3.1 Explain the term reactance with reference to an alternating current circuit - NSC Electrical Technology Power Systems - Question 3 - 2024 - Paper 1

The circuit is predominantly capacitive as the value of the voltage across the capacitor (VC = 15 V) is greater than the voltage across the inductor (VL = 10 V). In RLC circuits, if the voltage across the capacitor is greater, it indicates that the reactive behavior of the capacitance dominates, leading to a phase shift where the current leads the voltage.

To calculate the supply voltage (VT), we use the formula:
$V_T = \\sqrt{V_R^2 + (V_C - V_L)^2}$
Given:
VR = 18V, VL = 10V, VC = 15V, we calculate:
$V_T = \\sqrt{18^2 + (15 - 10)^2} = \\sqrt{324 + 25} = \\sqrt{349} = 18.68 V$.
Thus, the supply voltage is approximately 18.68 V.

The phase angle (θ) can be calculated using the relationship:
$\cos(θ) = \frac{V_R}{V_T}$
Substituting the values gives:
$\cos(θ) = \frac{18}{18.68}$,
Thus,
$θ = \cos^{-1} \left( \frac{18}{18.68} \right) ≈ 15.51°$.
Therefore, the phase angle is approximately 15.51 degrees.

The current through the inductor (IL) can be found using Ohm’s law:
$I_L = \frac{V_T}{X_L}$
Where XL = 62.83 Ω and VT = 230 V. Thus,
$I_L = \frac{230}{62.83} ≈ 3.66 A.$
Hence, the current through the inductor is approximately 3.66 A.

The total current flow (IT) in the circuit can be calculated using the formula:
$I_T = \sqrt{I_R^2 + (I_L - I_C)^2}$
Given IR = 1.15 A and IL previously calculated as 3.66 A, substitute the values:
$I_T = \sqrt{1.15^2 + (3.66 - 1.59)^2} ≈ \sqrt{1.32 + 4.28} = \sqrt{5.6} ≈ 2.37 A.$
Thus, the total current flow is approximately 2.37 A.

The power factor (pf) is calculated as:
$\cos(θ) = \frac{I_R}{I_T}$
Substituting the values gives:
$\cos(θ) = \frac{1.15}{2.37} ≈ 0.49.$
Therefore, the power factor is approximately 0.49.

At resonance, the inductive reactance equals the capacitive reactance:
$X_C = X_L = \frac{1}{2\pi f C}$
Rearranging gives:
$C = \frac{1}{2\pi f X_C}.$
Given XL = 62.83 Ω and frequency f = 50 Hz, substituting these values yields:
$C = \frac{1}{2\pi(50)(62.83)} = 50.66 μF.$
Thus, the required capacitance that would cause resonance is approximately 50.66 μF.

A decrease in resistance increases the Q-factor of the circuit. The Q-factor is defined as the ratio of the inductive reactance (XL) to the resistance (R) in the circuit. As resistance decreases, the ratio increases, leading to a sharper resonance peak and higher efficiency at the resonant frequency.

The Q factor can be calculated using the formula:
$Q = \frac{X_L}{R}.$
Given XL = 2000 Ω and R = 50 Ω, substituting gives:
$Q = \frac{2000}{50} = 40.$
Thus, the Q factor is 40.

The resonant frequency (fr) can be calculated with the formula: $fr = f_1 + f_2 = 200 + 100 = 300 Hz.$
Therefore, the resonant frequency is 300 Hz.