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

Reactance is the opposition to the flow of alternating current (AC) caused by capacitance or inductance, measured in ohms and represented by either capacitive reactance (Xc) or inductive reactance (Xl). Impedance, on the other hand, is the total opposition to current flow in an AC circuit and includes both resistance (R) and reactance (X), represented as a complex number: Z = R + jX.

To draw the frequency/impedance characteristic curve, plot impedance (Z) on the vertical axis and frequency (f) on the horizontal axis. The curve should illustrate a minimum point at the resonant frequency (fr). At frequencies lower than fr, the impedance is dominated by inductive reactance (Xl > Xc), while at frequencies higher than fr, capacitive reactance (Xc > Xl) dominates. The resonant point occurs where Xl equals Xc.

The Q-factor (Q) can be calculated using the formula:

$Q = \frac{X_l}{Z}$

Where:

• $X_l = 4 kΩ$
• $Z = R = 50 Ω$

Plugging in the values, we get:

$Q = \frac{4000}{50} = 80$

The inductive reactance (Xl) of the coil can be calculated using the formula:

$X_l = 2\pi f L$

Substituting in the given values:

• $f = 50 Hz$
• $L = 400 mH = 0.4 H$

Thus:

$X_l = 2\pi(50)(0.4) \approx 125.66 Ω$

The capacitive reactance (Xc) can be calculated using the formula:

$X_c = \frac{1}{2\pi f C}$

Substituting in the given values:

• $f = 50 Hz$
• $C = 47 μF = 47 \times 10^{-6} F$

Thus:

X_c = \frac{1}{2\pi(50)(47 \times 10^{-6}) \approx 67.73 Ω

The resonant frequency (fr) can be calculated using the formula:

$f_r = \frac{1}{2\pi \sqrt{LC}}$

Substituting in the given values:

• $L = 400 mH = 0.4 H$
• $C = 47 μF = 47 \times 10^{-6} F$

Thus:

$f_r = \frac{1}{2\pi \sqrt{0.4 \times 47 \times 10^{-6}}} \approx 36.71 Hz$