In this circuit, a resistor and a capacitor are connected in parallel to a 400-Hz line. The power factor is 47.1%, the apparent power is 4086.13 VA, and the capacitance of the capacitor is 33.15 µF (Figure 23–8). Find the following unknown values: ∠θ—angle theta P—true power VARsC—capacitive VARs

Question

In this circuit, a resistor and a capacitor are connected in parallel to a 400-Hz line. The power factor is 47.1%, the apparent power is 4086.13 VA, and the capacitance of the capacitor is 33.15 µF (Figure 23–8). Find the following unknown values:

∠θ—angle theta

P—true power

[latex]VARs_{C}[/latex]—capacitive VARs

[latex]X_{C}[/latex]—capacitive reactance

[latex]E_{C}[/latex]—voltage drop across the capacitor

[latex]I_{C}[/latex]—capacitive current

[latex]E_{R}[/latex]—voltage drop across the resistor

[latex]I_{R}[/latex]—resistive current

R—resistance of the resistor

[latex]E_{T}[/latex]—applied voltage

[latex]I_{T}[/latex]—total circuit current

Z—impedance of the circuit

Accepted Answer

Angle Theta
The power factor is the cosine of angle theta. To find angle theta, change the power factor from a percentage into a decimal fraction by dividing by 100:

[latex]PF = \frac{47.1}{100}[/latex]

[latex]PF = 0.471[/latex]

cos ∠θ = 0.471
∠θ = 61.9º

True Power
The power factor is determined by the ratio of true power to apparent power:

[latex]PF = \frac{P}{VA}[/latex]

This formula can be changed to calculated the true power when the power factor and apparent power are known (refer to the Resistive-Capacitive Parallel Circuits Formula Section of Appendix B).

P = VA × PF

P = 4086.13 × 0.471

P = 1924.47 W

Reactive Power
The apparent power, true power, and reactive power form a right triangle as shown in Figure 23–9. Because these powers form a right triangle, the Pythagorean theorem can be used to find the leg of the triangle represented by the reactive power:

[latex]VARs_{C} = \sqrt{VA^{2} - P^{2}}[/latex]

[latex]VARs_{C} = \sqrt{4086.13^{2} - 1924.47^{2}}[/latex]

[latex]VARs_{C} = \sqrt{12,992,873.6}[/latex]

[latex]VARs_{C} = 3604.56[/latex]

A vector diagram showing the relationship of apparent power, true power, reactive power, and angle theta is shown in Figure 23–10.

Capacitive Reactance
Because the capacitance of the capacitor and the frequency are known, the capacitive reactance can be found using the formula

[latex]X_{C} = \frac{1}{ 2 \pi f C}[/latex]

[latex]X_{C} = \frac{1 Hz}{2 imes 3.1416 imes 400 Hz imes 0.00003315 F} [/latex]

[latex]X_{C} = 12.003 \Omega[/latex]

Voltage Drop Across the Capacitor

[latex]E_{C} = \sqrt{VARs_{C} imes X_{C}}[/latex]

[latex]E_{C} = \sqrt{3604.56 VARs imes 12.003 \Omega}[/latex]

[latex]E_{C} = \sqrt{43,265.53}[/latex]

[latex]E_{C} = 208 V[/latex]

[latex]\pmb{E_{R}}[/latex] and [latex]\pmb{E_{T}}[/latex]
The voltage must be the same across all branches of a parallel circuit. Therefore, if 208 V are applied across the capacitive branch, 208 V must be the total voltage of the circuit as well as the voltage applied across the resistive branch:

[latex]E_{T} = 208 V[/latex]

[latex]E_{R} = 208 V[/latex]

[latex]\pmb{I_{C}}[/latex]
The amount of current flowing in the capacitive branch can be calculated using the formula

[latex]I_{C} = \frac{E_{C}}{X_{C}}[/latex]

[latex]I_{C} = \frac{208 V}{12 \Omega}[/latex]

[latex]I_{C} = 17.331[/latex]

[latex]\pmb{I_{R}}[/latex]
The amount of current flowing through the resistor can be calculated using the formula

[latex]I_{R} = \frac{P}{E_{R}}[/latex]

[latex]I_{R} = \frac{1924.47 W}{208}[/latex]

[latex]I_{R} = 9.25 A[/latex]

Resistance
The amount of resistance can be calculated using the formula

[latex]R = \frac{E_{R}}{I_{R}}[/latex]

[latex]R = \frac{208 V}{9.25 A}[/latex]

[latex]R = 22.49 \Omega[/latex]

Total Current
The total current can be calculated using Ohm’s law or by vector addition because both the resistive and capacitive currents are known. Vector addition is used in this example:

[latex]I_{T} = \sqrt{I_{R} ^{2} + I_{C} ^{2}}[/latex]

[latex]I_{T} = \sqrt{(9.25 A)^{2} + (17.33 A)^{2}}[/latex]

[latex]I_{T} = \sqrt{385.89 A^{2}}[/latex]

[latex]I_{T} = 19.64 A[/latex]

Impedance
The impedance of the circuit is calculated using the formula

[latex]Z = \frac{R imes X_{C}}{\sqrt{R^{2} + X_{C} ^{2}}}[/latex]

[latex]Z = \frac{22.479 \Omega imes 12.003 \Omega}{\sqrt{(22.479 \Omega)^{2} + (12.003 \Omega)^{2}}}[/latex]

[latex]Z = \frac{269.815 \Omega}{\sqrt{649.377 \Omega^{2}}}[/latex]

[latex]Z = \frac{269.815 \Omega}{25.483 \Omega}[/latex]

[latex]Z = 10.588 \Omega[/latex]

The complete circuit with all values is shown in Figure 23–11.

Question 23.2: In this circuit, a resistor and a capacitor are connected in...

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