Question

An ac source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ω will be.

• A. $\sqrt{\frac{3}{5}}$
• B. $\sqrt{\frac{5}{3}}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

Answer: (A)

$\sqrt{\frac{3}{5}}$

Answer 2

According to given problem,

I = $\frac{V}{Z} = V$ / [R² + (1/Cω²)]^1/2 . . . (1)

and $\frac{I}{2} = \frac{V}{\lbrack R^{2} + (3/C\omega)^{2}\rbrack^{2}}$ …………(2)
substituting the value of I from equation (1) in (2),

4 $\left( R^{2} + \frac{1}{C^{2}\omega^{2}} \right) = R^{2} + \frac{9}{C^{2}\omega^{2}}$, i.e., $\frac{1}{C^{2}\omega^{2}} = \frac{3}{5}R^{2}$

so that $\frac{X}{R} = \frac{(1/C\omega)}{R} = \frac{\lbrack(3/5)R^{2}\rbrack^{1/2}}{R} = \sqrt{\frac{3}{5}}$

Hence (1) is correct.