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?
Show Hint
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
Show Solution
The Correct Option isC
Solution and Explanation
Step 1: Key idea
An inductor resists any change in current, as $\varepsilon=-L\frac{dI}{dt}$. So when the switch opens the current cannot vanish suddenly.
Step 2: Trace the path
With S open, start at $L$, go through $B_2$, up the vertical wire to the junction, through $B_1$, then through $R$ and back to $L$. This is one single loop.
Step 3: Conclusion
The same decaying current $I(t)=I_0e^{-t/\tau}$ passes through both bulbs, so they dim together. Both turn off with the same delay. Option (C) is correct.
Final Answer:
Both bulbs fade out together.
\[ \boxed{\text{Option (C)}} \]
