An oscillating LC circuit consists of a 75 mH inductor and a 1.2 μF capacitor. If the maximum charge to the capacitor is 2.7 μC. The maximum current in the circuit will b e __________ mA.

Question

G1 Physics · Accepted Answer

In an ideal LC oscillator, maximum electric energy equals maximum magnetic energy: Q(max)²/(2C) = LI(max)²/2 Therefore, I(max) = Q(max)/√(LC) Given: Q(max) = 2.7 μC = 2.7×10⁻⁶ C L = 75 mH = 75×10⁻³ H C = 1.2 μF = 1.2×10⁻⁶ F Now, √(LC) = √[(75×10⁻³)(1.2×10⁻⁶)] = √(9×10⁻⁸) = 3×10⁻⁴ Hence, I(max) = 2.7×10⁻⁶ / 3×10⁻⁴ = 9×10⁻³ A = 9 mA Final value: 9