In the network shown, initially the switch S is open. Then just after the switch S is closed, brightness of the bulb
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a
will remain unchanged
b
will increase
c
will decrease
d
may increase or decrease depending upon the value of inductance L
answer is B.
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Detailed Solution
Before S is closed, magnetic flux linked with the inductor = LiAfter S is closed, Le=L x LL+L=L2So magnetic flux = L2i'By Lenz's law, L2i'=Li⇒i'=2iSince just after S is closed the current is doubled, so brightness increases