Question

An oscillator circuit contains an inductor 0.05 H and a capacitor of capacity 80 $\mu$F. When the maximum voltage across the capacitor is 200 V, the maximum current (in amperes) in the circuit is

• A. 2
• B. 4
• C. 8
• D. 10
• E. 16

Hint:

Conservation of energy is the fastest way to solve LC circuit problems involving maximum values of voltage and current.

Answer 1

The Correct Option is C

Concept:
In an ideal LC oscillator, energy oscillates between the magnetic field of the inductor and the electric field of the capacitor. The maximum energy in both must be equal: \[ \frac{1}{2} L I_{max}^2 = \frac{1}{2} C V_{max}^2 \]

Step 1: Identify the given values.
$L = 0.05$ H
$C = 80 \mu\text{F} = 80 \times 10^{-6}$ F
$V_{max} = 200$ V

Step 2: Calculate the maximum current ($I_{max}$).
\[ L I_{max}^2 = C V_{max}^2 \] \[ 0.05 \times I_{max}^2 = (80 \times 10^{-6}) \times (200)^2 \] \[ 0.05 \times I_{max}^2 = 80 \times 10^{-6} \times 40000 \] \[ 0.05 \times I_{max}^2 = 3.2 \] \[ I_{max}^2 = \frac{3.2}{0.05} = 64 \] \[ I_{max} = \sqrt{64} = 8 \text{ A} \]