Question

A $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathit{\mu }\mathbf{F}$capacitor has a capacitive reactance of $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\Omega }$. (a) What must be its operating frequency? (b) What will be the capacitive reactance if the frequency is doubled?

Short answer

1. The operating frequency is 8.84 kHz.
2. The capacitive reactance, if the frequency is doubled, is $6.00\Omega$.

Step-by-step solution

Determining the concept of capacitive reactance

The resistance caused by the capacitors in an AC circuit is called capacitive reactance. It depends on the frequency of AC current and the capacitance of the capacitor.

Use the capacitive reactance formula to find the circuit's operative frequency. Using the relation between the capacitive reactance and the operative frequency, find the capacitive reactance if the frequency is doubled.

The formula is as follows:

${\mathrm{X}}_{\mathrm{c}}=\frac{1}{2\mathrm{\pi fC}}$

Here, $X_c$is the capacitive reactance, f is the frequency, C and is the capacitance of the capacitor.

(a) Determining the operating frequency

The capacitive reactance is given by,

$X_c=\frac{1}{2\mathrm{\pi }fC}$

Rearrange this equation f,

$\begin{array}{c}f=\frac{1}{2\pi X_{c}C}\\ f=\frac{1}{2\pi \left(12.0\Omega \right)\left(1.50\times {10}^{-6}\mathrm{F}\right)}\\ f=8.84\times {10}^3\mathrm{Hz}\\ f=8.84\mathrm{kHz}\end{array}$

Therefore, the operating frequency is$8.84\mathrm{kHz}.$

(b) Determining the capacitive reactance, if the frequency is doubled  

Thecapacitive reactance is inversely proportional to the operating frequency, i.e.,

$X_c\propto \frac{1}{f}$

As the frequency is doubled, the capacitive reactance will decrease by a factor of 2.

Therefore,

$\begin{array}{c}{X}_c=\frac{12.0\Omega }{2}\\ =6\Omega \end{array}$

Thus,the capacitive reactance, if the frequency is doubled, is $6\Omega$.