Question

When $100\text{V}$ $\text{DC}$ is applied across a solenoid, a current of $1\text{A}$ flows in it. When $100\text{V}$ $\text{AC}$ is applied across the same coil, the current drops to $0.5\text{A}$. If the frequency of the $\text{AC}$ source is $50\text{Hz}$, the impedance and inductance of the solenoid are

• A. $100\Omega ,0.93\text{H}$
• B. $200\Omega ,1.0\text{H}$
  
• C. $10\Omega ,0.86\text{H}$
• D. $200\Omega ,0.55\text{H}$
  

Answer 1

The correct option is D $200\Omega ,0.55\text{H}$


$\text{Case 1:}$ When $100\text{V}$ $\text{DC}$ is applied across the solenoid


Given, $i=1\text{A}$

$\Rightarrow \frac{100}{R}=1$ $\Rightarrow R=100\Omega$

$\text{Case 2:}$ When $100\text{V}$ $\text{AC}$ is applied across the solenoid


Given, $i_{rms}=0.5\text{A}$

So, $Z=\frac{100}{0.5}=200\Omega$

Also, for $\text{LR}$ circuit, impedence

$Z^2=R^2+(\omega L{)}^2$

$\Rightarrow (200{)}^2=(100{)}^2+(\omega L{)}^2$

$\Rightarrow (\omega L{)}^2=30000$

$\Rightarrow \omega L=173.2\Omega$

$\Rightarrow 2\pi fL=173.2$

$\Rightarrow 2\times 3.14\times 50\times L=173.2$

$\Rightarrow L=0.55\text{H}$

Hence, option $\left(D\right)$ is correct.