1. CHAPTER 5
DC TRANSIENT ANALYSIS

2. Objectives
Investigate the behavior of currents and voltages when energy is either released or acquired by inductors and capacitors when there is an abrupt change in dc current or voltage source.
To do an analysis of natural response and step response of RL and RC circuit.

3. Lecture’s contents 5-1 NATURAL RESPONSE OF RL CIRCUIT
5-2 NATURAL RESPONSE OF RC CIRCUIT
5-3 STEP RESPONSE OF RL CIRCUIT
5-4 STEP RESPONSE OF RC CIRCUIT

4. First – Order Circuit R i + vs – Vs L C
A circuit that contains only sources, resistor and inductor is called and RL circuit.
A circuit that contains only sources, resistor and capacitor is called an RC circuit.
RL and RC circuits are called first – order circuits because their voltages and currents are describe by first order differential equations. vs R + – C i L Vs
An RL circuit
An RC circuit

5. Review (conceptual)
Any first – order circuit can be reduced to a Thévenin (or Norton) equivalent connected to either a single equivalent inductor or capacitor.
In steady state, an inductor behave like a short circuit.
In steady state, a capacitor behaves like an open circuit. L RN IN + –
VTh C
RTh

6. The natural response of an RL and RC circuit is its behavior (i. e
The natural response of an RL and RC circuit is its behavior (i.e., current and voltage ) when stored energy in the inductor or capacitor is released to the resistive part of the network (containing no independent sources)
The steps response of an RL and RC circuits is its behavior when a voltage or current source step is applied to the circuit, or immediately after a switch state is changed.

7. 5-1 Natural Response of an RL circuit
Consider the following circuit, for which the switch is closed for t<0, and then opened at t = 0:
The dc voltage V, has been supplying the RL circuit with constant current for a long time L Ro R Is
t = 0 i + V –

8. Solving the circuit L Ro R Io i + v – For t ≤ 0, i(t) = Io
For t ≥ 0, the circuit reduce to
At t = 0, the inductor has initial current Io, hence i(0) = Io
The initial energy stored in the inductor is, L Ro R Io i + v –

9. Cont. (1) (2) (3) (4) (5) Applying KVL to the circuit:
From equation (4), let say;
(5)

10. Cont. (6) (7) Integrate both sides of equation (5); Therefore,
hence, the current is
(6)
(7)

11. Cont. From the Ohm’s law, the voltage across the resistor R is:
And the power dissipated in the resistor is:
Energy absorb by the resistor is:

12. Time Constant, τ for RL circuit
Time constant, τ determines the rate at which the current or voltage approaches zero.
The time constant of a circuit is the time required for the response to decay to a factor of 1/e or 36.8% of its initial current
Natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in Fig. 5-1
Time constant for RL circuit is
And the unit is in seconds.
Figure 5-1

13. The expressions for current, voltage, power and energy using time constant concept:

14. Switching time
For all transient cases, the following instants of switching times are considered.
t = 0- , this is the time of switching between -∞ to 0 or time before.
t = 0+ , this is the time of switching at the instant just after time t = 0s (taken as initial value)
t = ∞ , this is the time of switching between t = 0+ to ∞ (taken as final value for step response)
The illustration of the different instance of switching times is: -∞ ∞

15. Example 1
For the circuit below, find the expression of io(t) and Vo(t). The switch was closed for a long time, and at t = 0, the switch was opened. 2H
0.1Ω
10Ω
20A
t = 0 i0 + V – iL
40Ω 2Ω

16. Current through the inductor remains the same (continuous)
Solution :
When t < 0, switch is closed and the inductor is short circuit.
When t > 0, the switch is open and the circuit become;
0.1Ω
10Ω
20A
iL(0-)
40Ω 2Ω
Therefore; iL(0-) = 20A 2H
10Ω
20A
io(0+) +
vo(0+) –
iL(0+)
40Ω 2Ω
Hence; iL(0+) = iL(0-) = 20A
Current through the inductor remains the same (continuous)

