Q113 / 150 · MHT CET 2026 (11 April Shift 1) · Physics
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?
- A$B_1$ becomes off promptly but $B_2$ with some delay
- B$B_2$ becomes off promptly but $B_1$ with some delay
- Correct answerCBoth $B_1$ and $B_2$ becomes off with same delay
- DBoth $B_1$ and $B_2$ becomes off promptly
Solution
When the switch S is turned off, the main voltage source is disconnected from the circuit.
The inductor L has stored energy in its magnetic field. According to Faraday's and Lenz's laws, it opposes the sudden drop in current and acts as a source of emf.
A closed loop is formed containing the inductor L, bulb $B_2$, bulb $B_1$, and resistor R in series.
The induced current from the inductor flows through this series loop, meaning the same decaying current passes through both bulbs $B_1$ and $B_2$.
Since the current through both bulbs is identical at any instant after the switch is opened, both $B_1$ and $B_2$ become off with the same delay.
The inductor L has stored energy in its magnetic field. According to Faraday's and Lenz's laws, it opposes the sudden drop in current and acts as a source of emf.
A closed loop is formed containing the inductor L, bulb $B_2$, bulb $B_1$, and resistor R in series.
The induced current from the inductor flows through this series loop, meaning the same decaying current passes through both bulbs $B_1$ and $B_2$.
Since the current through both bulbs is identical at any instant after the switch is opened, both $B_1$ and $B_2$ become off with the same delay.
Hence, the correct answer is option: 'C'.