Question

An oscillator circuit contains an inductor $0.05H$ and a capacitor of capacity $80\mu F$ . When the maximum voltage across the capacitor is $200V$ , the maximum current (in amperes) in the circuit is

• A. $4$
• B. $8$
• C. $10$
• D. $16$

Answer 1

Answer: (B)

Given $L=0.05H,C=80\mu F$
$V_{max}=200V$
$\because$ Voltage equation $V\left(t\right)=V_{m}sin\omega t$
$\therefore$ Current (i) $=\frac{cdv}{dt}=c\frac{d}{dt}{V}_{m}sin\omega t=C{V}_m\omega cos\omega t$
$\because \omega =\frac{1}{\sqrt{LC}}$
$\therefore i=V_m\sqrt{\frac{C}{L}}cos\omega t$
$\therefore$ Maximum current $\left(i_m\right),V_m\sqrt{\frac{C}{L}}=200\times \sqrt{\frac{80\mu }{0.05}}$
$i_m=8A$