© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the known quantities
• The resistance in the circuit is R = 100\,\Omega .
• The angular frequency of the AC source is \omega = 50\,\text{rad/s} .
• The initial power factor is \cos(\phi) = 0.6 , with the current lagging the voltage (indicating that the circuit is inductive).
Step 2: Relate power factor to the circuit impedance
The impedance Z of a series R - L circuit can be written as:
Z = \sqrt{R^{2} + X_{L}^{2}},
where X_{L} is the inductive reactance. The power factor \cos(\phi) for a series R - L circuit is given by:
\cos(\phi) = \frac{R}{Z}.
Hence,
Z = \frac{R}{\cos(\phi)}.
Step 3: Calculate the total impedance Z
Substitute R = 100\,\Omega and \cos(\phi) = 0.6 :
Z = \frac{100}{0.6} \approx 166.67\,\Omega.
Step 4: Find the inductive reactance X_{L}
From Z^{2} = R^{2} + X_{L}^{2} , we get:
X_{L} = \sqrt{Z^{2} - R^{2}}.
Substituting Z \approx 166.67\,\Omega and R = 100\,\Omega :
X_{L} = \sqrt{(166.67)^{2} - (100)^{2}}
= \sqrt{27777.78 - 10000}
= \sqrt{17777.78}
\approx 133.33\,\Omega.
Step 5: Condition for unity power factor with a capacitor
For the total power factor to become unity, the net reactance in the circuit must be zero, implying that the capacitive reactance X_{C} must cancel the inductive reactance X_{L} exactly:
X_{C} = X_{L}.
The capacitive reactance is given by:
X_{C} = \frac{1}{\omega C}.
Therefore,
\frac{1}{\omega C} = X_{L}.
Step 6: Calculate the required capacitance C
Solve for C :
C = \frac{1}{\omega \, X_{L}}.
Substituting \omega = 50\,\text{rad/s} and X_{L} \approx 133.33\,\Omega :
C = \frac{1}{50 \times 133.33}
\approx 1.5 \times 10^{-4}\,\text{F}
= 150\,\mu\text{F}.
Final Answer
The required series capacitance to achieve a unity power factor is approximately
150 μF.
