for given question correct answer is option 2. please explain.
Aman , 5 Years ago
Grade 12th pass
2 Answers
Arun
Last Activity: 5 Years ago
B1 dies out promptly but B2 with some delay because of presence of inductance in its circuit.
hence I think answer should be option 3
Sanjai
Last Activity: 5 Years ago
Yes even I think the answer should be 3. Because when the switch is closed the bulb 💡 b1 would light up early but do to infinite resistance from the inductor the bulb b2 would light late. Similarly when the switch is off the bubl b1 dies out immediately as there is no supply but now the inductor can act as a battery until it dies out of charges. So ultimately the bulb b2 would die a bit late.
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