Question

You connect an 8.00 MΩ resistor in a series with a 4.70μF capacitor and a battery with emf 9.00 V. Before you close the switch at t=0 to complete the circuit, the capacitor is uncharged.

1. Find the final charge

2. Find the Initial current

3. Find the time constant

4. How long does it take for the capac8to acquire half the final charge?

Answer 1

R = Resistance = 8 x 10⁶, C = Capacitance = 4.7 x 10^-6 F, V = Emf of the battery = 9 V.

(1) Final charge in the capacitor = Q = CV = 4.7 x 10^-6 F x 9 V = 4.23 x 10^-5 C.

(2) Initially the capacitor was discharged, hence, there was no initial current in the circuit.

(3) Time constant of the series RC circuit = = RC = 8 x 10⁶ x 4.7 x 10^-6 F = 37.6 seconds.

(4) Charge in the capacitor at time t is given by :

q ( t ) = Q [ 1 - exp ( - t /   ) ].

For a charge q = Q / 2 in the capacitor,

Q / 2 = Q [ 1 - exp ( - t /   ) ]

or, 1 - exp ( - t /   ) = 1 / 2

or, exp ( - t /   ) = 1 / 2

or, exp ( t /   ) = 2

or, t / = ln 2

or, t = ln 2 = ( 37.6 s ) x ln 2

or, t ~ 26.06 s.

Hence, in 26.06 seconds, the capacitor will acquire half of its maximum charge.