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

A

100 Hz

B

75 Hz

C

60 Hz

D

50 Hz

Text Solution

Verified by Experts

The correct Answer is:
D
Resistance of coil `(R )=(200)/(I)=200Omega1`
Current `I=(200)/(sqrt(R^(2)+X_(L)^(2)))`
or `0.5=(200)/(sqrt(R^(2)+X_(L)^(2)))`
or `R^(2)+(2pifL)^(2)=(400)^(2)`
or `(2pifxx(2sqrt(3))/(pi))^(2)=(400)^(2)-(200)^(2)`
`=120000`
or `4fsqrt(3)=200sqrt(3)`
or `f=50Hz`
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