HW5solns - HW #5 Solutions E245B 10/11/01 Problem 5.1 A 12F...

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Unformatted text preview: HW #5 Solutions E245B 10/11/01 Problem 5.1 A 12F capacitor has an accumulated charge of 480C. Determine the voltage across the capacitor after 4 s. Suggested Solution t2 v(t 2) - v(t1) = 1 C t1 v(t 1) = 0. v(t 2) = 40V i(t )dt C = 100 F i (t ) = 1mA = I so, I v(t 2) = C (t 2 - t1) where t 2 - t1 = 4sec Problem 5.16 Draw the waveform for the current in a 12-F capacitor when the capacitor voltage is as described in Figure 5.16 v(t)(V) 12 10 0 -8 6 16 t(s) Figure P5.16 Suggested Solution C = 12 F Time(s) i (t ) = C dv dt dv(V/s) dt i(t)(A) 0t 6 t 10 10 t 16 t > 16 v(t)(V) 2 -5 1.33 0 24 - 60 16 0 10 5 0 -5 -10 t(s) 0 2 4 6 8 10 12 14 16 i(t)(mA) 30 10 -10 -30 -50 -70 0 2 4 6 8 10 12 14 16 t(s) v(t)(V) 10 5 0 -5 -10 t(s) 0 2 4 6 8 10 12 14 16 i(t)(mA) 30 10 -10 -30 -50 -70 0 2 4 6 8 10 12 14 16 t(s) Problem 5.31 Draw the waveform for the voltage across a 10-mH inductor when the inductor current is given by the waveform shown in Figure P5.31 i(t)(A) 4 2 6 -1 3 Figure P5.31 9 11 t(s) Suggested Solution L = 10mH Time(s) v(t ) = L di dt di(A/s) dt v(t)(mV) 0t 3 3t 6 6t 9 9 t 11 t > 11 2/3 -1 0 2.5 0 6.67 - 10 0 25 0 4 2 i(t)(A) 0 -2 0 3 6 time(s) 30 v(t)(mV) 15 0 9 11 12 -15 0 3 6 time(s) 9 11 12 Problem 5.36 The capacitor in Figure P5.36a is 51 nF with a tolerance of 10%. Given the voltage waveform in Figure P5.36b graph the current i(t) for the minimum and maximum capacitor values. i(t) -60 v(t)(V) v(t) C 0 -60 0 1 2 3 4 time (ms) (b) Figure P5.36 5 6 (a) Suggested Solution Maximum capacitor value = 1.1C = 56.1 nF Minimum capacitor value = 0.9C = 45.9 nF The capacitor voltage and current are related by the equation ( i (t ) = C dvdtt ) Problem 5.44 Two capacitors are connected in series as shown in Figure P5.44. Find Vo i(t) + + 24V - Vo C1 C2 C1 = 12F C2 = 6F Suggested Solution i(t) + + 24V - Vo C1 C2 C1 = 12F C2 = 6F vc = 1 C i(t )dt Since same current charged both caps, C1Vo = C2 (24) Vo = 24(C2 / C1 ) Vo = 12V ...
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