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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. |
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Answer» <P> Solution :(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. 
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