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Fig. 2-23 (a) R 1 M; (b) R 100 k; (c) R 10 k. Hint: Compute the charge lost during the 1-ms period. Ans. (a) 0.1 V; (b) 1 V; (b) 10 V 2.26 The actual discharge current in Problem 2.25 is i 100=R e 10 t=R A. Find the capacitor voltage drop at 1 ms after connection to the resistor for (a) R 1 M; (b) R 100 k; (c) R 10 k. Ans. (a) 0.1 V; (b) 1 V; (c) 9.52 V A 10-mF capacitor discharges in an element such that its voltage is v 2e 1000t . delivered by the capacitor as functions of time. Ans. i 20e 1000t mA, p vi 40e 1000t mJ Find the current and power
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Find voltage v, current i, and energy W in the capacitor of Problem 2.27 at time t 0, 1, 3, 5, and 10 ms. By integrating the power delivered by the capacitor, show that the energy dissipated in the element during the interval from 0 to t is equal to the energy lost by the capacitor. Ans. t 0 1 ms 3 ms 5 ms 10 ms v 2V 736 mV 100 mV 13.5 mV 91 mV i 20 mA 7.36 mA 1 mA 135 mA 0.91 mA W 20 mJ 2.7 mJ 0.05 mJ % 0:001 mJ %0
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The current delivered by a current source is increased linearly from zero to 10 A in 1-ms time and then is decreased linearly back to zero in 2 ms. The source feeds a 3-k resistor in series with a 2-H inductor (see Fig. 2-24). (a) Find the energy dissipated in the resistor during the rise time W1 and the fall time W2 . (b) Find the energy delivered to the inductor during the above two intervals. (c) Find the energy delivered by the current source to the series RL combination during the preceding two intervals. Note: Series elements have the same current. The voltage drop across their combination is the sum of their individual voltages. Ans. a W1 100; W2 200; (b) W1 200; W2 200; (c) W1 300; W2 0, all in joules The voltage of a 5-mF capacitor is increased linearly from zero to 10 V in 1 ms time and is then kept at that level. Find the current. Find the total energy delivered to the capacitor and verify that delivered energy is equal to the energy stored in the capacitor. Ans. i 50 mA during 0 < t < 1 ms and is zero elsewhere, W 250 mJ.
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A 10-mF capacitor is charged to 2 V. A path is established between its terminals which draws a constant current of I0 . (a) For I0 1 mA, how long does it take to reduce the capacitor voltage to 5 percent of its initial value (b) For what value of I0 does the capacitor voltage remain above 90 percent of its initial value after passage of 24 hours Ans. (a) 19 ms, (b) 23.15pA Energy gained (or lost) by an electric charge q traveling in an electric eld is qv, where v is the electric potential gained (or lost). In a capacitor with charge Q and terminal voltage V, let all charges go from one plate to the other. By way of computation, show that the total energy W gained (or lost) is not QV but QV=2 and explain why. Also note that QV=2 is equal to the initial energy content of the capacitor. Ans. W qvdt Q V 0 QV=2 1 CV 2 . The apparent discrepancy is explained by the following. 2 2 The starting voltage vetween the two plates is V. As the charges migrate from one plate of the capacitor to the other plate, the voltage between the two plates drops and becomes zero when all charges have moved. The average of the voltage during the migration process is V=2, and therefore, the total energy is QV=2. Lightning I. The time pro le of the discharge current in a typical cloud-to-ground lightning stroke is modeled by a triangle. The surge takes 1 ms to reach the peak value of 100 kA and then is reduced to zero in 99 mS. (a) Find the electric charge Q discharged. (b) If the cloud-to-ground voltage before the discharge is 400 MV, nd the total energy W released and the average power P during the discharge. (c) If during the storm there is an average of 18 such lightning strokes per hour, nd the average power released in 1 hour. Ans. (a) Q 5 C; (b) W 109 J; P 1013 W; (c) 5 MW Lightning II. Find the cloud-to-ground capacitance in Problem 2.33 just before the lightning stroke. Ans. 12.5 mF Lightning III. The current in a cloud-to-ground lightning stroke starts at 200 kA and diminishes linearly to zero in 100 ms. Find the energy released W and the capacitance of the cloud to ground C if the voltage before the discharge is (a) 100 MV; (b) 500 MV. Ans. (a) W 5 108 J; C 0:1 mF; (b) W 25 108 J; C 20 nF The semiconductor diode of Example 2.4 is placed in the circuit of Fig. 2-25. (a) Vs 1 V, (b) Vs 1 V. Ans. (a) 14 mA; (b) 0 Find the current for
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The diode in the circuit of Fig. 2-26 is ideal. The inductor draws 100 mA from the voltage source. A 2-mF capacitor with zero initial charge is also connected in parallel with the inductor through an ideal diode such that the diode is reversed biased (i.e., it blocks charging of the capacitor). The switch s suddenly disconnects with the rest of the circuit, forcing the inductor current to pass through the diode and establishing 200 V at the capacitor s terminals. Find the value of the inductor. Ans. L 8 H Compute the static and dynamic resistances of the diode of Example 2.4 at the operating point v 0:66 V. Ans: R% 0:66 0:67 0:65 550  and r % 21:7  1:2 10 3 1:7 0:78 10 3
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