When 100 V DC is applied across a solenoid, a steady current of 1 A flows in it. when 100V AC is applied across the same solenoid, the current drops to 0.5 A. If the frequency of the Ad source is `150// sqrt3pi Hz`, the impedance and inductance of the solenoid are
A. `200 Omega` and `1//3 H`
B. `100 Omega` and `1//16 H`
C. `200 Omega` and `1.0 H`
D. `1100 Omega` and `3//11 H`


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Correct Answer - A
`R=100/1=100 Omega , x=sqrt(Z^(2)-R^(2))`
`Z=100/0.5=200 Omega , L=x/omega=1/3H`.

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