Question

A capacitor of capacitance 500 μF is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in the LC circuit will be ______A.

Answer 1

Correct Answer: 10

To solve the problem of finding the maximum current in an LC circuit formed by a capacitor and an inductor, we apply the principles of energy conservation. The energy stored in the capacitor is fully transferred to the inductor, reaching the maximum current. The energy initially stored in the capacitor is given by the formula: U = ¹⁄₂CV², where C = 500 μF = 500 × 10^-6 F and V = 100 V. Therefore, U = ¹⁄₂ × 500 × 10^-6 × (100)² = 2.5 joules.

In the LC circuit, this energy converts to magnetic energy stored in the inductor at maximum current, expressed as U = ¹⁄₂LI², where L = 50 mH = 50 × 10^-3 H. Solving for maximum current I gives 2.5 = ¹⁄₂ × 50 × 10^-3 × I². Simplifying, I²= 100 and thus I = √100 = 10 A.

The computed maximum current is 10 A, which neatly fits within the expected range of 10 to 10, confirming the solution's accuracy. 

Answer 2

q₀ = CV
= 500 × 100 × 10^-6 C
= 5 × 10^-2 C
For i_max,
\(\frac{1}{2}Li_m^2=\frac{1}{2}\frac{q_0^2}{C}\)
\(50×10^{−3}×I_m^2\)

\(=\frac{(5×10^{−2})^2}{500×10^{−6}}\)
\(⇒i_m=\frac{5×10^−2}{5×10^{−3}}\)
im=10 A
So, the answer is 10 A.