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3.1 Describe what is meant by the term in phase with reference to the relationship between the applied voltage and current in an RLC circuit connected to an AC supply - NSC Electrical Technology Power Systems - Question 3 - 2023 - Paper 1

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3.1 Describe what is meant by the term in phase with reference to the relationship between the applied voltage and current in an RLC circuit connected to an AC suppl... show full transcript

Worked Solution & Example Answer:3.1 Describe what is meant by the term in phase with reference to the relationship between the applied voltage and current in an RLC circuit connected to an AC supply - NSC Electrical Technology Power Systems - Question 3 - 2023 - Paper 1

Step 1

Describe what is meant by the term in phase with reference to the relationship between the applied voltage and current in an RLC circuit connected to an AC supply.

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Answer

The term 'in phase' refers to the condition where the voltage and current waveforms reach their maximum and minimum values at the same time. This occurs when the phase difference between the two waveforms is zero degrees, meaning they oscillate together harmoniously without any lag or lead, resulting in maximum power transfer in the circuit.

Step 2

Refer to the AC waveforms in FIGURE 3.2 below and draw the phasor diagram that represents them.

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Answer

In the phasor diagram, the voltage phasor is shown at an angle of 0 degrees, representing the reference voltage. The current phasor lags the voltage phasor due to the inductive nature of the circuit, forming a specific angle which relates to the values given in the figure. This angle can be represented as θ, where θ indicates the phase shift.

Step 3

Name TWO applications of resonance as applicable to tuned circuits.

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Answer

  1. Radio tuning circuits - used to select a specific frequency from a range of frequencies.
  2. Television tuning circuits - used to amplify signals at particular frequencies for better clarity.

Step 4

Calculate the value of the supply voltage.

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Answer

The supply voltage ( V_T ) can be calculated using the formula:

VT=sqrt(VR)2+(VLVC)2V_T = \\sqrt{(V_R)^2 + (V_L - V_C)^2}

Substituting the given values:

VT=sqrt(30)2+(6020)2=sqrt302+402=sqrt900+1600=sqrt2500=50VV_T = \\sqrt{(30)^2 + (60 - 20)^2} = \\sqrt{30^2 + 40^2} = \\sqrt{900 + 1600} = \\sqrt{2500} = 50 \, V

Step 5

Calculate the phase angle.

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Answer

The phase angle (θ) can be determined using the formula:

tan(θ)=VLVCVR\tan(θ) = \frac{V_L - V_C}{V_R}

Substituting the values:

θ = \tan^{-1}\left(\frac{60 - 20}{30}\right) = \tan^{-1}\left(\frac{40}{30}\right) ≈ 53.13^

Step 6

State whether the phase angle is leading or lagging.

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Answer

The phase angle is lagging because the voltage across the inductor ( V_L ) is greater than the voltage across the capacitor ( V_C ), indicating that current lags the voltage.

Step 7

Calculate the value of the inductance for the circuit to resonate at 3 kHz.

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Answer

At resonance, the inductive reactance equals the capacitive reactance:

XL=XCX_L = X_C

Thus,

L=XL2pifL = \frac{X_L}{2 \\pi f}

Substituting in the given values:

L=113.122pi(3000)6mHL = \frac{113.12}{2 \\pi (3000)} ≈ 6 \, mH

Step 8

Calculate the Q-factor of the circuit at resonance.

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Answer

The Q-factor (Q) is defined as:

Q=XCRQ = \frac{X_C}{R}

Substituting the given values:

Q=113.12100=1.13Q = \frac{113.12}{100} = 1.13

Step 9

Calculate the bandwidth of the circuit.

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Answer

Bandwidth (BW) can be calculated using:

BW=fRQBW = \frac{f_R}{Q}

Substituting the values:

BW=30001.132654.87HzBW = \frac{3000}{1.13} ≈ 2654.87 \, Hz

Step 10

Explain how the value of the total current would be influenced if R is halved when the circuit is at resonance.

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Answer

If the resistance (R) is halved at resonance, the total current in the circuit would double. This happens because the Q-factor increases, leading to a higher peak current for the same voltage.

Step 11

Calculate the current through the inductor.

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Answer

The current through the inductor is given by:

IL=VTXLI_L = \frac{V_T}{X_L}

Substituting in the values:

IL=120300=0.4AI_L = \frac{120}{300} = 0.4 \, A

Step 12

Calculate the active power.

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Answer

The active power (P) in the circuit can be calculated using:

P=I2RP = I^2 R

Substituting the current through the resistor:

P=(1.2)2×100=144WP = (1.2)^2 \times 100 = 144 \, W

Step 13

State, with a reason, if the circuit is capacitive or inductive.

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114 rated

Answer

The circuit is inductive because the inductive current ( I_L ) is greater than the capacitive current, indicating that the total current lags behind the voltage.

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