2.1 Define the following terms with reference to RLC circuits: 2.1.1 Phase angle 2.1.2 Capacitance 2.2 Explain the effect Lenz's law on an inductor in an RLC circuit connected across an alternating supply voltage - NSC Electrical Technology Electronics - Question 2 - 2021 - Paper 1

Capacitance is the ability of a component to store electrical charge. In an RLC circuit, the capacitor allows the circuit to store energy in the form of an electric field, measured in farads (F).

Lenz's law states that the direction of induced emf (electromotive force) in an inductor opposes the change in current that created it. In an RLC circuit with alternating current, this results in energy being stored in the magnetic field of the inductor and then released back into the circuit, affecting the overall impedance and phase angle.

The inductive reactance, Xl, can be calculated using the formula: $X_L = 2 \pi f L$ Where:

• f = 60 Hz
• L = 300 mH = 0.3 H

Thus, $X_L = 2 \pi (60)(0.3) = 113.1 \text{ Ω}$

Total impedance, Z, in a series RLC circuit is calculated as: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ Substituting values:

• R = 30 Ω
• Xl = 113.1 Ω
• Xc = 30.32 Ω

Thus, $Z = \sqrt{30^2 + (113.1 - 30.32)^2} = \sqrt{30^2 + 82.78^2} ≈ 94.3 \text{ Ω}$

To determine whether the circuit is capacitive or inductive, compare the inductive reactance (Xl) with the capacitive reactance (Xc). Since Xl (113.1 Ω) > Xc (30.32 Ω), the circuit is inductive.

Using Ohm's law, the current through the capacitor, Ic, can be calculated as: $I_C = \frac{V}{X_C}$ Where:

• V = 300 V
• Xc = 50 Ω

Thus, $I_C = \frac{300}{50} = 6 \text{ A}$

Using Kirchhoff's current law, the total current can be calculated as: $I_T = I_R + I_C + I_L$ Given:

• I_R = 4 A
• I_L = 3 A
• I_C (calculated) = 6 A

Thus, $I_T = 4 + 3 + 6 = 13 \text{ A}$

Using the relation for inductive reactance: $X_L = 2 \pi f L$ Given:

• Xl = 3 A (from inductive current formula)
• f = 300 V

Rearranging, the inductance L can be calculated as: $L = \frac{X_L}{2 \pi f}$

The phase angle θ can be calculated using: $\theta = \cos^{-1} \left( \frac{I_R}{I_T} \right)$ Replacing variables: $\theta = \cos^{-1} \left( \frac{4}{13} \right) = 36.87^\circ$

The response curve represented by A is known as the resonance curve or frequency response curve.

Below the resonant frequency, the capacitive reactance (Xc) exceeds the inductive reactance (Xl), resulting in a net capacitive behavior of the circuit, which leads to current leading voltage.

The inductive reactance is represented by a straight line because it increases linearly with frequency, whereas the capacitive reactance is represented by a curved line as it decreases with increasing frequency.

The resonant frequency, fr, is calculated using the formula: $f_r = \frac{1}{2 \pi \sqrt{LC}}$ Given values:

• R = 20 Ω
• C = 1.47 μF
• L = 2.12 H

Calculating: $f_r = \frac{1}{2 \pi \sqrt{2.12 * 1.47 \times 10^{-6}}} ≈ 90.16 Hz$

A higher Q-factor indicates that the circuit is more selective, resulting in greater current flow at the resonant frequency. This is due to reduced energy losses in the circuit.

Half power points are frequencies at which the output power drops to half of its maximum value, indicating the bandwidth of the circuit.

The two factors that determine the quality factor Q of the circuit are:

1. The value of the series resistor (R), which affects energy loss.
2. The inductor/capacitor (L/C) ratio, which influences resonance characteristics.

As the Q-factor is lowered, the selectivity decreases, leading to a broader bandwidth around the resonant frequency, thus affecting the bandpass frequencies negatively.