Question

The reactance of a capacitor of capacitance $C$ connected to an AC source of frequency $\omega$ is $X$. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become:

• A. $\frac{X}{6}$
• B. $6X$
• C. $\frac{2}{3}X$
• D. $\frac{3}{2}X$

Hint:

For capacitive reactance, remember that $X_C = \frac{1}{\omega C}$. If the capacitance or frequency changes, calculate the new reactance proportionally.

Answer 1

The Correct Option is A

The capacitive reactance $X_C$ is given by: \[ X_C = \frac{1}{\omega C}, \] where $\omega$ is the angular frequency, and $C$ is the capacitance. Initially, the reactance is: \[ X = \frac{1}{\omega C}. \] When the capacitance is doubled ($C' = 2C$) and the frequency is tripled ($\omega' = 3\omega$), the new reactance becomes: \[ X_C' = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6 \omega C}. \] Compare with the initial reactance: \[ X_C' = \frac{X}{6}. \] Thus, the new reactance is: \[ \boxed{\frac{X}{6}}. \]