Question

In the given circuit, the value of resistance effect of the coil L is exactly equal to the resistance R. Bulbs B1 and B2 are exactly identical. Answer the following question.
a) Which one of the two bulbs lights up earlier, when key k is closed and why?
b) What will be the comparative brightness of the two bulbs after sometime if the key K is kept closed and why?

Answer 1

a) As given in the question a coil of resistance $R$ and inductance $L$ connected in parallel to a resistor $R$ and in turn the combination connected to a battery. So the bulb which is connected with the arm of resistor $R$ lights up earlier because the induced emf generated by the coil opposes the current which produced it. it takes time for bulb in the$r$ of inductor to light up.

b) Both bulbs will h up with same brightness because an inductor does not offer resistance in the pat of $D.C$ current. Inductor only produces induced emf at the time of make or break circuit.