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:A capacitor is an electrical system made up of terminals, two in number. The device stores energy as electric charge. Resistance is considered to be an electrical circuit's objection to flow of current. The power factor is a measure of efficiency of energy.

Complete step by step answer:
This is a numerical question, so we need to read the question and note down the values given in the question. A resistor of resistance is given to be $R = 10\;\Omega $
Power factor represented in this manner is ($\cos \phi $) $ = 0.5$
They have mentioned the supply voltage to be ${E_V} = 80V$
The frequency at which this takes place is $\nu = 100Hz$
Capacitance refers to the capacitor's potential to retain charges.
Let capacitance be denoted as $C$, then we need to find $C = ?$

Note that the power factor is found by the equation:
$\cos \phi = \dfrac{R}{Z}$, here $Z = $ impedance which is an unknown value
This equation can be rearranged to give:
$Z = \dfrac{R}{{\cos \phi }}$
If this is the case, then we can substitute the known values of resistance and power factor to get impedance value.
$Z = \dfrac{{10\;\Omega }}{{0.5}}$
Simplifying we get:
$Z = 20\;\Omega $

Now there is a formula to find the capacitive reactance (${X_C}$) that looks like this:
$Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} $, here ${X_L} = 0$
$ \Rightarrow Z = \sqrt {{R^2} + {X_C}^2} $
Substitute the known values of impedance and resistance:
$20 = \sqrt {{{10}^2} + {X_C}^2} $
$ \Rightarrow 20 = \sqrt {100 + {X_C}^2} $
Squaring both sides we get the following equation:
$400 = 100 + {X_C}^2$
Rearranging the equation:
${X_C}^2 = 300$
Now we take square root on both sides to compute the value of capacitive reactance;
${X_C} = 10\sqrt 3 $

With the help of capacitive reactance, we have an equation that has an association to capacitance ($C$);
${X_C} = \dfrac{1}{{2\pi \nu C}}$
So substituting the values that we have, which are the capacitive reactance and frequency:
$10\sqrt 3 = \dfrac{1}{{2 \times 3.14 \times 100 \times C}}$
Rearranging the above equation to find capacitance:
$C = \dfrac{1}{{2 \times 3.14 \times 100 \times 10\sqrt 3 }}$
Simplifying we can write:
$ \therefore C = 9.2 \times {10^{ - 5}}\,F$

Therefore the required value of capacitance at the given values of resistance, power factor and frequency is $9.2 \times {10^{ - 5}}F$.

Note:Capacitors may be used for a variety of purposes: batteries are used to store electric potential energy with the help of capacitors. Capacitors help in removing unwanted frequency signals. When combined with resistors, capacitors can cause voltage variations which are delayed. Also they are used to distinguish between alternating current and direct current.