Physics 230 Spring 2000

Take C = 5 mF, L = 5 mH, E = 12 V and R = 6 W .

The switch has been closed in the circuit shown for a long time. What is the current through the inductor? What is the charge on the capacitor?

After a long time, the current is constant. That means there is no potential difference across the inductor. All the battery potential difference is across the resistor, and so the current is

I = E/R = (12 V)/(6 W ) = 2 A.

The potential difference across the capacitor equals the potential difference across the inductor, which is zero. Therefore the charge on the capacitor is zero.

At time t = 0 the switch is opened. What happens?

When the switch is opened the current can no longer flow through the loop on the right. But the inductor does not allow the current to drop instantly to zero. So the current flows in the loop on the left, charging the capacitor. As the capacitor charges, the potential difference across the inductor drops and the current decreases more slowly, until it becomes zero. Then all the energy is stored in the capacitor. The current then begins to increase in the opposite direction. The system oscillates.

What is the maximum charge on the capacitor, and when does it occur?

The charge is maximum when the current is zero. Using energy conservation:

Q2max/2C = LI2max/2

The maximum current is the current immediately after the switch is opened: 2 A.

Qmax = IÖ (LC) = (2 A) Ö (5 mH)(5 mF) = (2 A)(5 ms) = 10 mC

The maximum charge is on the capacitor after 1/4 period, that is at time

t = p /2w = (LC) /2= (5 ms)p /2 = 7.85 ms