Consider the following circuit The switch is closed How long

Consider the following circuit. The switch is closed. How long does it take until the charge on the capacitor reaches 75% of its maximum value? The switch id then opened after it had been closed for a long time. How long does it take for the charge on the capacitor to drop to 50% of its initial value.

Solution

Here, we need to keep the following concepts into consideration:

1.) The voltage difference around two parallel circuits is same. As in the above case, the voltage difference across the arm containing the capacior and the one containing 125 ohms resistor would be same.

2.) Once the capacitor is connected to an emf source it starts getting charged with a time constant of RC where R is the resistance of the circuit and C is the capacitor and equation for charge at any time t is given as:

Q(t) = Qo (1 - e^-t/RC) where Q = VC

3.) When the emf source is removed, the capacitor starts to discharge and the relation for charge at any time t is

Q(t) = Qo e^-t/RC

We will make use of the above to solve the given problems as:

Part A.) For the given circuit we have: R =57 Ohms, C = 745 uF and V = 24 V

For charge to reach 75% of its maximum charge would be:

0.75Q = Q(1 - e^-t/RC)

or, e^-t/RC = 0.25

or t/RC = 1.3863

or, T = 1.3863 x 57 x 745 x 10^-6 = 5.8869 x 10^-2 seconds is the required time

Part B.) For discharging the net resistance in the circuit would 57 + 125 = 182Ohms,

Hence we have:

Q(0.5) = Qe^-t/RC

or, t/RC = 0.69315

or, T = 0.69315 x 182 x 745 x 10^-6 = 9.3984 x 10^-2 seconds is the required time.