An AC source with emf E = E₀ sin(ωt) is connected to a capacitor. What is the expression for current?
Q1. An AC source with emf E = E₀ sin(ωt) is connected to a capacitor. What is the expression for current?
Answer: I = (E₀ / Xc) sin(ωt + π/2)
Explanation: The current in a capacitive circuit leads the voltage by 90°, so I = (E₀ / Xc) sin(ωt + π/2).