5.1 A capacitor with a capacitive reactance of 250 Ω, an inductor with an inductive reactance of 300 Ω and a resistor with a resistance of 500 Ω are all connected in series to a 220 V/50 Hz supply - NSC Electrical Technology Power Systems - Question 5 - 2017 - Paper 1

The power factor (cos θ) can be calculated using the formula:

$\cos \theta = \frac{R}{Z}$

Substituting the known values:

$\cos \theta = \frac{500}{502.49} \approx 0.995$

Since the power factor is positive and close to 1, the circuit is lagging.

To find the resistance (R) of the 60 watt lamp, we use:

$R = \frac{V^2}{P}$

Substituting the given values:

$R = \frac{110^2}{60} = \frac{12100}{60} \approx 201.67 \ \Omega$

The total current (I) can be calculated using:

$I = \frac{P}{V_R}$

Substituting the values:

$I = \frac{60}{110} \approx 0.545 \ A$

Using the formula for impedance:

$Z = \frac{V_S}{I}$

Substituting the values:

$Z = \frac{220}{0.545} \approx 403.67 \ \Omega$

To find the inductance (L), we use the formula:

$L = \frac{Z^2 - R^2}{2 \pi f}$

Substituting the known values:

$L = \frac{403.67^2 - 201.67^2}{2 \pi (50)}$

Calculating:

$L \approx 1.11 \ H$