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 :

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).

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