CBSE Class 12 · 2003 · Electromagnetic InductionVery Short Answer

In a given circuit, an inductor

L

and a resistor

R

have identical resistance. Two similar electric lamps

B_{1}

and

B_{2}

are connected as shown. When the switch

S

is closed:
(i) Which one of the lamps lights up earlier?
(ii) Will the lamps be equally bright after some time? Justify your answer.

Answer: (i) $B_{2}$ lights up earlier.
(ii) Yes, both lamps become equally bright after some time.

Solution

The two branches are identical in resistance, so in the steady state each lamp carries the same current and glows equally.
At the instant

S

is closed the branch with

L

opposes the growth of current through it (a back emf

- L d i / d t

appears), so the current in

B_{1}

rises slowly while

B_{2}

(pure

R

) reaches its final value at once.
Hence

B_{2}

lights up earlier.
After a long time

d i / d t \to 0

, the inductor behaves as a plain resistance equal to

R

, the two branches become identical, and

both lamps are equally bright

CBSE Class 12 Board · 2003 · Subjective