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. $200\, \Omega$ and $0.55\, H$
• B. $100 \, \Omega$ and $0.86\, H$
• C. $200\, \Omega$ and $1.0\, H$
• D. $100 \, \Omega$ and $0.93 \, H$

Answer 1

Answer: (A)

$R=\frac{E}{I}=\frac{100}{1}=100 \,\Omega$
for ac: $Z=\left[R^{2}+(2 \pi f L)^{2}\right]^{1 / 2}$
$=\frac{E_{ v }}{I_{ v }}=\frac{100}{0.5}=200\, \Omega$
or $200=\left[(100)^{2}+(100 \pi L)^{2}\right]^{1 / 2}$
Solving, we get $L=0.55 \,H$