ch07_ssm - 7 Response of First-Order RL and RC Circuits...

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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 2Ω resistor from the circuit. First combine the 30 Ω and 6Ω resistors in parallel: 30 k 6=5Ω Use voltage division to ±nd the voltage drop across the parallel resistors: v = 5 5+3 (120) = 75 V Now ±nd the current using Ohm’s law: i (0 )= v 6 = 75 6 = 12 . 5 A [b] w (0) = 1 2 Li 2 (0) = 1 2 (8 × 10 3 )(12 . 5) 2 = 625 mJ [c] To ±nd the time constant, we need to ±nd the equivalent resistance seen by the inductor for t> 0 . When the switch opens, only the resistor remains connected to the inductor. Thus, τ = L R = 8 × 10 3 2 =4 ms [d] i ( t i (0 ) e t/τ = 12 . 5 e t/ 0 . 004 = 12 . 5 e 250 t A ,t 0 [e] i (5 ms 12 . 5 e 250(0 . 005) = 12 . 5 e 1 . 25 = 3 . 58 A 7–1
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7–2 CHAPTER 7. Response of First-Order RL and RC Circuits So w (5 ms )= 1 2 Li 2 (5 ms 1 2 (8) × 10 3 (3 . 58) 2 =51 . 3 mJ w ( dis ) = 625 51 . 3 = 573 . 7 mJ % dissipated = ± 573 . 7 625 ² 100=91 . 8% AP 7.2 [a] First, use the circuit for t< 0 to ±nd the initial current in the inductor: Using current division, i (0 10 10+6 (6 . 4)=4 A Now use the circuit for t> 0 to ±nd the equivalent resistance seen by the inductor, and use this value to ±nd the time constant: R eq =4 k (6 + 10) = 3 . 2Ω , · .. τ = L R eq = 0 . 32 3 . 2 =0 . 1 s Use the initial inductor current and the time constant to ±nd the current in the inductor: i ( t i (0 ) e t/τ e t/ 0 . 1 e 10 t A ,t 0 Use current division to ±nd the current in the 10 Ω resistor: i o ( t 4 4+10+6 ( i 4 20 ( 4 e 10 t 0 . 8 e 10 t A 0 + Finally, use Ohm’s law to ±nd the voltage drop across the 10 Ω resistor: v o ( t )=10 i o = 10( 0 . 8 e 10 t 8 e 10 t V 0 + [b] The initial energy stored in the inductor is w (0) = 1 2 Li 2 (0 1 2 (0 . 32)(4) 2 =2 . 56 J Find the energy dissipated in the 4Ω resistor by integrating the power over all time: v 4Ω ( t L di dt . 32( 10)(4 e 10 t 12 . 8 e 10 t V 0 +
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Problems 7–3 p 4Ω ( t )= v 2 4Ω 4 =40 . 96 e 20 t W ,t 0 + w 4Ω ( t Z 0 40 . 96 e 20 t dt =2 . 048 J Find the percentage of the initial energy in the inductor dissipated in the 4Ω resistor: % dissipated = ± 2 . 048 2 . 56 ² 100 = 80% AP 7.3 [a] The circuit for t< 0 is shown below. Note that the capacitor behaves like an open circuit. Find the voltage drop across the open circuit by ±nding the voltage drop across the 50 k resistor. First use current division to ±nd the current through the 50 k resistor: i 50k = 80 × 10 3 80 × 10 3 +20 × 10 3 +50 × 10 3 (7 . 5 × 10 3 )=4 mA Use Ohm’s law to ±nd the voltage drop: v (0 ) = (50 × 10 3 ) i 50k = (50 × 10 3 )(0 . 004) = 200 V [b] To ±nd the time constant, we need to ±nd the equivalent resistance seen by the capacitor for t> 0 . When the switch opens, only the 50 k resistor remains connected to the capacitor. Thus, τ = RC = (50 × 10 3 )(0 . 4 × 10 6 )=20 ms [c] v ( t v (0 ) e t/τ = 200 e t/ 0 . 02 = 200 e 50 t V 0 [d] w (0) = 1 2 Cv 2 = 1 2 (0 . 4 × 10 6 )(200) 2 =8 mJ [e] w ( t 1 2 2 ( t 1 2 (0 . 4 × 10 6 )(200 e 50 t ) 2 e 100 t mJ The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains: 8 × 10 3 e 100 t
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This note was uploaded on 04/20/2011 for the course EECE 222 taught by Professor Amsddasgf during the Spring '11 term at American University in Bulgaria.

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ch07_ssm - 7 Response of First-Order RL and RC Circuits...

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