A series combination of resistor (R) and capacitor (C) is connected to an AC source of angular frequency ω . Keeping the voltage same, if the frequency is changed to (ω)/(3) , the current becomes half of the original

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Correct answer: A. 0.6

In the first case, capacitive reactance, X_C=(1)/(ω C) Impedance of the circuit, Z= R^2+X_C^2 Current in the circuit, I=(V)/(Z)= V R^2+X_C^2 (i) In the second case, ω^ =(ω)/(3) Capacitive reactance, X_C^ = 1 ω^ C =(1)/((ω)/(3) C)=(3)/(ω C)=3 X_C Impedance of the circuit, Z^ = R^2+X_C^ 2 = R^2+9 X_C^2 Current in the circuit, (I)/(2)= V Z^ = V R^2+9 X_C^2 Divide (ii) by (i), we get aligned & (1)/(2)= R^2+X_C^2 R^2+9 X_C^2 (1)/(4)=(R^2+X_C^2)/(R^2+9 X_C^2) & 5 X_C^2=3 R^2 or (X_C)/(R)= (3)/(5) = 0.6 aligned

A series combination of resistor (R) and capacitor (C) is connected...

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