Question

An ac source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I . If now the frequency of the 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 the reactance to resistance at the original frequency ω is given as $\sqrt{\frac{x}{5}}$. Find x .

Answer 1

Step 1: Given

Current at angular frequency $\omega$: $I$

Current at angular frequency $\frac{\omega }{3}$: $\frac{I}{2}$

Reactance to resistance ratio at $\omega$: $\frac{X_c}{R}=\sqrt{\frac{x}{5}}$

Step 2: Formula Used

$X_c=\frac{1}{C\omega }$

$I=\frac{V}{\sqrt{R^2+{X_c}^2}}$

Step 3: Calculate the value of x

Find the capacitive reactance at $\frac{\omega }{3}$ by using the formula.

$$\begin{gathered} X_c\text{'}\text{'}=\frac{1}{C{\displaystyle \frac{\omega }{3}}} \\ =\frac{3}{C{\displaystyle \omega }} \\ =3\frac{1}{C{\displaystyle \omega }} \\ =3X_c \end{gathered}$$

Equate the current equations and substitute the values in the formula.

$$\begin{gathered} I\text{'}\text{'}=\frac{I}{2} \\ \Rightarrow \frac{V}{\sqrt{R^2+{\left(3X_c\right)}^2}}=\frac{V}{2\sqrt{R^2+{X_c}^2}} \\ \Rightarrow R^2+{9X_c}^2=4R^2+4{X_c}^2 \\ \Rightarrow {9X_c}^2-4{X_c}^2=4R^2-R^2 \end{gathered}$$

Find the ratio of the reactance to resistance by solving the obtained equation further.

$$\begin{gathered} 5{X_c}^2=3R^2 \\ \frac{{X_c}^2}{R^2}=\frac{3}{5} \\ \frac{X_c}{R}=\sqrt{\frac{3}{5}} \\ \sqrt{\frac{x}{5}}=\sqrt{\frac{3}{5}} \end{gathered}$$

Hence, the value of x is 3.