Question

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

• A. $4$
• B. $8$
• C. $10$
• D. $16$

Answer 1

Answer: (B)

Given $L=0.05H, \, C=80μF$
$V_{m a x}=200\,V$
$\because $ Voltage equation $V\left(t\right)=V_{m}sin \omega t$
$\therefore $ Current (i) $=\frac{c d v}{d t}=c\frac{d}{d t}V_{m}sin \omega t = C V_{m} \omega cos ⁡ \omega t$
$\because \omega =\frac{1}{\sqrt{L C}}$
$\therefore \, i=V_{m}\sqrt{\frac{C}{L}}cos \omega t$
$\therefore $ Maximum current $\left(i_{m}\right), \, V_{m}\sqrt{\frac{C}{L}}=200\times \sqrt{\frac{80 \mu }{0.05}}$
$i_{m}=8A$