Two identical capacitors A and B , charged to the same potential 5 V are connected in two different circuits as shown below at time t = 0 . If the charge on capacitors A and B at time t = C R is Q A and Q B respectively, then (Here e is the base of natural logarithm)

Question

Two identical capacitors A and B , charged to the same potential 5 V are connected in two different circuits as shown below at time t = 0 . If the charge on capacitors A and B at time t = C R is Q A and Q B respectively, then (Here e is the base of natural logarithm), Q A = V C e , Q B = C V 2, Q A = V C , Q B = C V, Q A = V C , Q B = V C e, Q A = C V 2 , Q B = V C e

MARKS App · Accepted Answer

Charge on capacitor after charging with a battery of potential V , Q = C V Now, capacitor A is connected as shown in the figure. As diode is in reverse bias, so capacitor charge remains same after time t = C R . ∴ Q A = C V In case B , capacitor is connected as shown in the figure. Here, diode is in forward bias, so capacitor discharges and its charge decays as Q B = Q · e - t R C where, t = R C

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