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. i1> i2

B. i1 = i2

C. i1< i2

D. The information is insufficient to find the relation between i1 and i2.

Answer :



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).

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