DeCarlo Ch7 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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L&C Probs, 11/15/01 P:7-1 © R. A. DeCarlo, P. M. Lin PROBLEM SOLUTIONS CHAPTER 7 S OLUTION 7.1 . Given the coil has 48 turns and 12 turns/cm, we know that the length of the coil is 4 cm. Since the length of the coil is greater than 0.4 times its diameter, the formula given in the question can be used: L = 4 × 10 - 5 × 0.02 ( 29 2 × 48 ( 29 2 18 × 0.02 ( 29 + 40 × 0.04 ( 29 = 18.81 H S OLUTION 7.2 . Part 1. Applying (7.1) The voltage v L (t) can be computed using the inductor v-i relationship: v L ( t ) = L di L ( t ) dt The calculations for v L (t) for t = 0s to 5s are summarized in the following table: Time Interval d/dt (i in (t)) v L (t) 0s< t 1s 2 As -1 1V 1s< t 3s -2 As -1 -1V 3s< t 4s 2 As -1 1V 4s< t 5s 0 As -1 0V Below is the plot of V L vs time. Part 2. Applying (7.4) w L ( t ) = 1 2 (0.5 i in 2 ( t )) In the time interval 0s < t 1s, i in ( t ) = 2 t . Thus w L ( t ) = 1 2 0.5 i in 2 ( t ) ( 29 = t 2 In the time interval 1s < t 3s, i in ( t ) = 4 - 2 t . Thus
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L&C Probs, 11/15/01 P:7-2 © R. A. DeCarlo, P. M. Lin w L ( t ) = 1 2 0.5 i in 2 ( t ) ( 29 = 1 4 (4 - 2 t ) 2 = t 2 - 4 t + 4 In a similar way, w L ( t ) can be computed in the remaining intervals. The calculations for w L (t) for t = 0s to 5s are summarized in the following table: Time Interval i in (t) w L (t) 0s< t 1s 2t t 2 1s< t 3s 4-2t t 2 -4t+4 3s< t 4s -8+2t t 2 -8t+16 4s< t 5s 0 0 Below is the plot of w L vs t. S OLUTION 7.3 . Applying (7.2) i L ( t ) = i L (0) + 1 L v L ( ) d 0 t It is assumed that i L (0) = 0 A. Using the preceding formula and the fact that v in (t) = t in the time interval 0s < t 2s, the current is i L ( t ) = 1 0.5 d = t 2 0 t in the time interval 0s < t 2s In the time interval 2s < t 3s v in ( t ) = 2 . Hence i L ( t ) = 1 0.5 v L ( ) d 0 t = i L (2) + 1 0.5 2 dt 2 t = 4 + 4 2 t = 4 t - 4 In a similar way i L ( t ) can be computed in the remaining intervals. The calculations for i L (t) for t = 0s to 6s are summarized in the following table:
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L&C Probs, 11/15/01 P:7-3 © R. A. DeCarlo, P. M. Lin Time Interval v in (t), V i L (t), A 0s< t 2s t t 2 2s< t 3s 2 4t-4 3s< t 5s -2 20-4t 5s< t 6s 0 0 Below is the plot of i L vs t. S OLUTION 7.4 . Part 1. Using (7.1), we have v in ( t ) = 0.2 × 10 - 3 d dt i s ( t ) [ ] = 0.2 × 10 - 3 × 1000cos(1000 t ) = 0.2cos(1000 t ) mV For the 2mH inductor i out ( t ) = i out (0) + 1 L 10 v in 0 t ( ) d = 1 2 × 10 - 3 2cos 0 t (1000 ) d = sin(1000 t ) mA Below is a sketch of i out vs t. Part 2. Instantaneous power delivered by the dependent source is given by p L ( t ) = v L ( t ) × i L ( t ) = 10 v in ( t ) × i out ( t ) = 2cos(1000 t ) × sin(1000 t ) = sin(2000 t ) W
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L&C Probs, 11/15/01 P:7-4 © R. A. DeCarlo, P. M. Lin Part 3. The energy stored in the 2mH inductor is given by W L ( t ) = 1 2 Li L 2 ( t ) = 1 2 Li out 2 ( t ) = sin 2 (1000 t ) nJ Below is a sketch of W L vs t S OLUTION 7.5 . Part 1 For the excitation in Figure P7.5b, i 1 ( t ) = i 1 (0) + 1 0.5 v in 0 t ( ) d , i 2 ( t ) = i 2 (0) + 1 0.25 v in 0 t ( ) d = 2 i 1 ( t ) It is assumed that i 1 (0) = i 2 (0) = 0 A . In the interval 0s < t 1s, v in ( t ) = - 10V . Hence, in this interval i 1 ( t ) = 1 0.5 v in ( ) d 0 t = 1 0.5 ( - 10) d 0 t = - 20 t i 2 ( t ) = 1 0.25 v in ( ) d 0 t = 1 0.25 ( - 10) d 0 t = - 40 t In the interval 1s < t 2s, v in ( t ) = - 5V . Hence, in this interval, i 1 ( t ) = i 1 (1) + 1 0.5 v in ( ) d 1 t = - 20 + 1 0.5 ( - 5) d 1 t = - 10 t - 10 i 2 ( t ) = i 2 (1) + 1 0.25 v in ( ) d 1 t = - 40 + 1 0.25 ( - 5) d 1 t = - 20 t - 20 In a similar fashion i 1 (t) and i 2 (t) can be computed in other intervals.
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