(a) To determine Find the expression for v o ( t ) and v o 1 ( t ) for the time intervals 0 < t ≤ 0.5 s and 0.5 s < t ≤ t sat . Explanation Given data: Refer to Figure given in the textbook. The value of the resistors is, R 1 = 100 k Ω R 2 = 400 k Ω The value of the capacitors is, C 1 = 500 nF C 2 = 200 nF The voltage v g is 80 mV. Formula used: Write the general expression differential equation that shows the relation between v o and v g as follows, d 2 v o d t 2 = ( 1 R 1 C 1 ) ( 1 R 2 C 2 ) v g (1) Write the general expression differential equation that shows the relation between v o 1 and v g as follows, d v o 1 ( t ) d t = − ( 1 R 1 C 1 ) v g (2) Calculation: Substitute 100 k for R 1 , 400 k for R 2 , 500 n for C 1 , and 200 n for C 2 in equation (1) as follows, d 2 v 0 d t 2 = ( 1 100 × 10 3 × 500 × 10 − 9 ) ( 1 400 × 10 3 × 200 × 10 − 9 ) v g { ∵ 1 k = 10 3 , 1 n = 10 − 9 } d 2 v 0 d t 2 = ( 20 ) ( 12.5 ) v g (3) At 0 < t ≤ 0.5 − , Substitute 80 m for v g in equation (3) as follows, d 2 v 0 d t 2 = ( 20 ) ( 12.5 ) ( 80 × 10 − 3 ) d 2 v 0 d t 2 = 20 Consider, g ( t ) = d v o d t Therefore, d g ( t ) d t = d 2 v o d t 2 Substitute 20 for d 2 v o d t 2 in the above expression as follows, d g ( t ) d t = 20 d g ( t ) = 20 d t Integrating the above equation with the limits t → 0 to t , ∫ g ( 0 ) g ( t ) d g ( t ) = ∫ 0 t 20 d t g ( t ) − g ( 0 ) = 20 t By considering the initial conditions, g ( 0 ) = d v o d t ( 0 ) = 0 Therefore, g ( t ) = d v o d t = 20 t d v o = 20 t d t Integrating the above equation with the limits t → 0 to t , ∫ v o ( 0 ) v o ( t ) d v o = 20 ∫ 0 t t d t v o ( t ) − v o ( 0 ) = 20 [ t 2 2 ] v o ( t ) − v o ( 0 ) = 10 t 2 v o ( t ) = 10 t 2 V, 0 ≤ t ≤ 0.5 − { ∵ v o ( 0 ) = 0 } Substitute 100 k for R 1 , 500 n for C 1 , and 80 m for v g in equation (2) as follows, d v o 1 ( t ) d t = − ( 1 100 × 10 3 × 500 × 10 − 9 ) ( 80 × 10 − 3 ) d v o 1 ( t ) d t = − 1.6 d v o 1 ( t ) = − 1.6 d t Integrating the above equation with the limits t → 0 to t , ∫ v o 1 ( 0 ) v o 1 ( t ) d v o 1 ( t ) = − 1.6 ∫ 0 t d t v o 1 ( t ) − v o 1 ( 0 ) = − 1.6 [ t ] v o 1 ( t ) − 0 = − 1.6 t { v o 1 ( 1 ) = 0 } v o 1 ( t ) = − 1.6 t V, 0 ≤ t ≤ 0.5 − At 0.5 + ≤ t ≤ t sat , Substitute − 40 m for v g in equation (3) as follows, d 2 v 0 d t 2 = ( 20 ) ( 12.5 ) ( − 40 × 10 − 3 ) d 2 v 0 d t 2 = − 10 Consider, g ( t ) = d v o d t Therefore, d g ( t ) d t = d 2 v o d t 2 Substitute − 10 for d 2 v o d t 2 in the above expression as follows, d g ( t ) d t = − 10 d g ( t ) = − 10 d t Integrating the above equation with the limits t → 0 to t , ∫ g ( 0.5 + ) g ( t ) d g ( t ) = ∫ 0.5 + t − 10 d t g ( t ) − g ( 0.5 + ) = [ − 10 t ] 0.5 + t g ( t ) − g ( 0.5 + ) = − 10 t + 5 g ( t ) = − 10 t + 5 + g ( 0 (b) To determine Find the values of saturation time t sat for voltages v o 1 ( t ) and v o ( t ) .

11th Edition

ISBN: 9780134746968

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 8, Problem 63P

(a)

To determine

Find the expression for vo(t) and vo1(t) for the time intervals 0<t≤0.5 s and 0.5 s<t≤tsat.

(b)

To determine

Find the values of saturation time tsat for voltages vo1(t) and vo(t).

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Chapter 8 Solutions

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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