Explanation Given data: Refer to Figure given in the textbook. The value of the inductor is L = 20 mH . The value of the capacitor is C = 500 nF The value of the voltage source is 36 V. The value of the current source is 20 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) Write the general expression for current through the inductor as follows, i L ( t ) = L d i L ( t ) d t (4) Calculation: In the given circuit, no energy is stored in the capacitor initially. Therefore, v o ( 0 + ) = 0 V . The switch in the given circuit has been open a long time before closing at t < 0 . Therefore, the capacitor acts as an open circuit and inductor behaves like a short circuit The initial current through the inductor i L ( 0 − ) is calculated as follows, i L ( 0 − ) = 36 300 = 0.12 A As no energy is stored in the capacitor initially, therefore v C ( 0 − ) = 0 V . The given circuit is modified as shown in Figure 2 for t > 0 . In the given circuit, the voltage source with series resistance is converted into current source with parallel resistance using source transformation method. The value of the current source is calculated as follows, I = 36 300 = 0.12 A The modified circuit is shown in Figure 3. In Figure 3, the 300 ohms and 150 ohms resistor are connected in parallel. Therefore, the equivalent resistance for the parallel connection is, 300 Ω | | 150 Ω = 300 Ω × 150 Ω 300 Ω + 150 Ω = 100 Ω And, The equivalent current source is, I 1 = 0.12 A − 20 × 10 − 3 A { ∵ 1 m = 10 − 3 } = 0.1 A The modified circuit is shown in Figure 4. Substitute 100 for R and 500 n for C in equation (2) as follows, α = 1 2 × ( 100 ) × ( 500 × 10 − 9 ) { ∵ 1 n = 10 − 9 } = 10000 rad s Squaring the above expression as follows, α 2 = 10000 2 rad s Substitute 500 n for C and 20 m for L in equation (1) as follows, ω 0 = 1 ( 500 × 10 − 9 ) ( 20 × 10 − 3 ) { ∵ 1 n = 10 − 9 , 1 m = 10 − 3 } ω 0 = 10000 rad s Squaring the above expression as follows, ω 0 2 = 10000 2 rad s As α 2 = ω 0 2 , the output response is critically damped. Hence, the output equation will be as follows, i L ( t ) = I f + D 1 t e − α t + D 2 e − α t (5) In Figure 4, the value of I f is, I f = 0.1 A Substitute 0.1 for I f and 10000 for α in equation (5) as follows, i L ( t ) = 0

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ISBN: 9780134746968

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 8, Problem 38P

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

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Chapter 8, Problem 37P

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During the lab you will design and measure a differential amplifier, made with an opamp. inside generator R5 ww 500 V1 0.1Vpk 1kHz 0° R6 w 50Ω R1 ww 10ΚΩ VCC C1 balanced wire R3 w 15.0V signal+ 100nF U1A TL082CP ground 2 signal- R4 w C2 Question5: Calculate R3 and R4 for a 20dB. 100nF VEE -15.0V R2 ww 10ΚΩ

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3. Consider the system described by the transfer function Gp(s) polynomial controller to satisfy the below specifications: 1) The settling time is t = 1 second, 2) 0.1% peak overshoot, 3) and zero steady-state error for a ramp input. The sampling period is T = 0.01 second. 1 = Design a discrete-time s(s+5)*

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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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