FIGURE 2.4 below shows a parallel RLC circuit that consists of a 75 Ω resistor, an inductor with unknown inductance value and a capacitor with a capacitive reactance of 50 Ω, all connected across 300 VAC supply voltage - NSC Electrical Technology Power Systems - Question 2 - 2021 - Paper 1

Using Ohm's law for the inductor, we can find the inductive reactance ($X_L$) using the formula:

$X_L = \frac{V}{I_L}$

Where:

• $V = 300 V$ (voltage across the inductor)
• $I_L = 3 A$
  We can calculate:

$X_L = \frac{300 V}{3 A} = 100 \Omega$

So, the inductive reactance is 100 Ω.

The total current through the circuit can be calculated by using:

$I_T = \sqrt{I_R^2 + I_C^2 + I_L^2}$

Substituting the values: $I_T = \sqrt{(4)^2 + (1)^2 + (3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26} ≈ 5.1 A$

Thus, the total current flowing in the circuit is approximately 5.1 A.

The phase angle ($θ$) can be calculated using:

$θ = \cos^{-1} \left( \frac{I_R}{I_T} \right)$

Substituting the known values: $θ = \cos^{-1} \left( \frac{4 A}{5.1 A} \right) \approx 36.87°$

Therefore, the phase angle is approximately 36.87 degrees.