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An inductor is connected in series with ...

An inductor is connected in series with a bulb to an a.c source. What happens to brighness of bulb when number of turns in the inductor is reduced ?

Text Solution

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Current through the bulb
`I_(v) = (E_(v))/(Z) = (E_(V))/(sqrt(R^(2) + X_(L)^(2)))`, where `X_(L) = omega L`
As `L prop N^(2)`, therefore, on decreasing number of turns, L decreases , Z decreases, `I_(v)` increases.
Hence brightness of bulb increases.
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Knowledge Check

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