5.1 A capacitor with a capacitive reactance of 250 Ω, an inductor with an inductive reactance of 300 Ω and a resistor with a resistance of 500 Ω are all connected in series to a 220 V/50 Hz supply. Given: Xc = 250 Ω Xl = 300 Ω R = 500 Ω Vs = 220 V f = 50 Hz 5.1.1 Calculate the total impedance of the circuit. 5.1.2 Calculate the power factor of the circuit and state whether it is leading or lagging. 5.2 Calculate: 5.2.1 The resistance of the 60 watt 110 V lamp. 5.2.2 The total current flowing through the circuit. 5.2.3 The impedance of the circuit. 5.2.4 The inductance of the inductor.

NSC Electrical Technology Electronics Effect of Alternating Current: 5.1 A capacitor with a capacitiv

5.1 A capacitor with a capacitive reactance of 250 Ω, an inductor with an inductive reactance of 300 Ω and a resistor with a resistance of 500 Ω are all connected in series to a 220 V/50 Hz supply - NSC Electrical Technology Electronics - Question 5 - 2017 - Paper 1

Question 5

5.1 A capacitor with a capacitive reactance of 250 Ω, an inductor with an inductive reactance of 300 Ω and a resistor with a resistance of 500 Ω are all connected in... show full transcript

Worked Solution & Example Answer:5.1 A capacitor with a capacitive reactance of 250 Ω, an inductor with an inductive reactance of 300 Ω and a resistor with a resistance of 500 Ω are all connected in series to a 220 V/50 Hz supply - NSC Electrical Technology Electronics - Question 5 - 2017 - Paper 1

Step 1

5.1.1 Calculate the total impedance of the circuit.

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Answer

To calculate the total impedance (Z) of the circuit, we use the formula:

Z = \sqrt{R^2 + (X_L - X_C)^2}

Substituting the given values:

Z = \sqrt{500^2 + (300 - 250)^2}

This simplifies to:

Z = \sqrt{500^2 + 50^2} = \sqrt{250000 + 2500} = \sqrt{252500}

Therefore, the total impedance is approximately:

Z \approx 502.49 \ \Omega

Step 2

5.1.2 Calculate the power factor of the circuit and state whether it is leading or lagging.

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Answer

The power factor (cos θ) can be calculated using:

\cos \theta = \frac{R}{Z}

Substituting the values we have:

\cos \theta = \frac{500}{502.49} \approx 0.995

Since the circuit is in a series configuration with an inductor, it results in a lagging power factor. Thus, the power factor is approximately 0.995 and it is Lagging.

Step 3

5.2.1 The resistance of the 60 watt 110 V lamp.

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Answer

Using the power formula:

R = \frac{V^2}{P}

Substituting the known values:

R = \frac{110^2}{60} = \frac{12100}{60} \approx 201.67 \ \Omega

Step 4

5.2.2 The total current flowing through the circuit.

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We can find the total current (I) through the circuit using:

I = \frac{P}{V_R}

Substituting the values:

I = \frac{60}{110} \approx 0.545 \text{ A}

Step 5

5.2.3 The impedance of the circuit.

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Answer

To calculate the impedance (Z), we use:

Z = \frac{V}{I}

Substituting the values:

Z = \frac{220}{0.545} \approx 403.67 \ \Omega

Step 6

5.2.4 The inductance of the inductor.

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Answer

The inductive reactance (X_L) can be calculated using:

L = \frac{X_L}{2\pi f}

First, rearranging:

X_L = \sqrt{Z^2 - R^2}

Then substituting the calculated values:

X_L = \sqrt{400^2 - 201.67^2}

Now, substituting into the inductance formula yields:

L = \frac{X_L}{2\pi f} \approx \frac{100.69}{2\pi (50)} \approx 1.1 \text{ H}

5.1 A capacitor with a capacitive reactance of 250 Ω, an inductor with an inductive reactance of 300 Ω and a resistor with a resistance of 500 Ω are all connected in series to a 220 V/50 Hz supply - NSC Electrical Technology Electronics - Question 5 - 2017 - Paper 1

5.1.1 Calculate the total impedance of the circuit.

5.1.2 Calculate the power factor of the circuit and state whether it is leading or lagging.

5.2.1 The resistance of the 60 watt 110 V lamp.

5.2.2 The total current flowing through the circuit.

5.2.3 The impedance of the circuit.

5.2.4 The inductance of the inductor.

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