Question

In a series LR circuit, power of 400W is dissipated from a source of 250 V, 50 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 (n / 3π) μF, then the value of n is ______.

Answer 1

400

Answer 2

Solution:

1. Determine R and XL:

   • Given: Source voltage, $V=250\;V$, power $P=400\;W$, and power factor $\cos\phi=0.8$.
   • The current, $$I=\frac{P}{V\cos \phi}=\frac{400}{250\times0.8}=2\;A.$$
   • The impedance magnitude, $$|Z|=\frac{V}{I}=\frac{250}{2}=125\;\Omega.$$
   • Since $\cos\phi=\frac{R}{|Z|}$, we get $$R=0.8\times125=100\;\Omega.$$
   • Now, $$|Z|=\sqrt{R^2+X_L^2}\Rightarrow 125^2=100^2+X_L^2.$$ Thus, $$X_L=\sqrt{125^2-100^2}=\sqrt{15625-10000}=\sqrt{5625}=75\;\Omega.$$

2. Capacitor for PF Correction:

   To correct the power factor to unity, the total reactive component must be zero:

   $$X_L - X_C = 0 \Rightarrow X_C = X_L = 75\;\Omega,$$

   where

   $$X_C=\frac{1}{2\pi f C}.$$

   Hence,

   $$C=\frac{1}{2\pi f X_C}=\frac{1}{2\pi\times50\times75}=\frac{1}{7500\pi}\;F.$$

3. Expressing in the Given Form:

   The capacitor is given as

   $$C=\frac{n}{3\pi}\;\mu F=\frac{n}{3\pi}\times10^{-6}\;F.$$

   Equate this with the computed value:

   $$\frac{n}{3\pi}\times10^{-6}=\frac{1}{7500\pi}.$$

   Cancel $\pi$ and solve for $n$:

   $$\frac{n\times10^{-6}}{3}=\frac{1}{7500} \quad\Rightarrow\quad n\times10^{-6}=\frac{3}{7500}=\frac{1}{2500}.$$

   Thus,

   $$n=\frac{10^{6}}{2500}=400.$$

Minimal Explanation:

• Compute current $I=400/(250\times0.8)=2\;A$.
• Find impedance $Z=250/2=125\;\Omega$ and $R=0.8\times125=100\;\Omega$.
• Determine $X_L=\sqrt{125^2-100^2}=75\;\Omega$.
• For unity PF, set $X_C=X_L$ leading to $C=1/(2\pi\times50\times75)=1/(7500\pi)\;F$.
• Equate to given form $n/(3\pi)\mu F$ to find $n=400$.