Question

A coil is connected to an alternating emf of voltage $24\text{V}$ and ffrequency $50\text{Hz}$. The reading on the ammeter conected to the coil in series is $10\text{mA}$. If a $1\mu F$ capacitor is connected to the coil in series the ammeter shows $10\text{mA}$ again, What would be the approx. reading on a $dc$ ammeter $\text{(in A)}$ if the coil was connected to $180\text{V}$ $dc$ voltage supply ? $(\text{Take}{\pi }^2=10)$

Answer 1

Given
AC Voltage =$24\text{V}$
Frequency$f=50\text{Hz}$
$C=1\mu F$

Impedence of $RL$ circuit is given by
$Z=\sqrt{{\left(\omega L\right)}^2+R^2}=\frac{\text{Ac Voltage}}{\text{Ac current}}=\frac{24}{10\times {10}^{-3}}\dots 1$
RLC circuit also produces same current
$Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}=\frac{\text{Ac Voltage}}{\text{Ac current}}=\frac{24}{10\times {10}^{-3}}$

$\sqrt{R^2+{\left(\omega L\right)}^2}=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$
$\left(\omega L\right)=-\omega L+\frac{1}{\omega C}$
$L=\frac{1}{2{\omega }^{2}C}=\frac{1}{2\times 100\pi \times 100\pi \times {10}^{-6}}=5\text{H}$
From equation $1$
${\left(2400\right)}^2={\left(500\pi \right)}^2+R^2$
$R=\sqrt{{\left(2400\right)}^2-{(5\pi \times 100)}^2}=100\sqrt{{\left(24\right)}^2-25{\pi }^2}$
$=100\times \sqrt{326}\cong 1800$
current when DC Voltage=$180\text{V}$ is applied is
$I=\frac{V}{R}=\frac{180}{1800}=0.1\text{A}$