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 voltage across the capacitor (Vc = 15 V) is greater than the voltage across the inductor (Vl = 10 V). This indicates that the reactive component is larger and leads to a capacitive behavior.

The supply voltage can be calculated using the formula:

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

Substituting in the values:

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

The phase angle can be calculated using:

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

By substituting the values:

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

A phasor diagram should be drawn showing:

1. V_R along the horizontal axis.
2. V_C leading V_R, indicating a capacitive circuit.
3. V_T as the resultant vector drawn from the origin.

In a capacitive circuit, the supply current (I_R) is always in phase with the voltage across the resistance (V_R), while the supply voltage (V_T) lags behind. Thus, it is safe to assume that I_R leads V_T due to the inherent nature of capacitive reactance.

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

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

Substituting the given values:

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

The total current flow (I_T) can be calculated by:

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

Substituting the values:

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

The power factor (pf) can be calculated using:

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

Substituting the values:

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

At resonance, the inductive reactance equals the capacitive reactance (X_L = X_C). Using:

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

Rearranged to find C:

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

Where:

• $X_L = 62.83\, \Omega$
• $f = 50\, Hz$

Substituting the values:

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

A decrease in resistance increases the Q-factor of the circuit, which indicates improved selectivity and greater energy stored relative to energy dissipated.

The Q factor (Q) is calculated as:

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

The resonant frequency occurs at the point where the inductive and capacitive reactances are equal. However, specific calculations would depend on additional provided formulas or values.