Question 21.4: Three capacitors with values of 10 μF, 30 μF, and 15 μF are ......

Since the frequency and the capacitance of each capacitor are known, the capacitive reactance for each capacitor can be found using the formula

X_C=\frac{1}{2πfC}

Recall that the value for C in the formula is in farads and the capacitors in this problem are rated in microfarads.

X_{C1}=\frac{1}{2πfC}

X_{C1}=\frac{1}{377×0.000010}

X_{C1}=265.252\ Ω

X_{C2}=\frac{1}{2πfC}

X_{C2}=\frac{1}{377×0.000030}

X_{C2}=88.417\ Ω

X_{C3}=\frac{1}{2πfC}

X_{C3}=\frac{1}{377×0.000015}

X_{C3}=176.835\ Ω

Since there is no phase angle shift between any of the three capacitive reactances, the total capacitive reactance is the sum of the three reactances (Figure 21–12):

X_{CT}=X_{C1}+X_{C2}+X_{C3}

X_{CT}=265.252\ Ω+88.417\ Ω+176.835\ Ω

X_{CT}=530.504\ Ω

The total capacitance of a series circuit can be computed in a manner similar to that used for computing parallel resistance. Refer to the Pure Capacitive Circuits Formula section of Appendix D. Total capacitance in this circuit is computed using the formula

C_T=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}}

C_T=\frac{1}{\frac{1}{10\ μF}+\frac{1}{30\ μF}+\frac{1}{15\ μF}}

C_T=\frac{1}{0.2\frac{1}{μF}}

C_T=5\ μF

The total current can be found by using the total capacitive reactance to substitute for R in an Ohm’s law formula:

I_T=\frac{E_{CT}}{X_{CT}}

I_T=\frac{480\ V}{530.504\ Ω}

I_T=0.905\ A

Since the current is the same at any point in a series circuit, the voltage drop across each capacitor can now be computed using the capacitive reactance of each capacitor and the current flowing through it.

E_{C1}=I_{C1}×X_{C1}

E_{C1}=0.905×265.25

E_{C1}=240.051\ V

E_{C2}=I_{C2}×X_{C2}

E_{C2}=0.905×88.417

E_{C2}=80.017\ V

E_{C3}=I_{C3}×X_{C3}

E_{C3}=0.905×176.83

E_{C3}=160.031\ V

Now that the voltage drops of the capacitors are known, the reactive power of each capacitor can be found.

\text{VARs}_{C1}=E_{C1}×I_{C1}

\text{VARs}_{C1}=240.051×0.905

\text{VARs}_{C1}=217.246

\text{VARs}_{C2}=E_{C2}×I_{C2}

\text{VARs}_{C2}=80.017×0.905

\text{VARs}_{C2}=72.415

\text{VARs}_{C3}=E_{C3}×I_{C3}

\text{VARs}_{C3}=160.031×0.905

\text{VARs}_{C3}=144.828

Power, whether true power, apparent power, or reactive will add in any type of circuit. The total reactive power in this circuit can be found by taking the sum of all the VARs for the capacitors or by using total values of voltage and current and Ohm’s law:

\text{VARs}_{CT}=\text{VARs}_{C1}+\text{VARs}_{C2}+\text{VARs}_{C3}

\text{VARs}_{CT}=217.248+72.415+144.828

\text{VARs}_{CT}=434.491

The circuit with all computed values is shown in Figure 21–13.