Question

You have a circuit with four elements. Traversing the circuit in clockwise fashion, each is reached in turn. First is the emf. which varies in such a way that it always produces a constant current i if the switch is closed. Second is the switch which is initially open so that there is no current. Third and fourth are reached at the same time because they are a resistor (of resis­tance R) and an inductor (of inductance L) connected in parallel. If the switch is suddenly closed at t = 0 so that the emf suddenly sends the current i across the switch, show that the current as a function of time through the resistor is given by iR = ie-Rt/L.(Hint: You must use the junction rule, the loop rule, and the derivative of the junction rule.)

Answer 1

Answer 2

To see why, apply Kirchoff's loop rule. With the switch opened we have:

IR + L dI/dt = 0

or dI/dt = -IR/L

Solving this for current gives I(t) = I_o e^-t/t

The potential difference across the resistor has a similar form:

DV_R = e e^-t/t

The potential difference across the inductor is negative because it's acting like a battery hooked up in the opposite direction:

DV_L = -e e^-t/t

Once again, the graph of current as a function of time in the RL circuit has the same form as the graph of the capacitor voltage as a function of time in the discharging RC circuit, while the graph of the inductor voltage as a function of time in the RL circuit has the same form as the graph of current vs. time in the discharging RC circuit.