2.1 Define the term impedance with reference to RLC circuits - NSC Electrical Technology Electronics - Question 2 - 2018 - Paper 1

In a pure capacitive circuit, the current leads the voltage by 90 degrees. The waveforms can be depicted as follows:

• The voltage waveform is a sine wave starting at zero.
• The current waveform is also a sine wave but reaches its peak 90 degrees ahead of the voltage waveform.

In a pure inductive circuit, the voltage leads the current by 90 degrees. The waveforms can be illustrated as follows:

• The current waveform is a sine wave starting at zero.
• The voltage waveform is a sine wave that reaches its peak 90 degrees ahead of the current waveform.

The capacitance can be calculated using the formula:

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

Substituting the values: $C = \frac{1}{2 \pi (60)(36)} \approx 73.68 \mu F$

The inductance can be calculated using the formula:

$L = \frac{X_L}{2 \pi f}$

Substituting the values: $L = \frac{22}{2 \pi (60)} \approx 58.35 \text{ mH}$

The total impedance can be calculated using:

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

Substituting the values: $Z = \sqrt{12^2 + (22 - 36)^2} \approx 18.44 \Omega$

The current can be calculated using Ohm's law:

$I = \frac{V_s}{Z}$

Substituting the values: $I = \frac{60}{18.44} \approx 3.25 A$

The reactive power can be calculated using:

$Q = V_s \cdot I \cdot \sin(\theta)$

Where $heta$ is the phase angle. Since the circuit is RLC, we can assume (\theta \approx 50^{\circ}) leading to: $Q = 60 \cdot 3.25 \cdot \sin(50^{\circ}) \approx 149.38 VA$

The resonant frequency is the frequency at which the inductive reactance ($X_L$) is equal to the capacitive reactance ($X_C$), resulting in:

$X_L = X_C$

At resonance, the circuit behaves purely resistive.