5.1 Name the TWO factors that influence the reactance of a capacitor - NSC Electrical Technology Power Systems - Question 5 - 2016 - Paper 1

Reactance is the opposition to the flow of alternating current (AC) due to the specific reactive components (capacitors and inductors) in the circuit. Impedance, on the other hand, is the total opposition to the flow of current in an AC circuit, which encompasses both resistive and reactive components.

The frequency/impedance characteristic curve is a graph where the vertical axis represents the impedance (Z) in ohms and the horizontal axis represents the frequency (f) in hertz. The curve typically shows:

• A decrease in impedance as the frequency approaches the resonant frequency (f_r).
• At resonant frequency, the impedance reaches its minimum value, which is equal to the resistance (R).
• The reactance of the capacitor (X_c) and inductor (X_l) are equal at this point, resulting in a Q-factor that indicates the circuit's selectivity.

To calculate the Q-factor (Q) of the circuit, we use the formula: $Q = \frac{X_l}{Z}$ Where:

• $X_l$ (inductive reactance) = 4 kΩ (given at resonance)
• $Z$ = 50 Ω (series resistance) Thus, $Q = \frac{4000}{50} = 80$ The Q-factor of the circuit is 80.

The inductive reactance ($X_l$) can be calculated using the formula: $X_l = 2\pi f L$ Given:

• $f = 50 Hz$
• $L = 400 mH = 0.4 H$ Thus, $X_l = 2\pi \times 50 \times 0.4 = 125.66 \Omega$

The capacitive reactance ($X_c$) can be calculated with the formula: $X_c = \frac{1}{2\pi f C}$ Where:

• $C = 47 μF = 47 \times 10^{-6} F$ and $f = 50 Hz$. Thus, $X_c = \frac{1}{2\pi \times 50 \times 47 \times 10^{-6}} = 67.73 \Omega$

The resonant frequency ($f_r$) can be calculated using the formula: $f_r = \frac{1}{2\pi\sqrt{LC}}$ Where:

• $L = 400 mH = 0.4 H$
• $C = 47 μF = 47 \times 10^{-6} F$. Thus, $f_r = \frac{1}{2\pi\sqrt{0.4 \times 47 \times 10^{-6}}} = 36.71 Hz$