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.
