2.1 Define the following terms: 2.1.1 Capacitive reactance 2.1.2 Inductive reactance 2.2 FIGURE 2.2 below represents an RLC series circuit that consists of a 25 Ω resistor, a 44 mH inductor and a 120 μF capacitor, all connected across a 120 V/60 Hz supply - NSC Electrical Technology Power Systems - Question 2 - 2019 - Paper 1

Inductive reactance is the opposition to an alternating current by the reactive component of an inductor in an AC circuit.

The formula for inductive reactance is given by:

X_L = 2 imes rac{rac{1}{1000}}{60} imes 44 imes 10^{-3}

Calculating:

X_L = 2 imes rac{1}{1000} imes 60 imes 44 imes 10^{-3} = 16.59 ext{ Ω}

The formula for capacitive reactance is:

X_C = rac{1}{2 imes imes rac{1}{1000}}{60} imes 120 imes 10^{-6}

Calculating:

X_C = rac{1}{2 imes imes rac{1}{1000} imes 60 imes 120 imes 10^{-6}} = 22.11 ext{ Ω}

The impedance of the circuit can be calculated using:

Z = oxed{ ext{R}} + oxed{ ext{(X_C - X_L)}}

Given:

$R = 25 ext{ Ω}, X_C = 22.11 ext{ Ω}, X_L = 16.59 ext{ Ω}$

Thus,

Z = rac{ ext{25}^2 + (22.11 - 16.59)^2}{ ext{25}^2 + (5.52^2)} = 25.6 ext{ Ω}

Using the formula:

I_C = rac{V_S}{X_C}

Calculating:

I_C = rac{220 ext{ V}}{60 ext{ Ω}} = 3.67 ext{ A}

The reactive current through the inductor is calculated as:

$I_L = I_C - I_R$

Substituting the calculated values:

$I_L = 6 ext{ A} - 5.5 ext{ A} = 2.33 ext{ A}$

In this case, since the current through the capacitor is greater than that through the resistor, the phase angle is leading due to the capacitive nature of the circuit.

At resonance, the capacitive reactance is equal to the inductive reactance, hence:

$X_C = X_L = 50.27 ext{ Ω}$

Using the formula:

C = rac{1}{2 imes rac{1}{1000} imes f}

At resonance:

C = rac{1}{2 imes imes 60 imes 50.27} = 1 ext{ μF}

At resonance, the voltage across the inductor and capacitor will effectively cancel each other, allowing maximum current to flow through the circuit. This results in the inductor drawing a greater current than the supply voltage.