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7 Response of First-Order RL and RC
Circuits
Assessment Problems
AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 Ω resistor from the circuit.

First combine the 30 Ω and 6 Ω resistors in parallel: 30 6 = 5 Ω Use voltage division to ﬁnd the voltage drop across the parallel resistors: 5 (120) = 75 V v= 5+3 Nowﬁnd the current using Ohm’s law: 75 v i(0− ) = − = − = −12.5 A 6 6 1 1 [b] w(0) = Li2 (0) = (8 × 10−3 )(12.5)2 = 625 mJ 2 2 [c] To ﬁnd the time constant, we need to ﬁnd the equivalent resistance seen by the inductor for t > 0. When the switch opens, only the 2 Ω resistor remains connected to the inductor. Thus, L 8 × 10−3 τ= = = 4 ms R 2 [d] i(t) = i(0− )et/τ = −12.5e−t/0.004 = −12.5e−250t A, t≥0[e] i(5 ms) = −12.5e−250(0.005) = −12.5e−1.25 = −3.58 A 7–1

7–2

CHAPTER 7. Response of First-Order RL and RC Circuits So w (5 ms) = 1 Li2 (5 ms) = 1 (8) × 10−3 (3.58)2 = 51.3 mJ 2 2 w (dis) = 625 − 51.3 = 573.7 mJ 573.7 % dissipated = 100 = 91.8% 625

AP 7.2 [a] First, use the circuit for t < 0 to ﬁnd the initial current in the inductor:

Using current division, 10 (6.4) = 4 A i(0− ) = 10+ 6 Now use the circuit for t > 0 to ﬁnd the equivalent resistance seen by the inductor, and use this value to ﬁnd the time constant:

L 0.32 = 0.1 s = Req 3.2 Use the initial inductor current and the time constant to ﬁnd the current in the inductor: i(t) = i(0− )e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0 Use current division to ﬁnd the current in the 10 Ω resistor: 4 4 (−i) = (−4e−10t ) = −0.8e−10t A,t ≥ 0+ io (t) = 4 + 10 + 6 20 Req = 4 (6 + 10) = 3.2 Ω, .·. τ= Finally, use Ohm’s law to ﬁnd the voltage drop across the 10 Ω resistor: vo (t) = 10io = 10(−0.8e−10t ) = −8e−10t V, t ≥ 0+ [b] The initial energy stored in the inductor is 1 1 w(0) = Li2 (0− ) = (0.32)(4)2 = 2.56 J 2 2 Find the energy dissipated in the 4 Ω resistor by integrating the power over all time: di t ≥ 0+ v4Ω (t) = L =0.32(−10)(4e−10t ) = −12.8e−10t V, dt

Problems p4Ω (t) = w4Ω (t) =
2 v4Ω = 40.96e−20t W, 4 ∞ 0

7–3

t ≥ 0+

40.96e−20t dt = 2.048 J

Find the percentage of the initial energy in the inductor dissipated in the 4 Ω resistor: % dissipated = 2.048 100 = 80% 2.56

AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like an open circuit.

Find the voltage dropacross the open circuit by ﬁnding the voltage drop across the 50 kΩ resistor. First use current division to ﬁnd the current through the 50 kΩ resistor: 80 × 103 (7.5 × 10−3 ) = 4 mA 80 × 103 + 20 × 103 + 50 × 103 Use Ohm’s law to ﬁnd the voltage drop: v(0− ) = (50 × 103 )i50k = (50 × 103 )(0.004) = 200 V i50k = [b] To ﬁnd the time constant, we need to ﬁnd the equivalent resistance seen by thecapacitor for t > 0. When the switch opens, only the 50 kΩ resistor remains connected to the capacitor. Thus, τ = RC = (50 × 103 )(0.4 × 10−6 ) = 20 ms [c] v(t) = v(0− )e−t/τ = 200e−t/0.02 = 200e−50t V, t ≥ 0 1 1 [d] w(0) = Cv 2 = (0.4 × 10−6 )(200)2 = 8 mJ 2 2 1 1 2 [e] w(t) = Cv (t) = (0.4 × 10−6 )(200e−50t )2 = 8e−100t mJ 2 2 The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains: 8 ×10−3 e−100t = 2 × 10−3 , e100t = 4, t = (ln 4)/100 = 13.86 ms

AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested voltage, vo , is the sum of the voltage drops for the two RC circuits. The circuit for t < 0 is shown below:

7–4

CHAPTER 7. Response of First-Order RL and RC Circuits

Find the current in the loop and use it to ﬁnd the initial voltage drops acrossthe two RC circuits: 15 = 0.2 mA, v5 (0− ) = 4 V, i= v1 (0− ) = 8 V 75,000 There are two time constants in the circuit, one for each RC subcircuit. τ5 is the time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time constant for the 1 µF – 40 kΩ subcircuit: τ5 = (20 × 103 )(5 × 10−6 ) = 100 ms; τ1 = (40 × 103 )(1 × 10−6 ) = 40 ms Therefore, v5 (t) = v5 (0− )e−t/τ5 = 4e−t/0.1 = 4e−10t...