Question

When the current in the inductor has half its maximum value, what is the charge on the capacitor?

• A. $1.33\times {10}^{6}C$
• B. $9.33\times {10}^{-6}C$
• C. $4.33\times {10}^{-6}C$
• D. $7.33\times {10}^{6}C$

Answer 1

The correct option is C $4.33\times {10}^{-6}C$

Given : $Q_i=5\times {10}^{-6}C$ $C=4\times {10}^{-4}F$ $L=0.09H$
Charge stored in capacitor as a function of time $Q=Q_i\cos wt$ ......(1)
Current as a function of time is given by $I=I_o\sin wt$ ...(2)
For $I=\frac{I_o}{2}$
We get $\frac{I_o}{2}=I_o\sin wt$
$⟹\sin wt=0.5$
So $\cos wt=\sqrt{1-{\sin }^{2}wt}=\sqrt{1-{0.5}^2}=0.866$
Charge on capacitor at that instant $Q=5\times {10}^{-6}\times 0.866=4.33\times {10}^{-6}C$