An AC source producing emf is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be

Question

Philoid · Accepted Answer

Given:

Emf

Steady state current

Formula used:

Charge in steady state will be given by

… (i),

where C = capacitance, ε = emf, t = time

Hence, current is given by … (ii), where Q = charge, t = time

⇒ , from (i)

Comparing this with _, we get and .

Hence, we find that i₁< i₂ (Ans).