- Home
- Standard 12
- Physics
7.Alternating Current
easy
A capacitor of capacitance $C$ has initial charge $Q_0$ and connected to inductor $'L'$ as shown, at $t = 0$ switch $S$ is pressed. The current through the inductor when energy in the capacitor is three times of the energy of the inductor is
A
$\frac{{{Q_0}}}{{2\sqrt {LC} }}$
B
$\frac{{{Q_0}}}{{\sqrt {LC} }}$
C
$\frac{{{2Q_0}}}{{\sqrt {LC} }}$
D
$\frac{{{4Q_0}}}{{\sqrt {LC} }}$
Solution
$\because$ total energy $=\frac{Q_{0}^{2}}{2 C}$
$\frac{Q_{0}^{2}}{2 C}=\frac{Q^{2}}{2 C}+\frac{L i^{2}}{2}=\frac{Q^{2}}{2 C}+3\left(\frac{Q^{2}}{2 C}\right)$
$\frac{Q_{0}^{2}}{2 C}=\frac{4 Q^{2}}{2 C} \Rightarrow Q=\frac{Q_{0}}{2}$
Current at that instant $=Q \omega$
$\mathrm{i}=\frac{\mathrm{Q}_{0}}{2} \omega=\frac{\mathrm{Q}_{0}}{2 \sqrt{\mathrm{LC}}}.$
Standard 12
Physics
