Q. 24.7( 3 Votes )

# 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

A. i_{1}> i_{2}

B. i_{1} = i_{2}

C. i_{1}< i_{2}

D. The information is insufficient to find the relation between i_{1} and i_{2}.

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 i_{1}< i_{2} (Ans).

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