One resistor and two capacitors are connected in series to a 42.5-V, 60-Hz line (Figure 22–8). Capacitor 1 has a reactive power of 3.75 VARs, the resistor has a true power of 5 W, and the second capacitor has a reactive power of 5.625 VARs. Find the following values: VA—apparent power IT—total

Question

One resistor and two capacitors are connected in series to a 42.5-V, 60-Hz line (Figure 22–8). Capacitor 1 has a reactive power of 3.75 VARs, the resistor has a true power of 5 W, and the second capacitor has a reactive power of 5.625 VARs. Find the following values:

VA—apparent power

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

Z—impedance of the circuit

PF—power factor

∠θ—angle theta

[latex]E_{C1}[/latex]—voltage drop across the first capacitor

[latex]X_{C1}[/latex]—capacitive reactance of the first capacitor

[latex]C_{1}[/latex]—capacitance of the first capacitor

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

R—resistance of the resistor

[latex]E_{C2}[/latex]—voltage drop across the second capacitor

[latex]X_{C2}[/latex]—capacitive reactance of the second capacitor

[latex]C_{2}[/latex]—capacitance of the second capacitor

Accepted Answer

Because the reactive power of the two capacitors is known and the true power of the resistor is known, the apparent power can be found using the formula

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

In this circuit, [latex]VARs_{C}[/latex] is the sum of the VARs of the two capacitors. A power triangle for this circuit is shown in Figure 22–9.

[latex]VA = \sqrt{(5 W)^{2} + (3.75 VARs + 5.625 VARs)^{2}}[/latex]

[latex]VA = \sqrt{25 W^{2} + 87.891 VARs^{2}}[/latex]

[latex]VA = 10.625[/latex]

Now that the apparent power and applied voltage are known, the total circuit current can be calculated using the formula

[latex]I_{T} = \frac{VA}{E_{T}}[/latex]

[latex]I_{T} = \frac{10.625 VA}{42.5 V}[/latex]

[latex]I_{T} = 0.25 A[/latex]

In a series circuit, the current must be the same at any point in the circuit. Therefore, [latex]I_{C1}[/latex], [latex]I_{R}[/latex], and [latex]I_{C2}[/latex] will all have a value of 0.25 A.

The impedance of the circuit can now be calculated using the formula

[latex]Z = \frac{E_{T}}{I_{T}}[/latex]

[latex]Z = \frac{42.5 V}{0.25 A}[/latex]

[latex]Z = 170 \Omega[/latex]

The power factor is calculated using the formula

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

[latex]PF = \frac{5 W}{10.625 VA}[/latex]

[latex]PF = 0.4706, or 47.06 \%[/latex]

The cosine of angle theta is the power factor:

cos ∠θ = 0.4706
∠θ = 61.93º

A vector diagram is shown in Figure 22–10 illustrating the relationship of angle theta to the reactive power, true power, and apparent power.

Now that the current through each circuit element is known and the power of each element is known, the voltage drop across each element can be calculated:

[latex]E_{C1} = \frac{VARs_{C1}}{I_{C1}}[/latex]

[latex]E_{C1} = \frac{3.75 VARs}{0.25 A}[/latex]

[latex]E_{C1} = 15 V[/latex]

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

[latex]E_{R} = \frac{5 W}{0.25 A}[/latex]

[latex]E_{R} = 20 V[/latex]

[latex]E_{C2} = \frac{VARs_{C2}}{I_{C2}}[/latex]

[latex]E_{C2} = \frac{5.625 VARs}{0.25 A}[/latex]

[latex]E_{C2} = 22.5 V[/latex]

The capacitive reactance of the first capacitor is

[latex]X_{C1} = \frac{E_{C1}}{I_{C1}}[/latex]

[latex]X_{C1} = \frac{15 V}{0.25 A}[/latex]

[latex]X_{C1} = 60 \Omega[/latex]

The capacitance of the first capacitor is

[latex]C_{1} = \frac{1}{2 \pi f X_{C1}}[/latex]

[latex]C_{1} = \frac{1}{377 imes 30 \Omega}[/latex]

[latex]C_{1} = 0.0000442 F, or 44.2 \mu F[/latex]

The resistance of the resistor is

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

[latex]R = \frac{20 V}{0.25 A}[/latex]

[latex]R = 80 \Omega[/latex]

The capacitive reactance of the second capacitor is

[latex]X_{C2} = \frac{E_{C2}}{I_{C2}}[/latex]

[latex]X_{C2} = \frac{22.5 V}{0.25 A}[/latex]

[latex]X_{C2} = 90 \Omega[/latex]

The capacitance of the second capacitor is

[latex]C_{2} = \frac{1}{2 \pi f X_{C2}}[/latex]

[latex]C_{2} = \frac{1}{377 imes 90 \Omega}[/latex]

[latex]C_{2} = 0.0000295 F, or 29.5 \mu F[/latex]

The completed circuit with all values is shown in Figure 22–11.

Question 22.1: One resistor and two capacitors are connected in series to a...

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