Question

A $50\mu$F capacitor, a $30\Omega$ resistor and a $0.7$H inductor are connected in series to an ac supply which generates an emf 'e' given by $e=300\sin$ $\left(200t\right)$ Volt. Calculate peak value of the current flowing through the circuit.

Answer 1

Given: $C=50\mu F=50\times {10}^{-6}$F
$R=30$ohm, $L=0.7$H
and $e=300\sin \left(200t\right)$Volt.
Comparing the above equation with $e=e_0\sin \omega t$, we get
$e_0=300$volt, $i_0=?$
$\omega =200$Hz.
$\therefore 2\pi f=200$
$\Rightarrow 2\times \frac{22}{7}\times f=200$
$f=\frac{200\times 7}{2\times 22}=31.82$Hz
Now, $X_L=\omega L=2\pi fL$
$=2\times \frac{22}{7}\times 31.82\times 0.7=140$ohm
and $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$
$=\frac{1\times 7}{2\times 22\times 31.82\times 50\times 10-6}$
$=100$ohm.
$\therefore Z=\sqrt{(X_L-X_C{)}^2+R^2}$
$=\sqrt{(140-100{)}^2+(30{)}^2}$
$=\sqrt{1600+900}$
$\sqrt{2500}=50$ohm
Now peak value of current $i_0=\frac{e_0}{Z}=\frac{300}{50}=6$ampere.