Figure shows a circuit consisting capacitor, inductor and a resistor connected in series with an AC source. Find the power factor of the circuit.
(Given R = 60$\Omega$, X\(_{L}\) = 150$\Omega$, X\(_{C}\) = 70$\Omega$)
Step 1: Understanding the Question:
We are given a series RLC circuit with the values of resistance (R), inductive reactance (X\(_{L}\)), and capacitive reactance (X\(_{C}\)). We need to calculate the power factor of this circuit. Step 2: Key Formula or Approach:
The power factor (\(\cos \phi\)) in a series RLC circuit is defined as the ratio of the resistance (R) to the total impedance (Z) of the circuit.
\[ \text{Power Factor} = \cos \phi = \frac{R}{Z} \]
The impedance Z is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Step 3: Detailed Explanation:
Given values are:
Resistance, \(R = 60 \Omega\)
Inductive reactance, \(X_L = 150 \Omega\)
Capacitive reactance, \(X_C = 70 \Omega\)
First, calculate the total impedance Z of the circuit.
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
\[ Z = \sqrt{(60)^2 + (150 - 70)^2} \]
\[ Z = \sqrt{60^2 + 80^2} \]
\[ Z = \sqrt{3600 + 6400} = \sqrt{10000} \]
\[ Z = 100 \Omega \]
This is a standard 6-8-10 right triangle relationship, scaled by 10.
Now, calculate the power factor.
\[ \cos \phi = \frac{R}{Z} = \frac{60 \Omega}{100 \Omega} \]
\[ \cos \phi = 0.6 \]
Since \(X_L>X_C\), the circuit is inductive, and the current lags the voltage. The power factor is 0.6 lagging. Step 4: Final Answer:
The power factor of the circuit is 0.6.
