## In This post you will get answer of Q.

An oscillator circuit contains an inductor $0.05, H$ and a capacitor of capacity $80, mu F$. When

the maximum voltage across the capacitor is $200, V,$ the maximum current (in amperes) in

the circuit is

Solution:

## Given $L=0.05, H ,, C=80, mu F$

$V_{max }=200, V$

$because$ Voltage equation $V(t)=V_{m} sin omega t$

$therefore$ Current $(i)=frac{c d v}{d t}=c frac{d}{d t} V_{m} sin omega t=C V_{m} omega cos omega t$

$because, omega=frac{1}{L C}$

$therefore, i=V_{m} sqrt{frac{C}{L}} cos omega t$

$therefore$ Maximum current $left(i_{m}right), V_{m} sqrt{frac{C}{L}}=200 times sqrt{frac{80 mu}{0.05}}$

$i_{m}=8, A$