The following figure represents two bulbs \(B_1\) and \(B_2\), resistor R and an inductor L. When the switch S is turned off, which of the following statement is true?
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After the switch opens the inductor keeps the current flowing around a series loop containing both bulbs.
\(B_1\) becomes off promptly but \(B_2\) with some delay
\(B_2\) becomes off promptly but \(B_1\) with some delay
Both \(B_1\) and \(B_2\) becomes off with same delay
Both \(B_1\) and \(B_2\) becomes off promptly
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The Correct Option isC
Solution and Explanation
Step 1: Understand the circuit
The battery and switch S sit in one branch. Resistor \(R\) and bulb \(B_1\) are in series in the top branch. Inductor \(L\) and bulb \(B_2\) are in series in the lower branch. The two branches share the same two junctions.
Step 2: While S is closed
The battery drives current through both branches. The inductor stores energy in its magnetic field.
Step 3: When S is opened
The battery is cut off. The inductor opposes the sudden fall of its current by producing an induced emf that keeps the current going in the same direction. The only closed path left is \(L\to B_2\to\) junction \(\to B_1\to R\to L\). In this loop all four parts are in series.
Step 4: Compare the bulbs
Because they are in series, the same current flows through \(B_1\) and \(B_2\). Both glow while the current decays and both go off together after the same delay.
Step 5: Why the other options fail
(A) and (B) say one bulb goes off at once, but both bulbs are inside the closed loop. (D) says both go off at once, but the inductor does not let the current drop to zero instantly.
Final Answer:
Both bulbs go off after the same delay, option (C).
\[ \boxed{\text{Option (C)}} \]
