Given the circuit shown in the figure above assume that the

Given the circuit shown in the figure above, assume that the switch S has been closed for a long time so that a steady current exists in the inductor. The inductor has negligible resistance. Assume L = 1 H, the larger resistance R_1 = 105 ohm, the smaller resistance R_2 = 13 ohm and the EMF E = 9 V. Find the current passing through the battery.

Solution

For an inductor places in an electric circuit, it acts like an open circuit, however, with time, the inductors acts like a closed circuit and offer no resistance, [Opposite happens with a capacitor]

As the switch is kept open for a very long time, we can say that the inductor will act as a closed circuit and will effectively short the resistor R2, thereby leaving R1 as the only resistor in the circuit with the potential difference of 9 Volts.

Therefore the current in the circuit with emf of V and resistance of R is given as I = V/R

That is the current through the battery = 9/13 = 0.692 amperes