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5.1 Distinguish between the reactance and impedance in an RLC circuit - NSC Electrical Technology Power Systems - Question 5 - 2016 - Paper 1

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5.1 Distinguish between the reactance and impedance in an RLC circuit. 5.2 Explain what the phase angle indicates. FIGURE 5.1 shows the relationship between the in... show full transcript

Worked Solution & Example Answer:5.1 Distinguish between the reactance and impedance in an RLC circuit - NSC Electrical Technology Power Systems - Question 5 - 2016 - Paper 1

Step 1

5.1 Distinguish between the reactance and impedance in an RLC circuit.

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Answer

Reactance is the opposition to the flow of alternating current caused by the inductance and capacitance in a circuit. It is frequency-dependent: inductive reactance ( XL=2πfL\mathbf{X_L = 2\pi f L}) increases with frequency, while capacitive reactance ( XC=12πfC\mathbf{X_C = \frac{1}{2\pi f C}}) decreases with frequency.

Impedance, on the other hand, is the total opposition to current flow in an RLC circuit and is a combination of resistive and reactance components, represented as:

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

where ZZ is the impedance, RR is resistance, XLX_L is inductive reactance, and XCX_C is capacitive reactance.

Step 2

5.2 Explain what the phase angle indicates.

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Answer

The phase angle in an RLC circuit indicates the phase difference between the voltage across and the current through the circuit. It affects the power factor, which is defined as:

Power Factor=cos(ϕ)\text{Power Factor} = \cos(\phi)

where ϕ\phi is the phase angle. A phase angle of 0° implies that current and voltage are in phase, while a phase angle of 90° means they are out of phase. This affects the efficiency of power transmission in the circuit.

Step 3

5.3.1 Explain the effect of frequency on the impedance of the circuit at point A.

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Answer

At point A on the frequency response curve, the impedance of the circuit varies depending on the relationship between inductive and capacitive reactance. As frequency increases, inductive reactance increases, while capacitive reactance decreases. At certain frequencies, this can lead to resonance, minimizing the overall impedance of the circuit, potentially to a very low value. Conversely, at lower frequencies, capacitive reactance dominates, increasing the impedance.

Step 4

5.3.2 Calculate the frequency at point A if the circuit included a 50 µF capacitor and a 0.1 H inductor.

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Answer

To find the resonant frequency, we use the formula for resonance in an RLC circuit:

f=12πLCf = \frac{1}{2\pi \sqrt{LC}} Substituting the values:

  • L=0.1HL = 0.1 H
  • C=50μF=50×106FC = 50 \mu F = 50 \times 10^{-6} F

Calculating:

= \frac{1}{2\pi \sqrt{5 \times 10^{-6}}} \ = \frac{1}{2\pi \cdot 0.002236} \ \approx 71.6 Hz$$

Step 5

5.4.1 Impedance of the circuit.

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Answer

The total impedance ZZ of the series RLC circuit is calculated using the formula:

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

Substituting the values:

  • R=30ΩR = 30 \Omega
  • XL=40ΩX_L = 40 \Omega
  • XC=20ΩX_C = 20 \Omega

Calculating:

= \sqrt{900 + 400} \ = \sqrt{1300} \ = 36.06 \Omega$$

Step 6

5.4.2 Phase angle of the circuit.

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Answer

The phase angle ϕ\phi is calculated using the formula:

ϕ=tan1(XLXCR)\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)

Substituting the values:

  • R=30ΩR = 30 \Omega
  • XL=40ΩX_L = 40 \Omega
  • XC=20ΩX_C = 20 \Omega

Calculating:

= \tan^{-1}\left(\frac{20}{30}\right) \ = \tan^{-1}(0.667) \ \approx 33.69^ ext{o}$$

Step 7

5.5 Supply frequency if the capacitor draws a current of 10 mA.

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Answer

Using the formula for capacitor current:

IC=Vimes2πfCI_C = V imes 2\pi f C

Where:

  • IC=10mA=0.01AI_C = 10 mA = 0.01 A
  • V=20VV = 20 V
  • C=1.47μF=1.47×106FC = 1.47 \mu F = 1.47 \times 10^{-6} F

Rearranging to solve for ff:

= \frac{0.01}{20 \times 2\pi (1.47 \times 10^{-6})} \ \approx 0.107 Hz$$

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