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When a DC voltage of 200 V is applied to...

When a DC voltage of 200 V is applied to a coil of self inductance (2/`sqrt 3`/`pi`)H a current of 1A flows through it . But by replacing DC source with AC source of 200 V , the current in the coil is reduced to 0.5A . Then the frequency of AC supply is

A

30 Hz

B

60 Hz

C

75 Hz

D

50 Hz

Text Solution

Verified by Experts

The correct Answer is:
D
Resistance of coil(R ) = `200V/1A = 200Omega`
With AC source , `I = 200/sqrt R^(2)` or `0.5 = 200/sqrt(R^(2)+X_(L)^(2))
`implies` R^(2) + (2pifL)^(2) = (400)^(2)`
`implies(2pif xx 2sqrt 3/pi )^(2)` = `(400)^(2)- (200)^(2)= 200 xx 600`
`impliessqrt 3 = 2sqrt 3 xx 100 implies f = 50 Hz`
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