Question

In an L-C circuit $L=0.75H$ and $C=18\mu F$. At the instant when the current in the inductor is changing at a rate of $3.40A/s$, what is the charge on the capacitor?

• A. $45.9\mu C$
• B. $15.9\mu C$
• C. $55.9\mu C$
• D. $35.9\mu C$

Answer 1

The correct option is A $45.9\mu C$

In L-C circuit, $V_L=V_C$

$\Rightarrow L\frac{di}{dt}=\frac{q}{C}$

$\therefore$ $q=\left(LC\right)\frac{di}{dt}$

$=(0.75\times 18\times {10}^{-6})\left(3.4\right)$

$=45.9\times {10}^{-6}C$

$=45.9\mu C$