17. RT = (2+10//40) = 10Ω
So, time constant, sec
By using current division, the current in the 40Ω resistor is:
Using ohm’s Law, the Vo is,
Hence:

18. Example 2
The switch in the circuit below has been closed for a long time.
At t = 0, the switch is opened. Calculate i(t) for t > 0. +  2H 2Ω
12Ω
t = 0
16Ω 4Ω
i(t)
40V

19. Since the current through an inductor cannot change instantaneously
Solution :
When t < 0, the switch is closed and the inductor is short circuit to dc. The 16Ω resistor is short circuit too.
calculate i1;
using current division, calculate i(0-) +  2Ω
12Ω 4Ω
i(0-)
40V i1
Since the current through an inductor cannot change instantaneously
Hence; i(0) = i (0-) = 6A

20. Because current through the inductor is continuously
When t > 0, the switch is open and the voltage source is disconnect. 2H
12Ω
16Ω 4Ω
i(t)
hence;
i(0+) = i(0) = i (0-) = 6A
Because current through the inductor is continuously
RT = Req = (12 + 4)// 16 = 8Ω
Time constant,
Thus,

21. 5-2 Natural Response of an RC Circuit
The natural response of RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor, C is released to the resistors, R.
Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0: C Ro R Vo
t = 0 +  v –

22. Solving the circuit + v – C Ro R Vo  i For t ≤ 0, v(t) = Vo
For t > 0, the circuit reduces to
At t = 0, the initial voltage v(0) = Vo
The initial value of the energy stored is + v – C Ro R Vo  i

23. Cont.
Applying KCL to the RC circuit:
(1)
(2)
(3)
(4)
(5)

24. Cont. (6) (7) (8) From equation (5), let say:
Integrate both sides of equation (6):
Therefore:
(6)
(7)
(8)

25. Cont. Hence, The voltage is: Using Ohm’s law, the current is:
The power dissipated in the resistor is:
The energy absorb by the resistor is:

26. Time Constant, τ for RC circuit
The time constant for the RC circuit equal the product of the resistance and capacitance,
Time constant, sec
The natural response of RC circuit illustrated graphically in Fig 5.2
Figure 5.2

27. The expressions for voltage, current, power and energy using time constant concept:

28. Example 3 t = 0 +  Vo – 5kΩ 10kΩ a b 18kΩ 0.1μF 12kΩ 60kΩ 90V
The switch has been in position a for a long time. At time t = 0,
the switch moves to b. Find the expressions for the vc(t), ic(t) and
vo(t).
t = 0 +  Vo –
5kΩ
10kΩ a b
18kΩ
0.1μF
12kΩ
60kΩ
90V

29. Solution +  Vc(0-) – 5kΩ 10kΩ 90V 18kΩ 0.1μF 12kΩ 60kΩ + Vo –
At t < 0, the switch was at a. the capacitor behaves like an open
circuit as it is being supplied by a constant source.
At t > 0, the instant when the switch is at b. + 
Vc(0-) –
5kΩ
10kΩ
90V
the voltage across capacitor remains the same at this particular instant
18kΩ
0.1μF
12kΩ
60kΩ + Vo –
Vc(0+)
vc(0+) = vc(0-) = 60V

30. RT = (18 kΩ + 12 kΩ) // 60 kΩ = 20 kΩ
time constant, τ = RTC = 20kΩ x 0.1 μF = 2ms
Vc(t) = 60e-500t V
Using voltage divider rule,
Hence,
Vo(t) = 24e-500t V

31. Example 4
The switch in the circuit below has been closed for a long time,
and it is opened at t = 0. Find v(t) for t ≥0. Calculate the initial
energy stored in the capacitor.
t = 0 +  v – 3Ω 1Ω
20mF 9Ω
20V

