Question

An RC circuit consists of a 3.00 V battery attached to a 50.0 uF capacitor in series with a 100.0 kilo-Ohm resistor. The circuit has a switch that is initially open (no initial charge on the capacitor).

A: find the current through the circuit immediately after the switch is closed

B: find the current 12 seconds after the switch is closed.

Answer 1

R = 100.0 kilo ohm = 100 x 10^3 ohm = 10^5 ohm

C = 50.0 F = 50 x 10^-6 F

So, time constant of the circuit, T = RC = 10^5 x 50 x 10^-6 = 5.0 s

(A) When the switch is closed, the capacitor will behave as short circuited.

So, current through the circuit immediately after the switch is closed,

Io = V/R = 3.00 / (10^5) = 3.00 x 10^-5 A (Answer)

(B) Current through the circuit after time t, is given as -

I = Io * e^-t/T

Given that, t = 12 s

So, I = (3.00 x 10^-5) * e^-12/5

= 3.00 x 10^-5 x 0.0907

= 2.72 x 10^-6 A (Answer)