CHAPTER 5 DC TRANSIENT ANALYSIS

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

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

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

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.

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 –

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 –

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

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

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: -∞ ∞

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

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)

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:

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

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

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,

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 –

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

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

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

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

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

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

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

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

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

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-)

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)

Cont. Rearrange the equation; (2) (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;

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

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

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 = 2 + 3e-20t A = -15e-20t V And the voltage is:

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)

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

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

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:

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