Question 401: AC fundamentals

Question

An a.c. source of angular frequency '$\omega$' 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 $\frac{\omega}{3}$ (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 '$\omega$' will be

Accepted Answer

Let the reactance of the capacitor at the original frequency '$\omega$' be $X_C$.
The impedance of the circuit at the original frequency is given by:
$Z = \sqrt{R^2 + X_C^2}$
The current in the circuit at the original frequency is given by:
$I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + X_C^2}}$
When the frequency is changed to $\frac{\omega}{3}$, the reactance of the capacitor becomes $3X_C$ (since reactance is inversely proportional to frequency). The impedance of the circuit at this new frequency is given by:
$Z' = \sqrt{R^2 + (3X_C)^2} = \sqrt{R^2 + 9X_C^2}$
The current in the circuit at this new frequency is given by:
$I' = \frac{V}{Z'} = \frac{V}{\sqrt{R^2 + 9X_C^2}}$
We are given that the current in the circuit is halved when the frequency is changed to $\frac{\omega}{3}$. Therefore, we can write:
$I' = \frac{I}{2}$
$\frac{V}{\sqrt{R^2 + 9X_C^2}} = \frac{1}{2} \cdot \frac{V}{\sqrt{R^2 + X_C^2}}$
Squaring both sides, we get:
$\frac{1}{R^2 + 9X_C^2} = \frac{1}{4(R^2 + X_C^2)}$
$4R^2 + 4X_C^2 = R^2 + 9X_C^2$
$3R^2 = 5X_C^2$
Therefore, the ratio of reactance to resistance at the original frequency '$\omega$' is:
$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$
So the correct answer is
Option D: $\sqrt{\frac{3}{5}}$

Answer

$\sqrt{\frac{2}{5}}$

Answer

$\sqrt{\frac{1}{5}}$

Answer

$\sqrt{\frac{4}{5}}$

Answer

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

AC fundamentals

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