Question

In an L-C circuit $L=0.75H$ and $C=18\mu F$. When the charge on the capacitor is $4.2\times {10}^{-4}C$, what is the induced emf in the inductor?

• A. $23.3V$
• B. $29.5V$
• C. $25.0V$
• D. $21.6V$

Answer 1

The correct option is A $23.3V$

$\text{Given:-}$ Inductance,L $=0.75H$

Capacitor,C $=18\mu F$

$\text{Solution:-}$

We know that,

$V_L=V_C$

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

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

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

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

$=45.9\mu C$

Now,

$V_L=V_C=\frac{q}{C}$

$=\frac{4.2\times {10}^{-4}}{18{\times }^{-6}}$

$=23.3V$

$\text{Hence the correct option is A}$