To determine Find the expression for the inductor current i L ( t ) for t ≥ 0 . Explanation Given data: Refer to Figure given in the textbook. The value of the inductor is L = 4 H . The value of the capacitor is C = 250 nF The value of the voltage source is 150 V. The value of the current source is 100 mA. Formula used: Write the general expression for resonant radian frequency as follows, ω 0 = 1 L C (1) Write the general expression for neper frequency as follows, α = 1 2 R C (2) Write the general expression for damoing frequency as follows, ω d = ω 0 2 − α 2 (3) Calculation: Consider the circuit for t < 0 . In Figure 1, the switch 1 is opened and switch 2 is closed. Therefore, in the circuit the capacitor acts as an open circuit and inductor as a short circuit. The capacitor voltage v C ( 0 − ) is calculated by using voltage division principle as follows, v C ( 0 − ) = ( 150 ) ( 3750 ( 7.5 × 10 3 ) + 3750 ) { ∵ 1 k = 10 3 } = 50 V The initial current in the inductor is, i L ( 0 − ) = 100 mA Consider the circuit for t > 0 . In Figure 2, the voltage source with series resistance is converted into current source with parallel resistance by using source transformation. The value of current source is calculated as follows, I = 150 7.5 × 10 3 = 0.02 A The modified circuit is shown in Figure 3. In Figure 3, 7.5 k Ω and 3750 Ω are connected in parallel. Therefore, the equivalent resistance for the parallel connection is, 7.5 k Ω | | 3750 Ω = 7.5 k Ω × 3750 Ω 7.5 k Ω + 3750 Ω = 2500 Ω The modified circuit is shown in Figure 4. Substitute 2500 for R and 250 n for C in equation (2) as follows, α = 1 2 × ( 2500 ) × ( 250 × 10 − 9 ) { ∵ 1 n = 10 − 9 } = 800 rad s Squaring the above expression as follows, α 2 = 800 2 rad s Substitute 250 n for C and 4 for L in equation (1) as follows, ω 0 = 1 ( 250 × 10 − 9 ) ( 4 ) { ∵ 1 n = 10 − 9 } ω 0 = 1000 rad s Squaring the above expression as follows, ω 0 2 = 10 6 rad s As α 2 < ω 0 2 , the output response is under damped. Hence, the output equation will be as follows, i L ( t ) = I f + D 1 cos ω d t e − α t + D 2 sin ω d t e − α t (4) Substitute 10 6 for ω 0 2 and 800 2 for α 2 in equation (3) as follows, ω d = 10 6 − 800 2 ω d = 600 rad s In Figure 4, the value of I f is, I f = 0

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Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 8, Problem 35P

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Find the expression for the inductor current iL(t) for t≥0.

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Chapter 8, Problem 34P

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Electric Circuits. (11th Edition)

Ch. 8.1 - The resistance and inductance of the circuit in...Ch. 8.2 - Use the integral relationship between iL and v to...Ch. 8.2 - Prob. 3AP Ch. 8.2 - Prob. 4AP Ch. 8.2 - Prob. 5AP Ch. 8.3 - Prob. 6AP Ch. 8.4 - Prob. 7AP Ch. 8.4 - Prob. 8AP Ch. 8.4 - Repeat Assessment Problems 8.7 and 8.8 if the 80 Ω...Ch. 8 - The resistance, inductance, and capacitance in a...

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