Question

In a given circuit, switch S was closed for a long time and is opened in an instant. Charge on capacitor after time $t=\frac{\pi }{1200}sec$ after opening the switch is

• A. $1.73mC$
• B. $14.1mC$
• C. $10mC$
• D. $5mC$

Answer 1

The correct option is C $10mC$


Initial current through the inductor is $\frac{V}{R}=\frac{200}{50}=4A$.
After the switch is open, an LC circuit is formed with an angular frequency of $w=\sqrt{\frac{1}{LC}}=200$.
Threfore the equation from current is $i=i_0\cos wt=4\cos 200t$.
Therfore the equation for charge is $q=\int idt=\frac{4}{200}\sin 200t$
at $t=\frac{\pi }{1200}sec,q=\frac{4}{200}\sin (200\times \frac{\pi }{1200})=10mC$