3.1 Describe what is meant by the term in phase with reference to the relationship between the applied voltage and current in an RLC circuit connected to an AC supply - NSC Electrical Technology Power Systems - Question 3 - 2023 - Paper 1

To represent the AC waveforms as phasors, draw a horizontal axis for the voltage and a vertical axis for the current. The phasor corresponding to the voltage waveform will be drawn as a line at an angle that represents the phase difference between voltage and current, usually with the current phasor leading or lagging the voltage phasor based on the circuit characteristics.

1. Radio tuning circuits, which allow for the selection of specific frequencies to receive radio signals.
2. Television tuning circuits, used for selecting the desired channels by resonating at specific frequencies.

Using the formula for the total voltage in an RLC circuit:

The phase angle can be calculated using:
$\tan(\theta) = \frac{V_L - V_C}{V_R}$
Substituting the values:
$\theta = \tan^{-1}\left(\frac{60 - 20}{30}\right) = \tan^{-1}(\frac{40}{30}) = 53.13^{\circ}$

The phase angle is lagging because in this scenario, the voltage across the inductor is greater than that across the capacitor.

At resonance, the inductive reactance equals the capacitive reactance:
$X_L = X_C$
Thus, using
$X_L = 2\pi f L$, we can find:
$L = \frac{X_L}{2\pi f} = \frac{113.12}{2\pi \cdot 3000} = 6 mH$

The Q-factor can be calculated using:
$Q = \frac{X_C}{R} = \frac{113.12}{100} = 1.13$

The bandwidth is given by:
$BW = \frac{f}{Q} = \frac{3000}{1.13} \approx 2654.87 Hz$

When the resistance R is halved, the total current in the circuit will increase because the Q-factor will result in a sharper peak at resonance, leading to higher current draw at the resonating frequency. Therefore, the overall impedance decreases, allowing more current to flow through the circuit.

The current through the inductor can be found using:
$I_L = \frac{V_T}{X_L} = \frac{120}{300} = 0.4 A$

The active power can be calculated using the formula:
$P = I_R^2 \cdot R = (1.2)^2 \cdot 100 = 144 W$

The circuit is inductive because the inductive current through the inductor is greater than the capacitive current, indicating that the overall behavior of the circuit is more influenced by the inductor.