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Key is in position $2$ for time $t .$ Thereafter, it is in position $1 .$ Resistances of the bulb and inductance are marked in the figure choose the correct alternative

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Bulb $2$ dies as soon as key is switched into position $1$

Time in which brightness of bulb $1$ becomes half its maximum brightness does not depend on $R_{2} .$

If $t = \infty,$ total heat produced in bulb $1$ is $\frac{L E^{2}}{2 R_{2}^{2}}$

Ratio of maximum power consumption of bulbs depends on time.

answer is A, B, C.

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Detailed Solution

The maximum current in the inductor

$i_{0} = \frac{E}{R_{2}}$

Energy stored in the inductor

$U = \frac{1}{2} L i_{0}^{2}$

$= \frac{1}{2} L {(\frac{E}{R_{2}})}^{2} = \frac{L E^{2}}{2 R_{2}^{2}}$

When key is in position $1,$ this energy will convert into heat energy through resistor.

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