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

There is a 90° phase shift between $V_L$ and $I_L$, where $I_L$ lags $V_L$ by 90°.

To calculate the inductance $L$, we use the formula: $L = \frac{X_L}{2 \pi f}$ Substituting the values gives: $L = \frac{150}{2 \pi (60)} = 0.40 \ H$

The impedance $Z$ in an RLC circuit is given by: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ With the given values: $Z = \sqrt{60^2 + (150 - 120)^2} = 67.08 \Omega$

The power factor is calculated as: $\cos \phi = \frac{R}{Z} = \frac{60}{67.08} = 0.89$

1. $R = Z$.
2. The phase angle will be $0$°.
3. $V_L = V_C$ and $I$ will be maximum.

The resonant frequency is 800 Hz.

When the frequency increases from 200 Hz to 1600 Hz, the inductive reactance increases and the capacitive reactance decreases.

The voltage drop is calculated using: $V_L = I_L \times X_L$ Substituting the values: $V_L = 0.66 \times 10^{-6} \times 750 = 495 \mu V$

The relationship for a capacitor is: $X_C = \frac{1}{2 \pi f C}$ From this, we calculate: $C = \frac{1}{2 \pi (600)(1333)} = 198.99 \times 10^{-9} F$

At resonance, $Z = R = 20 \Omega$. Thus, the current $I$ is given by: $I = \frac{V_T}{Z} = \frac{220}{20} = 11 A$

The voltage drop across the inductor can be found as: $V_L = I \times X_L = 11 \times 50 = 550 V$

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

The phase angle would be zero because $X_L$ is equal to $X_C$ and thus $V_L = V_C$ and out of phase with each other, resulting in a power factor of 1. This occurs when $R = Z$. The circuit is at resonance.