Question

A coil of inductive reactance \(31 \Omega\) has a resistance of \(8 \Omega\). It is placed in series with a condenser of capacitive reactance \(25 \Omega\). The combination is connected to an a.c. source of 110 volt. The power factor of the circuit is. (a) \(0.80\) (b) \(0.33\) (c) \(0.56\) (d) \(0.64\)

Short answer

The power factor of the circuit is approximately 0.80. The correct answer is (a) 0.80.

Step-by-step solution

Calculate the total impedance

To calculate the total impedance (Z) of the circuit, we first need to find the difference between the inductive reactance (X_L) and the capacitive reactance (X_C). The formula for calculating this value is: \(X = X_L - X_C\) where \(X\) is the difference between the inductive and capacitive reactance. Substituting the given values \(X_L = 31\Omega\) and \(X_C = 25\Omega\), we get: \(X = 31 - 25 = 6\Omega\)

Calculate the total impedance using the Pythagorean theorem

Now that we have the difference in reactance (X) and the resistance (R), we can calculate the total impedance (Z) using the Pythagorean theorem: \(Z = \sqrt{R^2 + X^2}\) Substituting the given values for resistance \(R = 8\Omega\) and the calculated value for \(X = 6\Omega\), we get: \(Z = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\Omega\)

Find the angle between impedance and applied voltage

To find the angle between the total impedance (Z) and the applied voltage (V), we must use the arctangent function: \(\theta = \arctan (\frac{X}{R})\) Substituting the values for \(X = 6\Omega\) and \(R = 8\Omega\), we get: \(\theta = \arctan (\frac{6}{8}) = \arctan (0.75) \approx 36.87°\)

Calculate the power factor

Now that we have the angle between the impedance and voltage (\(\theta\)), we can calculate the power factor (PF) by taking the cosine of this angle: \(PF = \cos(\theta)\) Substituting the value for \(\theta \approx 36.87°\), we get: \(PF = \cos(36.87°) \approx 0.80\) Hence, the power factor of the circuit is approximately 0.80. The correct answer is (a) 0.80.