Question

When $100vDC$ is applied across a solenoid , a steady current of $1A$ flows in it. When $100VAC$ is applied across the same solenoid, the current drops to $0.5A$. If the frequency of the AC source is $150/\sqrt{3}\pi$Hz, the impedance and inductance of the solenoid are:

• A. $200\Omega and1/3H$
• B. $100\Omega and1/16H$
• C. $200\Omega and1/10H$
• D. $1100\Omega and3/117H$

Answer 1

The correct option is C $200\Omega and1/10H$

Given :When 100vDC100vDC is applied across a solenoid , a steady current of 1A1A flows in it. When 100VAC100VAC is applied across the same solenoid, the current drops to 0.5A0.5A. If the frequency of the AC source is 150/3√π150/3πHz, the impedance and inductance of the solenoid are:

$\omega =2\pi f$, $f=\frac{150}{\sqrt{3}\pi }$

Solution :

Let us assume some of the terms first

Z= impedance , L= inductance, $\omega L=x$

Case 1

When source is DC

Using ohm's law

V=iR

100=1$\times$R

R=100 $\Omega$

Case 2

When source is AC

Here we have formula V=iZ

100=0.5 Z

Z=200$\Omega$

$Z=\sqrt{R^2+x^2}$

$200=\sqrt{R^2+(x{)}^2}$ ( sq both sides)

$4\times {10}^4$= ${10}^4+x^2$

$x=10\sqrt{3}$

$\omega L=x$

$\omega L=10\sqrt{3}$

$2\pi fL=10\sqrt{3}$

$L=\frac{1}{10}$