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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

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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 se... show full transcript

Worked Solution & Example Answer: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

The total impedance (Z) can be calculated using the formula:

Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Substituting the values:

  • R = 500 Ω
  • X_L = 300 Ω
  • X_C = 250 Ω

Thus,

Z=5002+(300−250)2=5002+502=250000+2500=252500≈502.49ΩZ = \sqrt{500^2 + (300 - 250)^2} = \sqrt{500^2 + 50^2} = \sqrt{250000 + 2500} = \sqrt{252500} \approx 502.49 Ω

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) is calculated using:

cos⁡θ=RZ\cos \theta = \frac{R}{Z}

Substituting the values:

cos⁡θ=500502.49≈0.995\cos \theta = \frac{500}{502.49} \approx 0.995

Since the impedance has a positive reactive component, the circuit is lagging.

Step 3

5.2.1 The resistance of the 60 watt 110 V lamp.

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Answer

The resistance (R) of the lamp can be calculated using:

R=V2PR = \frac{V^2}{P}

Substituting the values:

  • V = 110 V
  • P = 60 W Thus,
R=110260≈201.67ΩR = \frac{110^2}{60} \approx 201.67 Ω

Step 4

5.2.2 The total current flowing through the circuit.

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Answer

The total current (I) can be found using:

I=PVRI = \frac{P}{V_R}

Substituting the values:

  • P = 60 W
  • V_R = 110 V Thus,
I=60110≈0.545AI = \frac{60}{110} \approx 0.545 A

Step 5

5.2.3 The impedance of the circuit.

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Answer

The impedance (Z) can be calculated using:

Z=VIZ = \frac{V}{I}

Substituting the values:

  • V = 220 V
  • I = 0.545 A Thus,
Z=2200.545≈403.67ΩZ = \frac{220}{0.545} \approx 403.67 Ω

Step 6

5.2.4 The inductance of the inductor.

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Answer

The inductance (L) can be calculated using:

XL=2πfLX_L = 2 \pi f L

Where:

  • X_L is the inductive reactance,
  • f = 50 Hz. Thus,
L=XL2πf=3002π×50≈0.955HL = \frac{X_L}{2 \pi f} = \frac{300}{2 \pi \times 50} \approx 0.955 H

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