32. Solution
For t<0, switch is closed and capacitor is open circuit. + 
vc(0-) – 3Ω 1Ω 9Ω
20V
For t>0, the switch is open and the RC circuit is +
vc(0+) – 1Ω
20mF 9Ω

33. Because;
Req = 1+9 = 10Ω
So ;
Time constant, τ = ReqC = 0.2s
Because;
vc(0) = vc(0+) = vc(0-) = 15 V
Voltage across the capacitor;
v(t) = 15e-5t V
Initial energy stored in the capacitor is;
wc(0) = 0.5Cvc2 =2.25J

34. Summary No RL circuit RC circuit 1 2
Inductor behaves like a short circuit when being supplied by dc source for a long time
Capacitor behaves like an open circuit when being supplied by dc source for a long time 3
Inductor current is continuous
iL(0+) = iL(0-)
Voltage across capacitor is continuous
vC(0+) = vC(0-)

35. 5-3 Step Response of RL Circuit
The step response is the response of the circuit due to a sudden application of a dc voltage or current source.
Consider the RL circuit below and the switch is closed at time t = 0.
After switch is closed, using KVL R Vs
t = 0 +
v(t) –  i L
(1)

36. Cont.
Rearrange the equation;
(2)
(3)
(4)
(5)
Therefore:
(6)

37. Cont. Hence, the current is; Or may be written as;
Where i(0) and i(∞) are the initial and final values of i, respectively.
The voltage across the inductor is;
Or;

38. Example 5
The switch is closed for a long time at t = 0, the switch opens.
Find the expressions for iL(t) and vL(t). 2Ω
10V
t = 0 +  3Ω
1/4H i

39. Because inductor current cannot change instantaneously
Solution
When t < 0, the 3Ω resistor is short – circuit, and the inductor acts like short circuit. 2Ω
10V + 
iL(0-)
When t< 0, the switch is open and the both resister are in series. 2Ω
10V +  3Ω
1/4H
So; i(0) = i(0+) = i(0-) = 5A
iL(0+)
Because inductor current cannot change instantaneously

40. Time constant;
When t = ∞, the inductor acts as short circuit again. 2Ω
10V +  3Ω
iL(∞)
Thus: iL(t) = i(∞) +[i(0) – i(∞)]e-t/τ = e-20t A
= -15e-20t V
And the voltage is:

41. 5-4 Step Response of RC Circuit
Consider the RC circuit below. The switch is closed at time t = 0
From the circuit;
Division of Equation (1) by C gives; R
t = 0 +
vc(t) – i Is C
(1)
(2)

42. Cont. Same mathematical techniques with RL, the voltage is:
Or can be written as:
And the current is:
v(t) = v(∞) + [v(0) – v(∞)]e-t/τ

43. Example 6 3kΩ t = 0 +  Vc – 5kΩ a b 4kΩ 0.5mF 24V 30V
The switch has been in position a for a long time. At t = 0, the
switch moves to b. Find Vc(t) for t > 0 and calculate its value at
t=1s and t=4s
3kΩ
t = 0 +  Vc –
5kΩ a b
4kΩ
0.5mF
24V
30V

44. Since voltage across the capacitor remains same.
Solution
When t<0, the switch is at position A. The capacitor acts like an open circuit.
3kΩ + 
Vc (0-) –
5kΩ
24V
Using voltage division:
Since voltage across the capacitor remains same.
When t <0, the switch is at position B.
4kΩ + 
30V
0.5mF
And the time constant is:

45. 4kΩ Vc(∞) + – 30V  Since, v(t) = v(∞) + [v(0) – v(∞)]e-t/τ
At t = ∞, the capacitor again behaves like an open circuit.
4kΩ + 
30V
Vc(∞) –
Hence;
Since, v(t) = v(∞) + [v(0) – v(∞)]e-t/τ
So; vc(t) = 30-15e-0.5t V
And;
At , t = 1s, Vc(t) = 20.9V
At , t = 4s, Vc(t) = 28 V