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2.1 Define the following terms: 2.1.1 Capacitive reactance 2.1.2 Inductive reactance 2.2 FIGURE 2.2 below represents an RLC series circuit that consists of a 25 Ω resistor, a 44 mH inductor and a 120 μF capacitor, all connected across a 120 V/60 Hz supply - NSC Electrical Technology Power Systems - Question 2 - 2019 - Paper 1

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2.1-Define-the-following-terms:--2.1.1-Capacitive-reactance--2.1.2-Inductive-reactance--2.2-FIGURE-2.2-below-represents-an-RLC-series-circuit-that-consists-of-a-25-Ω-resistor,-a-44-mH-inductor-and-a-120-μF-capacitor,-all-connected-across-a-120-V/60-Hz-supply-NSC Electrical Technology Power Systems-Question 2-2019-Paper 1.png

2.1 Define the following terms: 2.1.1 Capacitive reactance 2.1.2 Inductive reactance 2.2 FIGURE 2.2 below represents an RLC series circuit that consists of a 25 Ω... show full transcript

Worked Solution & Example Answer:2.1 Define the following terms: 2.1.1 Capacitive reactance 2.1.2 Inductive reactance 2.2 FIGURE 2.2 below represents an RLC series circuit that consists of a 25 Ω resistor, a 44 mH inductor and a 120 μF capacitor, all connected across a 120 V/60 Hz supply - NSC Electrical Technology Power Systems - Question 2 - 2019 - Paper 1

Step 1

2.1.1 Capacitive reactance

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Answer

Capacitive reactance is defined as the opposition to an alternating current by the reactive component of a capacitor in an AC circuit. It can be calculated using the formula:

XC=12πfCX_C = \frac{1}{2 \pi f C}

Step 2

2.1.2 Inductive reactance

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Inductive reactance is defined as the opposition to an alternating current by the reactive component of an inductor in an AC circuit. It can be calculated using the formula:

XL=2πfLX_L = 2 \pi f L

Step 3

2.2.1 Inductive reactance

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To calculate the inductive reactance, we use:

XL=2πfLX_L = 2 \pi f L

Substituting the given values:

XL=2π(60)(44×103)=16.59ΩX_L = 2 \pi (60)(44 \times 10^{-3}) = 16.59 \, \Omega

Step 4

2.2.2 Capacitive reactance

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To calculate the capacitive reactance, we use:

XC=12πfCX_C = \frac{1}{2 \pi f C}

Substituting the given values:

XC=12π(60)(120×106)=22.11ΩX_C = \frac{1}{2 \pi (60)(120 \times 10^{-6})} = 22.11 \, \Omega

Step 5

2.2.3 Impedance of the circuit

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To find the impedance, we utilize the formula:

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

Substituting the calculated values:

Z=(25)2+(22.1116.59)2=25.6ΩZ = \sqrt{(25)^2 + (22.11 - 16.59)^2} = 25.6 \, \Omega

Step 6

2.3.1 Calculate the current through the capacitor

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The current through the capacitor can be calculated using:

IC=VSXCI_C = \frac{V_S}{X_C}

Substituting the values:

IC=22060=3.67AI_C = \frac{220}{60} = 3.67 \, A

Step 7

2.3.2 Calculate the reactive current

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To calculate the reactive current, we use:

IL=ITICI_L = I_T - I_C

Given that IT=6AI_T = 6 \, A:

IL=63.67=2.33AI_L = 6 - 3.67 = 2.33 \, A

Step 8

2.3.3 State whether the phase angle is leading or lagging

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The phase angle is leading because the current through the capacitor leads the voltage across it in an AC circuit. Capacitive circuits result in a leading phase angle.

Step 9

2.4.1 State the value of the capacitive reactance at resonance

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At resonance, capacitive reactance XCX_C equals inductive reactance XLX_L. Therefore, the value is simply equal to XLX_L.

Step 10

2.4.2 Calculate the value of the capacitor at resonance

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Answer

To calculate the value of the capacitor at resonance, we use:

C=1(2πfXL)C = \frac{1}{(2 \pi f X_L)}

Substituting for f=1000Hzf = 1000 \, Hz and XL=50.27ΩX_L = 50.27 \Omega:

C=12π(1000)(50.27)=3.16×105F(31.6μF)C = \frac{1}{2 \pi (1000)(50.27)} = 3.16 \times 10^{-5} \, F\, (31.6 \mu F)

Step 11

2.4.3 Explain how the value of the current at resonance will be affected by the supply voltage

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Answer

At resonance, the circuit behaves as a purely resistive circuit, and the current is maximum due to the impedance being minimized (only the resistance). If the supply voltage increases, the current will increase proportionally.

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