Question

A capacitor of capacity '$C$' is charged to a potential '$V$'. It is connected in parallel to an inductor of inductance '$L$'. The maximum current that will flow in the circuit is

• A. $V\sqrt{\frac{L}{C}}$
• B. $V\sqrt{LC}$
• C. $V\sqrt{\frac{C}{L}}$
• D. $\frac{V C^2}{L}$

Hint:

You can verify the expression using dimensional analysis! The term $\sqrt{\frac{1}{LC}}$ is angular frequency $\omega$. Since maximum charge is $Q_0 = CV$, the maximum current is $I_0 = \omega Q_0 = \frac{1}{\sqrt{LC}} \cdot CV = V\frac{C}{\sqrt{LC}} = V\sqrt{\frac{C}{L}}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A capacitor with capacitance $C$ is fully charged to a voltage potential $V$, accumulating an initial electrical energy. It is then connected directly across an ideal inductor of inductance $L$, forming an LC resonant circuit. We need to determine the maximum peak current $I_0$ that circulates during the resulting electromagnetic oscillations.

Step 2: Key Formula or Approach:
In an ideal LC circuit with no resistance, total energy is perfectly conserved. Energy continuously oscillates between the electric field of the capacitor and the magnetic field of the inductor: $$\text{Maximum Electric Energy (Capacitor)} = \text{Maximum Magnetic Energy (Inductor)}$$ $$\frac{1}{2} C V^2 = \frac{1}{2} L I_0^2$$ We can isolate $I_0$ from this energy equality.

Step 3: Detailed Explanation:
Let's set up the energy conservation equation and cancel out the common factor of $\frac{1}{2}$ from both sides: $$C V^2 = L I_0^2$$ Isolate the squared current term $I_0^2$: $$I_0^2 = \frac{C V^2}{L} = V^2 \left( \frac{C}{L} \right)$$ Take the principal square root of both sides to solve for the peak current $I_0$: $$I_0 = \sqrt{V^2 \left( \frac{C}{L} \right)} = V\sqrt{\frac{C}{L}}$$

Step 4: Final Answer:
The maximum current that will flow in the circuit is $V\sqrt{\frac{C}{L}}$, which corresponds to option (C).