NSC Electrical Technology Electronics Calculations: 3.1 Define capacitive reactance

Question

3.1 Define capacitive reactance with reference to RLC circuits.

Capacitive reactance is the opposition of the capacitor to alternating current in an AC circuit.

3.2 State the phase relationship between the current and voltage in a pure inductive AC circuit.

There is a 90° phase shift between $V_L$ and $I_L$ where $I_L$ lags $V_L$ by 90°.

3.3 Refer to FIGURE 3.3 below and answer the questions that follow.
Given:

$R = 60 \Omega$
$X_C = 120 \Omega$
$X_L = 150 \Omega$
$V_T = 110 V$
$f = 60 Hz$

3.3.1 Calculate the inductance of the inductor.

Using the formula:
$$L = \frac{X_L}{2\pi f}$$
$$L = \frac{150}{2 \pi \times 60} = 0.40 H$$

3.3.2 Calculate the impedance of the circuit.

The formula for impedance in an RLC series circuit is:
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
Substituting the values we have:
$$Z = \sqrt{60^2 + (150 - 120)^2} = 67.08 \Omega$$

3.3.3 Calculate the power factor.

The power factor can be calculated as:
$$\cos(\theta) = \frac{R}{Z}$$
So,
$$\cos(\theta) = \frac{60}{67.08} = 0.89$$

3.3.4 State THREE conditions that will occur if the power factor is at unity in a RLC series circuit.

1. $R = Z$
2. Phase angle = 0°
3. $V_L = V_C$ and $X_L = X_C$, I is maximum.

3.4 Refer to FIGURE 3.4 A and FIGURE 3.4 B below to answer the questions that follow.

3.4.1 Determine the resonant frequency in FIGURE 3.4 B.

The resonant frequency $ f $ is 800 Hz.

3.4.2 Compare the values of the inductive reactance and capacitive reactance when the frequency increases from 200 Hz to 1600 Hz.

As the frequency increases, the inductive reactance increases and the capacitive reactance decreases.

3.4.3 Calculate the voltage drop across the inductor when the frequency is 600 Hz.

Using the formula:
$$V_L = I \times X_L$$
$$V_L = 0.66 \times 10^{-6} \times 750 = 495 \mu V$$

3.4.4 Calculate the value of the capacitor using the reactance value at 600 Hz.

Using:
$$X_C = \frac{1}{2 \pi f C}$$
Rearranging gives:
$$C = \frac{1}{2 \pi f X_C}$$
Substituting the known values:
$$C = \frac{1}{2\pi(600)(1333)} = 198.99 \times 10^{-9} F$$

3.5 Refer to FIGURE 3.5 below and answer the questions that follow.

3.5.1 Calculate the total current flow through the circuit.

At resonance, $$Z = R = 20 \Omega$$ thus:
$$I = \frac{V_T}{Z} = \frac{220}{20} = 11 A$$

3.5.2 Calculate the voltage drop across the inductor.

$$V_L = I \times X_L = 11 \times 50 = 550 V$$

3.5.3 Calculate the Q-factor of the circuit.

$$Q = \frac{X_L}{R} = \frac{50}{20} = 2.5$$

3.5.4 Explain why the phase angle of the circuit in FIGURE 3.5 would be zero.

The phase angle would be zero because $X_L$ is equal to $X_C$ and thus $V_L = V_C$ and out of phase with each other, resulting in a power factor of 1. The circuit is at resonance.

3.1 Define capacitive reactance with reference to RLC circuits - NSC Electrical Technology Electronics - Question 3 - 2021 - Paper 1

Worked Solution & Example Answer:3.1 Define capacitive reactance with reference to RLC circuits - NSC Electrical Technology Electronics - Question 3 - 2021 - Paper 1

Define capacitive reactance with reference to RLC circuits.

State the phase relationship between the current and voltage in a pure inductive AC circuit.

Calculate the inductance of the inductor.

Calculate the impedance of the circuit.

Calculate the power factor.

State THREE conditions that will occur if the power factor is at unity in a RLC series circuit.

Determine the resonant frequency in FIGURE 3.4 B.

Compare the values of the inductive reactance and capacitive reactance when the frequency increases from 200 Hz to 1600 Hz.

Calculate the voltage drop across the inductor when the frequency is 600 Hz.

Calculate the value of the capacitor using the reactance value at 600 Hz.

Calculate the total current flow through the circuit.

Calculate the voltage drop across the inductor.

Calculate the Q-factor of the circuit.

Explain why the phase angle of the circuit in FIGURE 3.5 would be zero.

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