DC TRANSIENT ANALYSIS

DC TRANSIENT ANALYSIS CHAPTER 5

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.

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

Vs R R i i – + – + vs L C An RC circuit An RL circuit First – Order Circuit • 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.

RTh VTh – + IN RN L C 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.

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.

t = 0 i + V – Is Ro L R 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

i + v – Io Ro L R Solving the circuit • 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,

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

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

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:

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

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

∞ -∞ 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:

2Ω i0 t = 0 + V – 0.1Ω 10Ω 2H 20A 40Ω iL 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.

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

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:

+  2Ω 4Ω t = 0 i(t) 12Ω 40V 16Ω 2H 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.

2Ω 4Ω i1 12Ω i(0-) 40V +  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-) Since the current through an inductor cannot change instantaneously Hence; i(0) = i (0-) = 6A

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

t = 0 Ro + v – +  Vo R C 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:

i + v – Ro +  Vo C R Solving the circuit • 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

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

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

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:

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

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

5kΩ 18kΩ a b t = 0 + Vo – +  90V 10kΩ 12kΩ 60kΩ 0.1μF Example 3 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).

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. 5kΩ + Vc(0-) – 90V 10kΩ +  18kΩ + Vo – + Vc(0+) – 60kΩ 12kΩ 0.1μF Solution the voltage across capacitor remains the same at this particular instant vc(0+) = vc(0-) = 60V

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

t = 0 1Ω 3Ω + v – 20V 9Ω 20mF +  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.

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

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

Summary

+ v(t) – L t = 0 R Vs +  i 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 (1)

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

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;

t = 0 3Ω 2Ω 10V 1/4H +  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). i

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

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

+ vc(t) – t = 0 Is R C i 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; (1) (2)

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

3kΩ 4kΩ a b t = 0 + Vc – 24V 5kΩ 30V +  +  0.5mF Example 6 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 4kΩ + Vc (0-) – 24V 5kΩ 30V 0.5mF +  +  Solution When t<0, the switch is at position A. The capacitor acts like an open circuit. Using voltage division: Since voltage across the capacitor remains same. When t <0, the switch is at position B. And the time constant is:

+  4kΩ + Vc(∞) – 30V At t = ∞, the capacitor again behaves like an open circuit. 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

DC TRANSIENT ANALYSIS