Question

In a series LR circuit, power of \[400W\] is dissipated from a source of \[250{\text{ }}V,{\text{ }}50{\text{ }}Hz\]. The power factor of the circuit is $0.8$. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as $(\dfrac{n}{{3\pi }})\mu F$, then the value of $n$ is __

Answer 1

Hint: In order to solve this question, we will first find the value of inductive reactance of inductor using given information and general formulas and by the concept that when the power factor is unity the capacitive reactance value and inductive reactance value are equal, so we will find capacitive reactance and then compare it with given value to find the value of n.

Formula Used:
Power factor in LR series circuit is $\cos \phi = \dfrac{R}{{\sqrt {{R^2} + {X_L}^2} }}$ where, $R,{X_L}$ are resistance and inductive reactance.
Power is given by $P = \dfrac{{{V^2}}}{z}\cos \phi $ where, z is called impedance given by $z = \sqrt {{R^2} + {X_L}^2} $



Complete answer:
The power factor given to us is $\cos \phi = 0.8$ using the formula $\cos \phi = \dfrac{R}{{\sqrt {{R^2} + {X_L}^2} }}$ we get,
$0.8 = \dfrac{R}{z}$ where $z = \sqrt {{R^2} + {X_L}^2} $

Now, power given to us is $P = 400W$ using the formula $P = \dfrac{{{V^2}}}{z}\cos \phi $ and putting the values we get,
$
  400 = \dfrac{{{{(250)}^2}R}}{{{z^2}}} \\
  400 = 0.8\dfrac{{{{(250)}^2}}}{z} \\
   \Rightarrow z = 125\Omega \to (i) \\
 $
Now, using previous relation $0.8 = \dfrac{R}{z}$ we get, $R = 100\Omega \to (ii)$

Now, using the formula $z = \sqrt {{R^2} + {X_L}^2} $ and using value from equation (i) and (ii) we get
$
  {(100)^2} + {X_L}^2 = {(125)^2} \\
  {X_L}^2 = 5625 \\
  {X_L} = 75\Omega \to (iii) \\
 $

Now, in order to make power factor unity, the inductive reactance and capacitive reactance of the series LCR circuit must be equal so if ${X_C} = \dfrac{1}{{\omega C}}$ is the capacitive reactance then we have,
${X_L} = {X_C}$ using equation (iii) value and we have given that
$
  f = 50Hz \\
   \Rightarrow \omega = 2\pi f = 100\pi \\
 $

so, we get
$
  75 = \dfrac{1}{{2\pi fC}} \\
  C = \dfrac{1}{{7500\pi }} \\
 $

But we have given the capacitance in the form of $(\dfrac{n}{{3\pi }})\mu F$ so converting the capacitance in the same form as
$
  C = \dfrac{1}{{3\pi (2500)}} \\
  C = \dfrac{{4 \times {{10}^{ - 4}}}}{{3\pi }} \\
 $

Equating this value with $(\dfrac{n}{{3\pi }})\mu F$ we get,
$
  \dfrac{{4 \times {{10}^{ - 4}}}}{{3\pi }} = (\dfrac{n}{{3\pi }}){10^{ - 6}} \\
  n = 400 \\
 $

Hence, the value from the term $(\dfrac{n}{{3\pi }})\mu F$ is $n = 400$

Note: It should be remembered that when the power factor of a series LCR circuit is unity then the current and voltage are in the same phase, and in this condition, the inductive reactance and capacitive reactance become equal which is also the condition of resonance.