When 100 V d.c. is applied across a solenoid, a current of 1 A flows in it. When 100 a.c. is applied across…
When 100 V d.c. is applied across a solenoid, a current of 1 A flows in it. When 100 a.c. is applied across it, the current drops to 0.5 A . If the frequency is 50 Hz , the impedance and inductance is
$200 \Omega, \sqrt{3} / \pi \mathrm{H}$
$100 \Omega, \sqrt{3} \mathrm{H}$
$200 \Omega, 1 \mathrm{H}$
$100 \Omega, 1 \mathrm{H}$
Solution
When dc is applied
$i=\frac{V}{R} \Rightarrow 1=\frac{100}{R} \Rightarrow R=100 \Omega$
When ac is applied
$i=\frac{V}{Z} \Rightarrow 0.5=\frac{100}{Z} \Rightarrow Z=200 \Omega$
Hence $Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+4 \pi^2 v^2 L^2}$
$\begin{aligned}
& \Rightarrow(200)^2=(100)^2+4 \pi^2(50)^2 L^2 \\
& \Rightarrow L=0.55 H
\end{aligned}$