A capacitor of capacitance 100 μ F is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be :

Solution

Physics sir · 4 steps

No resistance means no loss: all the capacitor's stored energy turns into magnetic energy at peak current.

1. Energy is conserved. An ideal LC circuit has no resistor, so no energy is lost. The current is maximum when the capacitor is fully discharged, and at that moment all the energy sits in the inductor.

   1/2CV² = 1/2LIₘₐₓ² ⇒ Iₘₐₓ = V√(C/L)

2. Put in numbers. C/L = (100 × 10⁻⁶)/(6.4 × 10⁻³) = 1/64, so √(C/L) = 1/8 = 0.125. Then Iₘₐₓ = 12 × 0.125 = 1.5 A.
3. Check another way. ω = 1/√LC = 1/(8 × 10⁻⁴) = 1250 rad s⁻¹ and Q₀ = CV = 1.2 × 10⁻³ C. So Iₘₐₓ = Q₀ω = 1.5 A, the same answer.
4. So. The maximum current is 1.5 A, option (c).

The idea to keep. In an LC circuit, energy only swaps between C and L. Set 1/2CV² = 1/2LI², or use Iₘₐₓ = Q₀ω. Convert μF and mH to SI before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.