Question 21.1: One resistor and two capacitors are connected in series to a......
One resistor and two capacitors are connected in series to a 42.5-V, 60-Hz line (Figure 21–8). 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:
V_{A}—apparent power
\mathsf{I}_{T}—total circuit current
Z—impedance of the circuit
PF—power factor
\angle \theta—angle theta
\mathsf{E_{C1}}—voltage drop across the first capacitor
\mathsf{X_{C1}}—capacitive reactance of the first capacitor
\mathsf{C_{1}}—capacitance of the first capacitor
\mathsf{E_{R}}—voltage drop across the resistor
R—resistance of the resistor
\mathsf{E_{C2}}—voltage drop across the second capacitor
\mathsf{X_{C2}}—capacitive reactance of the second capacitor
\mathsf{C_{2}}—capacitance of the second capacitor
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
{\mathsf{VA}}=\sqrt{\mathsf{P}^{2}+{\mathsf{VARs}_{C}^{2}}}
In this circuit, {VARs}_{C} is the sum of the VARs of the two capacitors. A power triangle for this circuit is shown in Figure 21–9.
\mathsf{VA}=\sqrt{( \mathsf{5\ W})^{2}+(3.75 \ {\mathrm{VARs}}+5.625\ {\mathrm{VARs}})^{2}}
\mathsf{VA}=\sqrt{25\ {\mathsf{ W}}^{2}+87.891\ {\mathsf{VARs}}^{2}}
VA = 10.625
Now that the apparent power and applied voltage are known, the total circuit current can be calculated using the formula
\mathsf{I_{T}}={\frac{\mathsf{VA}}{\mathsf{E_{T}}}}
\mathsf{I_{T}}={\frac{\mathsf{10.625\ VA}}{\mathsf{42.5\ V}}}
\mathsf{I_{T}} = 0.25 A
In a series circuit, the current must be the same at any point in the circuit. Therefore, I_{C1}, I_{R} , and I_{C2} will all have a value of 0.25 A.
The impedance of the circuit can now be calculated using the formula
\mathsf{Z}={\frac{\mathsf{E}_{\mathsf{T}}}{\mathsf{I_{T}}}}
\mathsf{Z}={\frac{\mathsf{42.5\ V}}{\mathsf{0.25\ A}}}
Z = 170 Ω
The power factor is calculated using the formula
{\mathsf{PF}}={\frac{\mathsf{P}}{\mathsf{{VA}}}}
{\mathsf{PF}}={\frac{\mathsf{5\ W}}{\mathsf{{10.625\ VA}}}}
PF = 0.4706, or 47.06%
The cosine of angle theta is the power factor:
\cos\angle \theta = 0.4706
\angle \theta = 61.93°
The vector diagram shown in Figure 21–10 illustrates 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:
\mathsf{E}_{\mathsf{C1}}={\frac{\mathsf{VARs_{C1}}}{{\mathsf{I_{C1}}}}}
\mathsf{E}_{\mathsf{C1}}={\frac{\mathsf{3.75\ VARs}}{{\mathsf{0.25\ A}}}}
\mathsf{E}_{\mathsf{C1}} = 15 V
\mathrm{E_{R}}={\frac{\mathsf{P}}{\mathsf{I_{R}}}}
\mathrm{E_{R}}={\frac{\mathsf{5\ W}}{\mathsf{0.25\ A}}}
\mathrm{E_{R}} = 20 V
\mathsf{ E}_{\mathsf{C2}}={\frac{\mathrm{VARs}_{\mathrm{C2}}}{\mathsf{I_{C2}}}}
\mathsf{ E}_{\mathsf{C2}}={\frac{\mathrm{5.625\ VARs}}{\mathsf{0.25\ A}}}
\mathsf{ E}_{\mathsf{C2}} = 22.5 V
The capacitive reactance of the first capacitor is
{\mathsf{X_{C1}}}={\frac{\mathsf{E}_{\mathsf{C1}}}{\mathsf{I}_{\mathsf{C1}}}}
{\mathsf{X_{C1}}}={\frac{\mathsf{15\ V}}{\mathsf{0.25\ A}}}
{\mathsf{X_{C1}}} = 60 Ω
The capacitance of the first capacitor is
{\mathsf{C}}_{1}=\frac{1}{2\pi {\mathsf{f X_{C1}}}}
{\mathsf{C}}_{1}=\frac{1}{377\times 30\ \Omega}
{\mathsf{C}}_{1} = 0.0000442 F, or 44.2 μF
The resistance of the resistor is
\mathsf{R}={\frac{\mathsf{E_{R}}}{\mathsf{I_{R}}}}
\mathsf{R}={\frac{\mathsf{20\ V}}{\mathsf{0.25\ A}}}
R = 80 Ω
The capacitive reactance of the second capacitor is
\mathsf{X_{c2}}={\frac{\mathsf{E}_{C2}}{\mathsf{I_{C2}}}}
\mathsf{X_{c2}}={\frac{\mathsf{22.5\ V}}{\mathsf{0.25\ A}}}
\mathsf{X_{c2}} = 90 Ω
The capacitance of the second capacitor is
{\mathsf{C}}_{2}=\frac{1}{2\pi {\mathsf{f X_{C2}}}}
{\mathsf{C}}_{2}=\frac{1}{377\times 90\ \Omega}
{\mathsf{C}}_{2} = 0.0000295 F, or 29.5 μF
The completed circuit with all values is shown in Figure 21–11
