3.1 Define capacitive reactance with reference to RLC circuits - NSC Electrical Technology Electronics - Question 3 - 2021 - Paper 1

The impedance $Z$ in an RLC circuit can be calculated using: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{60^2 + (150 - 120)^2} = 67.08 \Omega.$

The power factor can be calculated using: $\cos(\theta) = \frac{R}{Z} = \frac{60}{67.08} = 0.89.$

If the power factor is at unity:

1. The circuit behaves as a purely resistive circuit.
2. The voltage across the inductor ($V_L$) and the voltage across the capacitor ($V_C$) will be equal and opposite.
3. The current will reach its maximum value, indicating resonance.

The resonant frequency ($f_r$) is calculated as follows: $f_r = \frac{1}{2\pi\sqrt{LC}}.$ This value will be determined from further calculations.

As the frequency increases, inductive reactance ($X_L$) increases while capacitive reactance ($X_C$) decreases, leading to a shift in the circuit's behavior.

The voltage drop across the inductor ($V_L$) is given by: $V_L = I_L \times X_L.$ Assuming adequate current is calculated, use the appropriate values.

The capacitive reactance at 600 Hz is used to find the capacitance: $X_C = \frac{1}{2\pi fC} \Rightarrow C = \frac{1}{2\pi f X_C}.$ Substitute appropriate values to find capacitance.

At resonance: $Z = R = 20 \Omega$ The total current flow is: $I = \frac{V_T}{Z} = \frac{220}{20} = 11 A.$

The voltage drop across the inductor is given by: $V_L = I \times X_L = 11 \times 50 = 550 V.$

The Q-factor can be calculated using: $Q = \frac{X_L}{R} = \frac{50}{20} = 2.5.$

The phase angle would be zero because $X_L$ is equal to $X_C$, resulting in $V_L = V_C$ and they cancel each other out, leading to a power factor of 1.