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

To calculate the impedance $Z$ of the circuit, use:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

Substituting the values: $Z = \sqrt{60^2 + (150 - 120)^2} \approx 67.08 \Omega$

To calculate the power factor, use:

$\text{Cos } \phi = \frac{R}{Z}\$

Substituting the values: $\text{Cos } \phi = \frac{60}{67.08} \approx 0.89$

If the power factor is at unity, the following conditions will occur:

1. $R = Z$
2. The phase angle will be $0^{ ext{o}}$,
3. The voltage across the inductive and capacitive reactive components will be equal, meaning $V_L = V_C$.

The resonant frequency $f_r$ can be determined by the condition where $X_L = X_C$. In the given data, that value occurs at: $f_r = 800 Hz$

As frequency increases, $X_L$ increases while $X_C$ decreases. Thus:

• At lower frequencies, capacitive reactance dominates.
• As frequency rises, inductive reactance becomes significant and eventually surpasses $X_C$.

Using the formula for voltage drop across the inductor: $V_L = I_L \times X_L$ First, assume $I_L = 0.66 mA$. Substituting, $V_L = 0.66 \times 10^{-6} \times 750 \approx 495 \mu V$

Using the formula for capacitive reactance: $X_C = \frac{1}{2 \pi f C}$ To find the capacitance $C$: $C = \frac{1}{2 \pi f X_C}$ Substituting: $C \approx 198.99 nF$

At resonance, $Z = R = 20 \Omega$, $I = \frac{V_T}{Z} = \frac{220}{20} = 11 A$

Using the formula: $V_L = I \times X_L$ Substituting: $V_L = 11 \times 50 = 550 V$

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

The phase angle is zero when $X_L$ equals $X_C$. Thus, $V_L$ and $V_C$ will be in phase, resulting in the cancellation of reactive power and essentially implying a pure resistive circuit at resonance.