Question 15.9: Series Capacitors Three capacitors with values of 10 µF, 30 ......
Series Capacitors
Three capacitors with values of 10 µF, 30 µF, and 15 µF are connected in series to a 480-V, 60-Hz line (Fig. 15-44). Find the following circuit values.
X_{C 1} — capacitive reactance of the first capacitor
X_{C 2} — capacitive reactance of the second capacitor
X_{C3} — capacitive reactance of the third capacitor
X_{CT} — total capacitive reactance for the circuit
C_{T} — total capacitance for the circuit
I_{T} — total circuit current
E_{C1} — voltage drop across the first capacitor
VARs_{C1} — reactive power of the first capacitor
E_{C2} — voltage drop across the second capacitor
VARs_{C2} — reactive power of the second capacitor
E_{C3} — voltage drop across the third capacitor
VARS _{ C 3} — reactive power of the third capacitor
\text { VARs }_{ CT } — total reactive power for the circuit
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}{2 \pi F C}
Recall that the value for C in the formula is in farads, and the capacitors in this problem are rated in microfarads.
\begin{aligned}& X _{ C 1}=\frac{1}{2 \pi FC } \\ \\& X _{ C 1}=\frac{1}{377 \times 0.000010}\\ \\& X _{ C 1}=265.25~ \Omega\\ \\& X _{ C 2}=\frac{1}{2 \pi FC }\\ \\& X _{ C 2}=\frac{1}{377 \times 0.000030}\\ \\& X _{ C2}=88.417~ \Omega\\ \\& X _{ C 3}=\frac{1}{2 \pi FC }\end{aligned}
\begin{aligned}& X_{C 3}=\frac{1}{377 ~\times~ 0.000015} \\ \\& X_{C 3}=176.83~ \Omega\end{aligned}
Since there is no phase angle shift between any of the three capacitive reactances, the total capacitive reactance will be the sum of the three reactances (Fig. 15-45).
X_{C T}=X_{C 1}+X_{C 2}+X_{C 3}
X_{C T}=265.25+88.417+176.83
X_{C T}=530.497 ~\Omega
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 the appendix. Total capacitance in this circuit will be 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}~+~\frac{1}{30}~+~\frac{1}{15}}
C _{ T }=\frac{1}{0.2}
C_T=5~ \mu F
The total current can be found by substituting the total capacitive reactance for R in an Ohm’s law formula.
I _{ T }=\frac{ E _{ CT }}{ X _{ CT }}
I_T=\frac{480}{530.497}
I _{ T }=0.905 ~A
Because 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 _{ C 1}= I _{ C 1} \times X _{ C 1}
E_{C 1}=0.905 \times 265.25
E_{C 1}=240.051 ~V
\begin{aligned}& E _{ C 2}= I _{ C 2} \times X _{ C 2} \\& E _{ C 2}=0.905 \times 88.417 \\& E _{ C 2}=80.017~ V \\& E _{ C 3}= I _{ C 3} \times X _{ C 3} \\& E _{ C 3}=0.905 \times 176.83 \\& E _{ C 3}=160.031 ~V\end{aligned}
Now that the voltage drops of the capacitors are known, the reactive power of each capacitor can be found.
\begin{aligned}\operatorname{VARs}_{ C 1} & = E _{ C 1} \times I _{ C 1} \\\operatorname{VARs}_{ C 1} & =240.051 \times 0.905 \\\operatorname{VARs}_{ C 1} & =217.246 \\\operatorname{VARs}_{ C 2} & = E _{ C 2} \times I _{ C 2} \\\operatorname{VARs}_{ C 2} & =80.017 \times 0.905 \\\operatorname{VARs}_{ C 2} & =72.415 \\\operatorname{VARs}_{ C 3} & = E _{ C 3} \times I _{ C 3} \\\operatorname{VARs}_{ C 3} &=160.031 \times 0.905 \\\operatorname{VARs}_{ C 3} & =144.828\end{aligned}
Power, whether true, apparent, 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.
\begin{aligned}& \text { VARs }_{ CT }=\text {VARs }_{C 1}+\text { VARs }_{C 2}+\text { VARs }_{C 3} \\& \text { VARs }_{ CT }= 2 1 7 . 2 4 6 +72.415+144.828 \\& \text { VARs }_{ CT }= 4 3 4 . 4 8 9\end{aligned}
The circuit with all computed values is shown in Figure 15-46.
