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

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

Hint:

To avoid confusion under exam pressure, note that since current dropped by half, the total impedance must have exactly doubled ($Z' = 2Z \implies Z'^2 = 4Z^2$). This sets up a direct algebraic comparison: $R^2 + 9X_C^2 = 4(R^2 + X_C^2) \implies 5X_C^2 = 3R^2$, leading you straight to the square root answer!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question describes a series $RC$ alternating current circuit connected to a source with variable angular frequency. When the frequency is dropped to a third of its original value, the total circuit current drops to half its initial value. We need to determine the ratio of capacitive reactance to pure resistance ($\frac{X_C}{R}$) at the initial frequency.

Step 2: Key Formula or Approach:
1. The total impedance ($Z$) of a series $RC$ circuit is: $$Z = \sqrt{R^2 + X_C^2}$$ 2. Capacitive reactance is inversely proportional to angular frequency: $$X_C = \frac{1}{\omega C}$$ When the frequency becomes $\omega' = \frac{\omega}{3}$, the new reactance becomes $X_C' = 3X_C$. 3. Alternating current obeys Ohm's law: $I = \frac{V}{Z}$. Since voltage $V$ remains constant, current is inversely proportional to impedance.

Step 3: Detailed Explanation:
Let's express the initial current $I$ and the new current $I'$ using the respective circuit impedances: $$I = \frac{V}{\sqrt{R^2 + X_C^2}}$$ $$I' = \frac{V}{\sqrt{R^2 + (3X_C)^2}} = \frac{V}{\sqrt{R^2 + 9X_C^2}}$$ We are given that the new current is half of the original current ($I' = \frac{I}{2}$): $$\frac{V}{\sqrt{R^2 + 9X_C^2}} = \frac{1}{2} \cdot \frac{V}{\sqrt{R^2 + X_C^2}}$$ Cancel out the constant voltage $V$ on both sides: $$\frac{1}{\sqrt{R^2 + 9X_C^2}} = \frac{1}{2\sqrt{R^2 + X_C^2}}$$ Square both sides of the equation to eliminate the radical signs: $$\frac{1}{R^2 + 9X_C^2} = \frac{1}{4(R^2 + X_C^2)}$$ Cross-multiplying gives: $$4(R^2 + X_C^2) = R^2 + 9X_C^2$$ $$4R^2 + 4X_C^2 = R^2 + 9X_C^2$$ Rearrange the terms by grouping the resistance variables on the left and reactance variables on the right: $$4R^2 - R^2 = 9X_C^2 - 4X_C^2$$ $$3R^2 = 5X_C^2$$ To find the required ratio $\frac{X_C}{R}$, rearrange the coefficients: $$\frac{X_C^2}{R^2} = \frac{3}{5}$$ Taking the square root on both sides yields: $$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$$

Step 4: Final Answer:
The ratio of reactance to resistance at the original frequency is $\sqrt{\frac{3}{5}}$, which matches option (A).