Consider the following circuit. The capacitor is initially uncharged, and at tim

Question

Consider the following circuit. The capacitor is initially uncharged, and at time t = 0 the switch is closed. Find an expression for Q(t), where Q is the charge on the capacitor plates, in terms of epsilon, R and C. Check that Q(t=0) and Q(t= infinity) are what you expect. Find an expression for I(t), where I is the current flowing in the circuit. Cheek that I(t=0) and I(t= infinity) arc what you expect. (i) How much energy is dissipated in resistors R1 and R2 individually when the capacitor is charging? (ii) How much energy is stored in the capacitor when it is fully charged? Check that the total energy dissipated in the two resistors and the energy stored in the capacitor is equal to the total energy supplied by the battery.

Explanation / Answer

a)

To analyze this circuit quantitatively, let apply Kirchhoff's loop rule to the circuit after the switch is closed.

- q/C - I (3R) = 0

Substitute I = dq/dt into the equation:

dq/dt = /3R - q/3RC = 0

Combine the terms on the right-hand side:

dq/dt = C/3RC - q/3RC = - (q-C/3RC)

So

dq/(q-C) = - (1/3RC) dt

Integrating this expression, using the fact that q=0 at t=0, we obtain

ln(q-C/(-C)) = -t/3RC

We can write this expression as

q(t) = C (1 - e-t/3RC)

b)

Using I=dq/dt, we find that

I(t) = dq/dt = d( C (1 - e-t/3RC))/dt = C (0 - (-1/3RC)e-t/3RC) = ( C/3RC)e-t/3RC) = (/3R) e-t/3RC

c)

The energy dissipated in the two resistors = E3R

E3R = (-3R) I2 dt = -3R ((/3R) e-t/3RC)2 dt = (-2/3R) e-2t/3RC dt = (-C2/2) (e-2t/3RC - 1)

The energy stored in the capacitor = EC

Ec = q2/2C = ( C (1 - e-t/3RC))2/2C = (2C/2) (1 - e-t/3RC)2

The total energy supplied by the battery = Eb

Eb = - I dt = - (/3R)e-t/3RC dt = -C2 (e-t/3RC - 1)

E3R + Ec = (-C2/2) (e-2t/3RC - 1) + (2C/2) (1 - e-t/3RC)2 = -C2 (e-t/3RC - 1) = Eb

So,

E3R + Ec = Eb

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