30°+jsin30°] =5 +

2 2

 j











(ii) V = 5.e^j⁽^{− °}⁶⁰⁾= 5∠−60°= 5 1 2

3 2

−j













Thus,

i t( )= RP[I⋅e^j^w^t]

= RP [ae^j^b ⋅e^j^w^t]

= RP[ae^j⁽^w^t⁺^b⁾]

=acos(wt+b)

=a w f w

cos − −tan⁻¹ −90 L R



 

 °











=a w +f w

sin t tan L R

− ⁻¹

 













i t

R L

L R

ss( )

( )

. sin tan

= m +

2 2



 













w w f− ⁻ w

Therefore, i t ke i t

R L t

( )= + ( )

− ss

By substituting t = 0

i ke

R L

L R ( )

( )

sin tan 0 ⁰ m

2 2

= 1

− + − − 

 













w f w

2 2

= + + k

R L

L R

( )

sin tan w f− ⁻ w

 













R L

L R

− ^m − ⁻

2 2



 













( )

sin tan

w f w

Suppose f−tan⁻¹ w 0 L R



 

=

⇒ k = 0 ⇒ i

tr(t) = 0 Then the transient free response

i(t) = i

ss(t)

(This perfect circuit is a hightly desirable state.) So, the condition for transient free response at t = 0 is

f w

= tan⁻¹ L R



 



There is no transient at the time of switching. So f depends upon circuit elements and applied frequency.

Note: From the above condition, if the total phase of the excitation at the time of switching is equal to tan⁻¹ wL ,



 

 then no transients will result in the system for the sinusoidal excitation at the time of switching.

Now if the switch is closed at t = t

0 instead of t = 0 then the condition for the transient free response is

w f w

L R

+ = tan⁻1



 



If f= 0°

w w

L R

= tan⁻



 



If the excitation is

V( )t V cos( t )

= m w +f

Then the complete response through the inductor is i(t) = i

tr(t) + i

ss(t) i t ke

R L

L R

R L t

( )

cos tan

= + m

− − −

2 2

w w f w

 













where, k

R L

L R

= +

− ^m − ⁻

( )

cos tan

2 2

w f w

 













In this, the condition for the transient-free response, f−tan⁻¹ w p

2 L R



 

= Therefore, at t = 0

f w p

=tan⁻¹ + 2 L R



 



Then the condition at t = t

w f w p

L R

2 + =tan⁻ +



 



If f= 0°

w w p

L R

=tan⁻ +



 



Problem 1.91: For the circuit shown in the figure given below, the source frequency is 50 Hz.

Determine the value of t

0 which results in a transient free response.

t = 0

5Ω

0.01H

Dalam dokumen wiley acing the gate - electrical engineering (Halaman 76-89)