An oscillating LC circuit consists of a 75 mH inductor and a 1.2 μ F capacitor. If the maximum charge to the capacitor is 2.7 μ C. The maximum current in the circuit will be ___________ mA

Solution

Physics sir · 4 steps

Energy just moves between C and L, so match the energy in each and Iₘₐₓ = ω Qₘₐₓ falls out.

1. Energy is conserved. With no resistance, all the energy sits in the capacitor when the charge is maximum. All of it sits in the inductor when the current is maximum. So Qₘₐₓ²/2C = 1/2LIₘₐₓ².

   Iₘₐₓ = Qₘₐₓ/√LC = ω Qₘₐₓ

2. Find the root. LC = (75 × 10⁻³ H) × (1.2 × 10⁻⁶ F) = 9 × 10⁻⁸ s², so √LC = 3 × 10⁻⁴ s.
3. Put in numbers. Iₘₐₓ = (2.7 × 10⁻⁶ C)/(3 × 10⁻⁴ s) = 9 × 10⁻³ A.
4. So. The maximum current is 9 mA.

The common slip: 1.43. Using f = 1/2π√LC in place of ω divides the answer by 2π and gives about 1.43 mA, not 9.

The idea to keep. In an LC circuit, charge oscillates as q = Qₘₐₓ cos ω t, so the peak current is always ω Qₘₐₓ with ω = 1/√LC. Use angular frequency, not f.

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