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

The circuit is predominantly capacitive because the capacitive voltage ($V_C = 15 ext{ V}$) is greater than the inductive voltage ($V_L = 10 ext{ V}$). This indicates that the reactive behavior leads to a net capacitive effect, resulting in the supply current leading the supply voltage.

To calculate the supply voltage ($V_T$), we use the formula:

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

Substituting the given values:

$V_T = \\sqrt{18^2 + (15 - 10)^2} = \\sqrt{324 + 25} = \\sqrt{349} \\approx 18.68 ext{ V}$

The phase angle ($θ$) can be calculated using:

$θ = \\cos^{-1} \left( \frac{V_R}{V_T} \right)$

Using the calculated supply voltage,

$θ = \\cos^{-1} \left( \frac{18}{18.68} \right) \approx 15.51°$

This step requires visual representation. Ensure the voltage vectors for $V_C$, $V_L$, and $V_R$ are drawn accurately considering their phase relationships.

In capacitive circuits, the supply current leads the supply voltage due to the nature of capacitive reactance. This means that the current flow begins before the voltage reaches its peak, indicating a leading phase relationship.

The current flowing through the inductor ($I_L$) can be calculated using:

$I_L = \frac{V_T}{X_L}$

Substituting the values:

$I_L = \frac{230}{62.83} \approx 3.66 ext{ A}$

The total current flow ($I_T$) is found using:

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

Substituting the known values:

$I_T = \sqrt{(1.15)^2 + (3.66 - 1.59)^2} \approx 2.37 ext{ A}$

The power factor is calculated using:

$\cos(\phi) = \frac{I_R}{I_T}$

So,

$\cos(\phi) = \frac{1.15}{2.37} \approx 0.49$

At resonance, the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$). The formula for capacitance is:

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

Set $X_C$ equal to $X_L$:

$C = \frac{1}{2\pi f X_L} = \frac{1}{2\pi (50)(62.83)} \approx 50.66 \, \mu F$

A decrease in resistance increases the Q-factor of the circuit. A higher Q-factor indicates a sharper resonance peak, meaning the circuit can store and release energy more efficiently.

The Q-factor can be calculated using:

$Q = \frac{X_L}{R} = \frac{2000}{50} = 40$

The resonant frequency can be calculated based on the given values. The midpoint between $f_1$ and $f_2$ gives:

$f_{resonance} = \frac{f_1 + f_2}{2} = \frac{200 + 2100}{2} = 1150 \, Hz$