1.

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

Answer»

Bulb 2 dies as soon as key is wsitched into position 1.
TIME in which brigthness of bulb 1 become half its maximum brightness does not depend on t.
If `t = oo`, total heat produced in bulb 1 is `(L epsilon^(2))/(2R_(2)^(2))`
Ratio of maximum power consumption of BULBS depends on time

Solution :The maximum current in the INDUCTOR
`i_(0)=(epsilon)/(R_(2))`
Energy stored in the inductor
`U=(1)/(2)Li_(0)^(2)=(1)/(2)L((epsilon)/(R_(2)))^(2)=(L epsilon^(2))/(2R_(2)^(2))`
When key is in position 1, this energy will convert into heat energy through RESISTOR.


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