Transient Analysis

CSC 23 Electric Circuits – Transient Analysis

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2. Transient Analysis 2.1 Natural Response of an RL circuit The natural response of the RL circuit can be described in terms of circuit shown in Fig. 2.1(a). We assume that the independent current source generates a constant current of IS amperes and that the switch has been in a closed position for a long time. All currents and voltages have reached a constant value. Only constant current can exist in the circuit just prior to the switch being opened. The inductor appears as a short circuit prior to the release of the stored energy.

i L

t=0 IS

R0

L

R v

Fig. 2.1(a)

R v

Fig. 2.1(b)

Note that no current flowing through both resistors. Now, the problem is to find the voltage and current at the terminals of the resistor after the switch is opened. For t ≥ 0 , the circuit is configured as Fig. 2.1(b). By using KVL, summing the voltages around the closed loop in 2.1(b) gives, L

di di R + iR = 0 ⇒ + i=0 dt dt L

(2.1)

How to solve (2.1)? There are two general approaches: Approach (1) Let’s consider a general linear first-order ordinary differential equation first: dx(t ) + ax(t ) = f (t ) dt

(2.2)

dx(t ) + ax(t ) = 0 + f (t ) dt

(2.3)

(2.2) can be rewritten as

CSC 23 Electric Circuits – Transient Analysis

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The addition of zero to RHS, along with the important property of linearity, allows us to see that the solution for x(t ) will be the sum of two parts: transient (or natural) solution and steady-state (or forced) solution. dx(t ) d [xT (t ) + xSS (t )] + a[xT (t ) + xSS (t )] = 0 + f (t ) (2.4) + ax(t ) = 0 + f (t ) ⇒ dt dt where xT (t ) is transient solution and xSS (t ) is steady-state solution. Due to the property of linearity, (2.4) can be divided into two portions: (a)

dxT (t ) + axT (t ) = 0 dt

and (b)

dxSS (t ) + axSS (t ) = f (t ) dt

(2.5)

First consider the transient solution. A possible form for xT (t ) is the exponential form. Let’s try a solution of the form

xT (t ) = Ae kt

(2.6)

Substituting (2.6) into (2.5)(a) to yield

kAekt + aAe kt = 0 ⇒ k = −a

(2.7)

The solution of (2.3) is the sum of the transient and steady-state solutions:

x(t ) = xT (t ) + xSS (t ) = Ae − at + xSS (t )

(2.8)

Suppose we wish to obtain the solution for x(t) for all times greater than some initial time t0; that is, we wish to find x(t) such that (2.8) is correct for t ≥ t0. Without loss of generality, we may select t0 = 0. Evaluating (2.8) immediately after t = 0 or t = 0+, we have

x(0 + ) = A + xSS (0 + ) ⇒ A = x(0 + ) − xSS (0 + )

(2.9)

We have the solution for x(t) for all t ≥ 0 in terms of initial value of x(t):

(

)

x(t ) = x(0 + ) − xSS (0 + ) e − kt + xSS (t )

(2.10)

CSC 23 Electric Circuits – Transient Analysis

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(2.10) is the general solution for any first order linear differential equation. Applying the above results to (2.1), k = − R/L, the solution for the inductor current is of the form of (2.10):

(

)

i (t ) = i (0 + ) − iSS (0 + ) e − kt + iSS (t )

(2.11)

The steady-state solution can be directly obtained from (2.1), by substituting di/dt = 0 and obtaining

iSS = 0A and iSS (0+ ) = 0A

(2.12)

To finish the solution, we need the value of the inductor current immediately after the switch is opened. We know that inductor current cannot change value instantaneously. Therefore, the inductor current immediately before the switch opens i(0−) and the inductor current immediately after the switch opens i(0+) must be the same:

i ( 0 − ) = i (0 + ) = I S

(2.13)

Thus, the total solution is

i (t ) = I S e −( R

L )t

= I S e −t τ

(2.14)

where τ = time constant = L/R. Fig. 2.1 shows this response graphically. The voltage across the resistor R is,

v(t ) = RI S e − ( R

L )t

(2.15)

and the power dissipated in the resistor is p (t ) = v(t )i (t ) = I S 2 R e −2( R

L )t

(2.16)

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

t

w(t ) = ∫ p ( x)dx = ∫ I S 2 R e − 2( R 0

0

L) x

dx

∴ w(t ) =

1 LI S 2 (1 − e − 2( R 2

L )t

) (2.17)

CSC 23 Electric Circuits – Transient Analysis

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Approach (2) (2.1) can be rewritten as di R = − dt i L

(2.18)

We obtain an explicit expression for i as a function of t by integrating both sides of (2.18). Using x and y as variable of integration yields i (t )

dx Rt ∫ x = − L ∫ dy i (t ) t 0

(2.19)

0

in which i(t0) is the current corresponding to time t0 and i(t) is the current corresponding to time t. Here, t0 = 0. Thus, (2.19) becomes ln

i (t ) R = − t ⇒ i (t ) = i (0)e − ( R i ( 0) L

L )t

(2.20)

Substituting (2.13) into (2.20) can obtain

i (t ) = I S e − t τ

(2.21)

The result is identical to (2.14). i(t) I0

i (t ) = I 0 e −t τ time

0

Fig. 2.2 The current response for the circuit shown in Fig. 2.1 It should be noted that the time constant τ gives the time required for a current to reach its final value if it continued to change at its initial rate. At t = 0+, the current continues to change at this rate: I di + R (0 ) = − I 0 = − 0 dt L τ

(2.22)

If the time constant τ is sufficient large, the expression for i can be

CSC 23 Electric Circuits – Transient Analysis

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approximated to

i = I 0 (1 − t τ )

(2.23)

(2.23) indicates that i would reach its final value of zero in τ seconds.

2.2 Natural Response of an RC circuit The natural response of an RC circuit is analogous to that of the RL circuit. The setup is shown in Fig. 2.3(a). The switch has been in position a for a long time, allowing circuit operating in steady-state condition. In this case the constant voltage source cannot sustain a current in the capacitor; therefore the capacitor must be charged to the source voltage Vg volts. When the switch is moved from position a to position b (at t = 0), the voltage on the capacitor is Vg volts. Because there can be no instantaneous change in the voltage at the terminals of a capacitor, the problem reduces to solving the circuit shown in Fig. 2.3(b). a b t=0 Vg

C

Vg

R

Fig. 2.3(a)

i C

R v

Fig. 2.3(b)

For the passive sign convention, the current flowing towards capacitor C opposes to i. Thus, we obtain C

dv v dv v + =0 ⇒ + =0 dt R dt RC

(2.24)

Recall (2.10), the solution of v can be directly obtained as

(

)

v(t ) = v(0 + ) − vSS (0 + ) e − (1 RC )t + vSS (t )

(2.25)

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For t > 0, the steady-state value of v is equal to 0. In addition, v(0 + ) = v(0 − ) because the voltage at the terminals of a capacitor cannot change immediately. The total solution of v is v(t ) = Vg e − t τ

(2.26)

where τ = time constant = RC. Fig. 2.4 shows this response graphically. v(t) Vg

v (t ) = Vg e− t τ time

0

Fig. 2.4 The natural response of the RC circuit After determining v(t), we can easily derive the dissipated power: p (t ) = v(t )i (t ) =

Vg 2 R

e − 2t

τ

(2.27)

The total dissipated energy is equal to t

1 w(t ) = ∫ p ( x)dx = CVg 2 (1 − e − 2t τ ) 2 0

(2.28)