Fig. 2-23 (a) R ¼ 1 M; (b) R ¼ 100 k; (c) R ¼ 10 k. Hint: Compute the charge lost during the 1-ms period. Ans. (a) 0.1 V; (b) 1 V; (b) 10 V 6
The actual discharge current in Problem 2.25 is i ¼ ð100=RÞe10 t=R A. Find the capacitor voltage drop at 1 ms after connection to the resistor for (a) R ¼ 1 M; (b) R ¼ 100 k; (c) R ¼ 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 ¼ 2e1000t . delivered by the capacitor as functions of time. Ans. i ¼ 20e1000t mA, p ¼ vi ¼ 40e1000t mJ
Find voltage v, current i, and energy W in the capacitor of Problem 2.27 at time t ¼ 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.
Find the current and power
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 ðW1 Þ and the fall time ðW2 Þ. (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. ðaÞ W1 ¼ 100; W2 ¼ 200; (b) W1 ¼ 200; W2 ¼ 200; (c) W1 ¼ 300; W2 ¼ 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 ¼ 50 mA during 0 < t < 1 ms and is zero elsewhere, W ¼ 250 mJ.