Question

A $1.50\mu F$capacitor has a capacitive reactance of $12.0\Omega$.
(a) What must be its operating frequency?
(b) What will be the capacitive reactance if the frequency is doubled?

Answer 1

In this question capacitive reactance of a capacitor is given so by using the capacitive reactance formula we will find the operating frequency, now it is said that the frequency is doubled so by keeping the value of the capacitor same we will find the capacitive reactance whose frequency is doubled.

Complete step by step answer:
Capacitance of the capacitor $C = 1.50\mu F = 1.5 \times {10^{ - 6}}F$
Capacitive reactance of the capacitor ${X_C} = 12\Omega$
We know the capacitive reactance of a capacitor is given by the formula
${X_C} = \dfrac{1}{{2\pi fC}} - - (i)$
(a) Since initially the capacitive reactance and the Capacitance of the capacitor are given, hence by substituting these value in equation (i) we get

$${X_C} = \dfrac{1}{{2\pi fC}} \\\ \Rightarrow 12 = \dfrac{1}{{2\pi f \times \left( {1.5 \times {{10}^{ - 6}}F} \right)}} \\\$$

By further solving for the frequency, we get

$$f = \dfrac{1}{{2 \times \left( {3.14} \right) \times \left( {1.5 \times {{10}^{ - 6}}F} \right) \times 12}} \\\ \Rightarrow f = \dfrac{1}{{75.36 \times 1.5 \times {{10}^{ - 6}}}} \\\ \Rightarrow f = \dfrac{1}{{1.13 \times {{10}^{ - 4}}}} \\\ \therefore f = 8.84kHz \\\$$

Hence the operating frequency $= 8.84kHz$.

(b) Now it is said that the frequency is doubled so the new frequency will be
$f' = 2f = 2 \times 8.84 = 17.69kHz$
Hence capacitive reactance if the frequency is double when the same capacitor is being operated, so we can write equation (i) as
${X_C} = \dfrac{1}{{2\pi f'C}}$
Now substitute the value of the frequency and the capacitance in the equation, so we get

$${X_C} = \dfrac{1}{{2\pi f'C}} \\\ \Rightarrow {X_C} = \dfrac{1}{{2 \times \left( {3.14} \right) \times \left( {17.69} \right) \times \left( {1.5 \times {{10}^{ - 6}}} \right)}} \\\ \Rightarrow {X_C} = \dfrac{1}{{1.6 \times {{10}^{ - 4}}}} \\\ \therefore{X_C} = 6\Omega \\\$$

So the capacitive reactance if the frequency is doubled $= 6\Omega$.

Additional Information:
The capacitive reactance of a capacitor is denoted by ${X_C}$, where capacitive reactance is given by the formula ${X_C} = \dfrac{1}{{2\pi fC}}$, where $C$ is the capacitance in farads whose reactance is to be found and $f$ is the frequency in hertz.

Note: It is interesting to note here that as the operating frequency of the capacitor increases then, the capacitive reactance decreases but at constant capacitance value. So, we always opt for the higher frequency so that the value of the capacitive reactance decreases.