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

Question

When 100 V dc is applied across a solenoid, a current of 1.0 A flows in it. When 100 V ac is applied across the same coil, the current drops to 0.5 A. If the frequency of the ac source is 50 Hz, the impedance and inductance of the solenoid are (A) 200 Ω and 0.55 H (B) 100 Ω and 0.86 H (C) 200 Ω and 1.0 H (D) 100 Ω and 0.93 H

Tardigrade Admin · Accepted Answer

R=(E/I)=(100/1)=100 Ω for ac: Z=[R2+(2 π f L)2]1 / 2 =(E v /I v )=(100/0.5)=200 Ω or 200=[(100)2+(100 π L)2]1 / 2 Solving, we get L=0.55 H

Q. When $100 V$ dc is applied across a solenoid, a current of $1.0 A$ flows in it. When $100 V$ ac is applied across the same coil, the current drops to $0.5 A$ . If the frequency of the ac source is $50 Hz$ , the impedance and inductance of the solenoid are

$200 Ω$ and $0.55 H$

$100 Ω$ and $0.86 H$

$200 Ω$ and $1.0 H$

$100 Ω$ and $0.93 H$

$R = \frac{E}{I} = \frac{100}{1} = 100 Ω$
for ac: $Z = [R^{2} + (2 π f L)^{2}]^{1/2}$
$= \frac{E _{v}}{I _{v}} = \frac{100}{0.5} = 200 Ω$
or $200 = [(100)^{2} + (100 π L)^{2}]^{1/2}$
Solving, we get $L = 0.55 H$

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

200Ω and 0.55H

100Ω and 0.86H

200Ω and 1.0H

100Ω and 0.93H