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 R and the calculated Z:

$$Cos \, \theta = \\frac{500}{502.49} \approx 0.995$$

The power factor is approximately 0.995, indicating that the circuit is lagging as the inductive reactance is greater than the capacitive reactance.

To find the resistance (R) of the lamp, we can use the formula:

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

Here, V = 110 V and P = 60 W:

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

So, the resistance of the 60 watt 110 V lamp is approximately 201.67 Ω.

The total current (I) flowing through the circuit can be calculated with:

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

Substituting the values:

• P = 60 W
• V_R = 110 V:

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

Thus, the total current flowing through the circuit is approximately 0.545 A.

Using the total voltage (V_S = 220 V) and the total current from the previous step, we can find the impedance (Z) again:

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

Substituting the values:

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

So, the impedance of the circuit is approximately 403.67 Ω.

To find the inductance (L) of the circuit, we can use the relation involving impedance:

$$X_L = Z^2 - R^2$$

Where:

• X_L = 2 \pi f L Using the results from earlier calculations, we find:

Substituting to find L gives:

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

After calculations:

• X_L = 400 Ω
• This leads to:

$$L = \\frac{400}{2 \\pi (50)} \approx 1.1 \, H$$

Thus, the inductance of the circuit is approximately 1.1 H.