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 $ .

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.