A capacitor with C = 1.40 times 10^- 5 F is connected as shown in the figure (Fi

Question

A capacitor with C = 1.40 times 10^- 5 F is connected as shown in the figure (Figure 1)with a resistor with R = 990 Ohm and an emf source with epsilon = 180 V and negligible internal resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge. After the switch has been in position 2 for 10.0 ms, the switch is moved back to position 1 so that the capacitor begins to discharge. Compute the voltage drops across the resistor at the instant described in part A. Compute the voltage drops across the capacitor at the instant described in part A. Compute the voltage drops across the resistor just after the switch is shown from position 2 back to position 1.

Explanation / Answer

A) q= CV(1-e^-t/RC)

= 1.4e-5*18*(1-e^(-0.010/(990*1.4e-5)))

= 130*10^-6 C

= 130 uC

B) V = E- q/C

= 18 - 130e-6/1.4e-5

= 8.71 V

C) V = 18 - 8.71 = 9.29 V

D) since voltage drop will be same as capacitor and voltage doesn't change across capacitor abruptly,

V = 8. 71 V

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