Question

An $L-C$ circuit consists of an inductor of $L=0.0900H$ and a capacitor of $C=4\times {10}^{-4}F$. The initial charge on the capacitor is $5.00\mu C$ and the initial current in the inductor is zero. When the current in the inductor has half its maximum value, what is the energy stored in the inductor?

• A. $7.8\times {10}^{-9}J$
• B. $9.8\times {10}^{7}J$
• C. $7.8\times {10}^{7}J$
• D. $9.8\times {10}^{-7}J$

Answer 1

Answer: (A)

$7.8\times {10}^{-9}J$

Given : Assuming an AC source powers the circuit, the instantaneous value of charge is given by $Q=Q_{i}sin\omega t.$. Differentiating, we get$I=dQ/dt=Q_i\omega cos\omega t$ .

The maximum values of this current will be $I_{max}=\omega Q_i$

$Q_i=5\times {10}^{-6}C$ $C=4\times {10}^{-4}F$ $L=0.09H$
Maximum current $I_o=Q_{i}w=\frac{Q_i}{\sqrt{LC}}$
$⟹I_o=\frac{5\times {10}^{-6}}{\sqrt{0.09\times 4\times {10}^{-4}}}=8.33\times {10}^{-4}A$
Current in the inductor $I=\frac{I_o}{2}=\frac{8.33\times {10}^{-4}}{2}=4.165\times {10}^{-4}A$
Energy stored $U=\frac{1}{2}L{I}^2=\frac{1}{2}\times 0.09\times (4.165\times {10}^{-4}{)}^2=7.8\times {10}^{-9}J$