When 100 volts dc is supplied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts ac is applied across the same coil, the current drops to 0.5 ampere. If the frequency of ac source is 50 Hz, then the impedance and inductance of the solenoid are

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200 Ω and 0.55 henry

100 Ω and 0.86 henry

200 Ω and 1.0 henry

100Ω and 0.93 henry

answer is A.

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Detailed Solution

For dc, R=Vi=1001=100Ω For ac, Z=Vi=1000.5=200Ω∵Z=R2+(ωL)2⇒200=(100)2+4π2(50)2L2∴L=0.55H

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