Question

In an ac circuit containing Resistance R and capacitance C, the current is I. Keeping the ac voltage constant, if the frequency is made \( \frac{1}{3} \), the current is \( \frac{I}{2} \). Then the ratio of initial reactance to the resistance is:

• A. \( \left(\frac{3}{5}\right)^{1/2} \)
• B. \( \left(\frac{2}{5}\right)^{1/2} \)
• C. \( \left(\frac{1}{5}\right)^{1/2} \)
• D. \( \left(\frac{4}{5}\right)^{1/2} \)

Hint:

Capacitive reactance is inversely proportional to frequency: \[ X_c=\frac{1}{2\pi fC}. \] If the frequency becomes one-third, the capacitive reactance becomes three times its original value.

Answer 1

The Correct Option is A

Step 1: Series RC current.
With a resistor $R$ and capacitor $C$ in series, the impedance is $Z = \sqrt{R^2 + X_c^2}$ and the current is $I = \dfrac{V}{Z}$, where $X_c = \dfrac{1}{2\pi f C}$. The voltage $V$ is held fixed.
Step 2: How reactance changes with frequency.
Because $X_c \propto \dfrac{1}{f}$, dropping the frequency to $\dfrac{f}{3}$ triples the reactance: the new value is $3X_c$.
Step 3: Write both current equations.
\[ I = \frac{V}{\sqrt{R^2 + X_c^2}}, \qquad \frac{I}{2} = \frac{V}{\sqrt{R^2 + 9X_c^2}}. \]
Step 4: Divide one by the other.
Dividing the first by the second cancels $V$: \[ 2 = \frac{\sqrt{R^2 + 9X_c^2}}{\sqrt{R^2 + X_c^2}}. \]
Step 5: Square and simplify.
\[ R^2 + 9X_c^2 = 4(R^2 + X_c^2) \;\Rightarrow\; 5X_c^2 = 3R^2 \;\Rightarrow\; \frac{X_c^2}{R^2} = \frac{3}{5}. \]
Step 6: Take the square root.
\[ \frac{X_c}{R} = \left(\frac{3}{5}\right)^{1/2}. \]
\[ \boxed{\left(\dfrac{3}{5}\right)^{1/2}} \]