Refer to Figure Z.2 below and answer the questions that follow - NSC Electrical Technology Electronics - Question 5 - 2017 - Paper 1

An increase in the capacitance of a capacitor leads to a decrease in its capacitive reactance. This is expressed mathematically as:

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

As capacitance increases, the denominator of this formula increases, hence reducing the capacitive reactance.

Resonance occurs when the capacitive reactance of a circuit equals its inductive reactance. This is mathematically expressed as:

$X_L = X_C$

The circuit conditions at resonance are also that the total impedance ( $Z$) equals the resistance ( $R$), which can be stated as $\theta = 0$.

Inductive reactance $X_L$ is given by:

$X_L = 2\pi f L$

Substituting the values, the calculation becomes:

$X_L = 2 \pi \times 50 \times 400 \times 10^{-3} = 125.66 \Omega$

The capacitive reactance $X_C$ can be calculated as:

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

Substituting the given values:

$X_C = \frac{1}{2 \pi \times 50 \times 47 \times 10^{-6}} = 67.72 \Omega$

The total impedance $Z$ of the circuit can be calculated using:

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

For the given values:

$Z = \sqrt{20^2 + (125.66 - 67.72)^2} = 61.29 \Omega$

The quality factor $Q$ is given by:

$Q = \frac{L}{R C}$

Substituting the values provides:

$Q = \frac{400 \times 10^{-3}}{20 \times 47 \times 10^{-6}} = 4.61$

In this circuit, the behavior is predominantly inductive. This is indicated by the fact that the inductive reactance ($X_L$) is greater than the capacitive reactance ($X_C$). This leads to describing the nature of the circuit as inductive.