3.1 Define a phasor diagram with reference to RLC circuits connected across an alternating voltage supply - NSC Electrical Technology Electronics - Question 3 - 2022 - Paper 1

To calculate the total supply voltage, we apply the formula:

$V_T = \\sqrt{( V_R )^2 + ( V_L - V_C )^2 }$

Where:

• $V_R = 150 \text{V}$
• $V_L = 180 \text{V}$
• $V_C = 90 \text{V}$

Calculating: $V_T = \\sqrt{(150)^2 + (180 - 90)^2} = \\sqrt{150^2 + 90^2} = \\sqrt{22500 + 8100} = \\sqrt{30600} \approx 174.93 \text{V}$

Thus, the total supply voltage applied to the circuit is approximately 174.93 V.

The power factor is lagging in this circuit, as the inductive voltage ($V_L$) is greater than the capacitive voltage ($V_C$). This means that the voltage leads the current in the circuit.

To find the total current ($I_T$) in a parallel circuit, use:

$I_T = \sqrt{(I_C)^2 + (I_L)^2 + (I_R)^2}$

Given:

• $I_C = 4\text{ A}$
• $I_L = 6\text{ A}$
• $I_R = 4\text{ A}$

Calculating: $I_T = \sqrt{(4)^2 + (6)^2 + (4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} \approx 8.25 \text{ A}$

The phase angle ($\theta$) can be calculated using:

$\theta = \cos^{-1}\left(\frac{I_R}{I_T}\right)$

Substituting in the values: $\theta = \cos^{-1}\left(\frac{4}{8.25}\right) \approx 26.49^\circ$

The phasor diagram would illustrate the relationship between the currents $I_C$, $I_L$, $I_R$, and total current $I_T$, with the respective angles represented accurately, showing the contributions to the total current.

The circuit is inductive because the inductive current ($I_L$) is greater than the capacitive current ($I_C$), leading to a net inductive effect.

At resonance, $X_L = X_C = 150 \Omega$. Therefore, the quality factor (Q) is given by:

$Q = \frac{R}{X_L}$

Substituting the values: $Q = \frac{2200}{150} \approx 14.67$

The bandwidth (BW) of the circuit is calculated using the formula:

$BW = f_r \times \frac{1}{Q}$

So, $BW = 2,387 \times 10^3 \times \frac{1}{14.67} \approx 162.82 \text{ Hz}$

Using the formula for capacitive reactance:

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

Rearranging gives: $C = \frac{1}{2 \pi f_r X_C}$ Substituting the values: $C = \frac{1}{2 \pi (2,387 \times 10^3)(150)} \approx 4.445 \times 10^{-7} F \approx 444.51 nF$

Selectivity is a measure of how well a resonant circuit responds to a range of frequencies while excluding others. It indicates the circuit's ability to become resonant at specific frequencies while effectively filtering out unwanted signals.

This section would require visual analysis of Figure 3.5 to answer any related questions. The specifics of currents and voltages in a resonating scenario would be discussed.