Question 414: 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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