The given Fig. 6.39, shows an inductor L and resistor R connected in parallel to a battery B through a switch S. The resistance of R is same as that of the coil that makes L. Two identical bulbs Pand Q are put in each arm of the circuit as shown.
(a) When S is closed, which of the two bulbs will light up earlier ?
(b) Will the bulbs be equally bright after some time? Justify your answer.

Explanation

(a) The bulb P joined in the arm containing resistor R will light up earlier when switch S is closed. On closing the switch, a self-induced back emf is developed in the inductor L in a direction opposite to the emf of battery B whereas no such emf is induced in resistor R. As a result, bulb P glows up earlier and the bulb Q glows up a bit late.
(b) Yes, after sometime both the bulbs will glow equally bright. It is because once the current has reached its maximum value, self-inductance plays no role. As resistance of both R and L is same and the bulbs are identical, both will finally glow with equal brightness.