Question 21.1: One resistor and two capacitors are connected in series to a......

Since 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:

\mathrm{VA}=\sqrt{\mathrm{P}^2+\mathrm{VARs}_{\mathrm{C}}{ }^2}

In this circuit, \operatorname{VARs}_{\mathrm{C}} is the sum of the \text {VARs} of the two capacitors. A power triangle for this circuit is shown in Figure 21–9.

V A=\sqrt{(5 \mathrm{~W})^2+(3.75 \text { VARs }+5.625 \text { VARs })^2}

V A=\sqrt{25 ~W^2+87.891 \text { VARs }^2}

V A=10.625

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

\mathrm{I}_{\mathrm{T}}=\frac{\mathrm{VA}}{\mathrm{E}_{\mathrm{T}}}

\mathrm{I}_T=\frac{10.625~ \mathrm{VA}}{42.5 \mathrm{~V}}

\mathrm{I}_{\mathrm{T}}=0.25 \mathrm{~A}

In a series circuit, the current must be the same at any point in the circuit. Therefore, \mathrm{I}_{\mathrm{C} 1}, \mathrm{I}_{\mathrm{R}} \text {, and } \mathrm{I}_{\mathrm{C} 2} will all have a value of 0.25 A.

The impedance of the circuit can now be computed using the formula:

\mathrm{Z}=\frac{\mathrm{E}_{\mathrm{T}}}{\mathrm{I}_{\mathrm{T}}}

Z=\frac{42.5 \mathrm{~V}}{0.25 \mathrm{~A}}

Z=170 ~\Omega

The power factor is computed using the formula:

P F=\frac{P}{V A}

\mathrm{PF}=\frac{5 \mathrm{~W}}{10.625 ~\mathrm{VA}}

P F=0.4706, \text { or } 47.06 ~\%

The cosine of angle theta is the power factor.

\cos \angle \theta=0.4706

\angle \theta=61.93^{\circ}

A vector diagram is shown in Figure 21–10 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 computed:

\mathrm{E}_{\mathrm{C} 1}=\frac{\mathrm{VARs}_{\mathrm{C} 1}}{\mathrm{I}_{\mathrm{C} 1}}

\mathrm{E}_{\mathrm{C} 1}=\frac{3.75 \text { VARs }}{0.25 \mathrm{~A}}

\mathrm{E}_{\mathrm{C} 1}=15 \mathrm{~V}

E_R=\frac{P}{I_R}

E_R=\frac{5 \mathrm{~W}}{0.25 \mathrm{~A}}

\mathrm{E}_{\mathrm{R}}=20 \mathrm{~V}

\mathrm{E}_{\mathrm{C} 2}=\frac{\mathrm{VARs}_{\mathrm{C} 2}}{\mathrm{I}_{\mathrm{C} 2}}

\mathrm{E}_{\mathrm{C} 2}=\frac{5.625 \text { VARs }}{0.25 \mathrm{~A}}

\mathrm{E}_{\mathrm{C} 2}=22.5 \mathrm{~V}

The capacitive reactance of the first capacitor is:

\mathrm{X}_{\mathrm{C} 1}=\frac{\mathrm{E}_{\mathrm{C} 1}}{\mathrm{I}_{\mathrm{C} 1}}

\mathrm{X}_{\mathrm{C} 1}=\frac{15 \mathrm{~V}}{0.25 \mathrm{~A}}

\mathrm{X}_{\mathrm{C1}}=60 ~\Omega

The capacitance of the first capacitor is:

C_1=\frac{1}{2 \pi \mathrm{f} \mathrm{X}_{\mathrm{C} 1}}

C_1=\frac{1}{377 \times 60 ~\Omega}

C_1=0.0000442 ~F, \text { or } 44.2 ~\mu \mathrm{F}

The resistance of the resistor is:

R=\frac{E_R}{I_R}

\mathrm{R}=\frac{20 \mathrm{~V}}{0.25 \mathrm{~A}}

\mathrm{R}=80 ~\Omega

The capacitive reactance of the second capacitor is:

X_{\mathrm{C} 2}=\frac{\mathrm{E}_{\mathrm{C} 2}}{\mathrm{I}_{\mathrm{C} 2}}

X_{\mathrm{C} 2}=\frac{22.5 \mathrm{~V}}{0.25 \mathrm{~A}}

X_{C 2}=90 ~\Omega

The capacitance of the second capacitor is

C_2=\frac{1}{2 \pi \mathrm{fX}_{\mathrm{C} 2}}

C_2=\frac{1}{377 \times 90~ \Omega}

\mathrm{C}_2=0.0000295 \mathrm{~F} \text {, or } 29.5 ~\mu \mathrm{F}

The completed circuit with all values is shown in Figure 21–11.