The circuit above shows a battery, two identical lightbulbs, an inductor and a s

Question

The circuit above shows a battery, two identical lightbulbs, an inductor and a switch. All of the components may be considered to be ideal. Initially, no currents are flowing. At t = 0, the switch is closed, it is left closed for a long time, then it is reopened, and another long time passes.

Just after the switch is first closed, which bulb will be brighter? What about just before it is reopened? Just after it is reopened? Finally, which bulb will emit more total light during the course of the experiment?

PLEASE EXPLAIN.

Explanation / Answer

here it was said all the components are ideal,so inductor will simply act as short circuit as ther was no resistance. 1.switch is first closed,the both light will be equal brighter. 2.after switch reopening both of them will be off,as they are cut off from power supply. 3.both the lights emit equal amount of light during the course of the experiment.

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