An alternating emf E=110 √2 sin 100 t volt is applied to a capacitor of 2 μ F, the rms value of current in the circuit is ________ mA.

Solution

Physics sir · 3 steps

Read Vᵣₘₛ and ω off the emf, find X_C, and Ohm's law gives the rms current.

1. Read the emf. Compare with E = E₀ sin ω t. The peak is E₀ = 110√2 V, so Eᵣₘₛ = E₀/√2 = 110 V. The angular frequency is ω = 100 rad s⁻¹.
2. Capacitive reactance. A capacitor opposes AC with reactance X_C = 1/(ω C) = 1/(100 × 2 × 10⁻⁶) = 5000 Ω.

   Iᵣₘₛ = Eᵣₘₛ/X_C = 110/5000 A = 0.022 A

3. So. Iᵣₘₛ = 0.022 A = 22 mA, so the answer is 22.

The common slip: 31.1. Dividing the peak emf 110√2 V by X_C gives the peak current, about 31.1 mA, not the rms current of 22 mA.

The idea to keep. In a pure AC element, Ohm's law works with reactance in place of resistance. Use rms voltage for rms current, and remember X_C = 1/(ω C) uses ω, not f.

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