A 40 resistoris connected in series with a 7 µFcapacitor and a battery. What is
ID: 1672584 • Letter: A
Question
A 40 resistoris connected in series with a 7 µFcapacitor and a battery. What is the maximum charge to which thiscapacitor can be charged when the battery voltage is 9 V? (b) Initially the capacitor has zero charge. At what timeduring the charging process will the charge on this capacitor be39.8 µC (c) The capacitor is now fully charged. The battery is removedand the capacitor is discharged through the resistor. What will bethe charge on the capacitor after 0.28ms?A 40 resistoris connected in series with a 7 µFcapacitor and a battery. What is the maximum charge to which thiscapacitor can be charged when the battery voltage is 9 V? (b) Initially the capacitor has zero charge. At what timeduring the charging process will the charge on this capacitor be39.8 µC (c) The capacitor is now fully charged. The battery is removedand the capacitor is discharged through the resistor. What will bethe charge on the capacitor after 0.28ms?
Explanation / Answer
(a)The maximum charge to which this capacitor can be chargedwhen the battery voltage is 9 V is Qo = C * V where C = 7 F = 7 * 10-6 F and V = 9 V or Qo = 7 * 10-6 * 9 = 63 *10-6 C = 63 C (b)The charge on the capacitor in a RC circuit is Q = Qo * (1 - e-t/RC)-----------(1) or (Q/Qo) = 1 - e-t/RC or e-t/RC = 1 - (Q/Qo) or (-t/RC) = ln{1 - (Q/Qo)} or (t/RC) = -ln{1 - (Q/Qo)} or (t/RC) = ln{1 - (Q/Qo)}-1 or (t/RC) = ln{1/1 - (Q/Qo)} or t = RC * ln{1/1 - (Q/Qo)} where R = 40 and Q = 39.8 C = 39.8 *10-6 C (c)The charge on the capacitor after 0.28 ms is Q = Qo * (1 - e-t/RC) where t = 0.28 ms = 0.28 * 10-3 sor (t/RC) = ln{1/1 - (Q/Qo)} or t = RC * ln{1/1 - (Q/Qo)} where R = 40 and Q = 39.8 C = 39.8 *10-6 C (c)The charge on the capacitor after 0.28 ms is Q = Qo * (1 - e-t/RC) where t = 0.28 ms = 0.28 * 10-3 s