Question

Find the Capacitance of a Capacitor which when connected in series with a $10\Omega$ resistance, makes the power factor equal to $0.5$ . The A.C. supply voltage is $80V - 100Hz$ .

Answer 1

Hint : In order to this question, to find the capacitance of a given capacitor, we will first rewrite the given facts and then we will apply the formula of angle between $R\,and\,Z$ and then we will find the capacitance in series.
Applying the formula that relates the angle between the $R\,and\,Z$ i.e.. $\cos \phi = \dfrac{R}{Z}$ and after that for capacitance, we use the formula: ${R^2} + {X_C}^2 = {Z^2}$ .
where, ${X_C}$ is the capacitance.

Complete Step By Step Answer:
Given that-
Resistance, $R = 10\Omega$
Power factor, $\cos \phi = 0.5$
Voltage, ${E_V} = 80V$
Frequency, $v = 100Hz$
We have to find the Capacitance, $C = ?$
As we know that the angle between the $R\,and\,Z$ :
$\because \cos \phi = \dfrac{R}{Z} \\\ \Rightarrow Z = \dfrac{R}{{\cos \phi }} \\\ \Rightarrow Z = \dfrac{{10}}{{0.5}} = 20$
Now, apply- ${R^2} + {X_C}^2 = {Z^2}$
$\Rightarrow {X_C} = \sqrt {{Z^2} - {R^2}} \\\ \Rightarrow {X_C} = \sqrt {{{20}^2} - {{10}^2}} \\\ \therefore {X_C} = 10\sqrt 3$
As we can write- ${X_C}$ as $\dfrac{1}{{\omega C}}$ or $\dfrac{1}{{\omega C}} = 10\sqrt 3$
Now, we can find the Capacitance:-
$\therefore C = \dfrac{1}{{\omega C}} = \dfrac{1}{{\omega 10\sqrt 3 }}$
As we know $\omega$ (omega) is a constant whose value is $2\pi \times 100$ .
$\Rightarrow C = \dfrac{1}{{2\pi \times 100 \times 10\sqrt 3 }} = 9.2 \times {10^{ - 5}}F$
Hence, the required capacitance is $9.2 \times {10^{ - 5}}F$ .

Note :
The ability of a component or circuit to gather and retain energy in the form of an electrical charge is known as capacitance. Capacitors are energy-storage devices that come in a variety of forms and sizes.