Question 15.10: Parallel Capacitors Three capacitors having values of 50 μF,......
Parallel Capacitors
Three capacitors having values of 50 μF, 75 μF, and 20 μF are connected in parallel to a 60-Hz line. The circuit has a total reactive power of 787.08 VARs (Fig. 15-47). Find the following unknown values.
X _{ C 1} — capacitive reactance of the first capacitor
X _{ C 2} — capacitive reactance of the second capacitor
X _{ C 3} — capacitive reactance of the third capacitor
X_{CT} — total capacitive reactance for the circuit
E_{T} — total applied voltage
I_{C1} — current flow through the first capacitor
VARs_{C1} — reactive power of the first capacitor
I_{C2} — current flow through the second capacitor
VARs_{C2} — reactive power of the second capacitor
I_{C3} — current flow through the third capacitor
VARs_{C3} — reactive power of the third capacitor
Since the frequency of the circuit and the capacitance of each capacitor are known, the capacitive reactance of each capacitor can be computed using the formula
X_C=\frac{1}{2 \pi F C}
(Note: Refer to the Pure Capacitive Circuits Formula section of the appendix.)
\begin{aligned}& X _{ C 1}=\frac{1}{377 \times 0.000050} \\ \\& X _{ C 1}=53.05 ~\Omega \\ \\ & X _{ C 2}=\frac{1}{377 \times 0.000075} \\ \\& X _{ C 2}=35.367~ \Omega \\ \\& X _{ C 3}=\frac{1}{377 \times 0.000020} \\ \\ & X _{ C 3}=132.63~\Omega\end{aligned}
The total capacitive reactance can be found in a manner similar to finding the resistance of parallel resistors.
X_{C T}=\frac{1}{\frac{1}{X_{C 1}}~+~\frac{1}{X_{C 2}}~+~\frac{1}{X_{C 3}}}
X_{C T}=\frac{1}{\frac{1}{53.05}~+~\frac{1}{35.367}~+~\frac{1}{132.63}}
\begin{aligned}& X_{C T}=\frac{1}{0.05466} \\& X_{C T}=18.295~ \Omega\end{aligned}
Now that the total capacitive reactance of the circuit is known and the total reactive power is known, the voltage applied to the circuit can be found using the formula
\begin{aligned}& E _{ T }=\sqrt{ VARs _{ CT } \times X _{ CT }} \\& E _{ T }=\sqrt{787.08 \times 18.295} \\& E _{ T }=120~ V\end{aligned}
In a parallel circuit the voltage must be the same across each branch of the circuit. Therefore, 120 V is applied across each capacitor.
Now that the circuit voltage is known, the total current for the circuit and the current in each branch can be found using Ohm’s law.
\begin{aligned}& I_{C T}=\frac{E_{C T}}{X_{C T}} \\ \\& I _{ CT }=\frac{120}{18.295} \\ \\& I _{ CT }=6.559 ~A \\& I _{ C 1}=\frac{ E _{ C 1}}{ X _{ C 1}}\\ \\& I_{C 1}=\frac{120}{53.05} \\ \\& I _{ C 1}= 2 . 2 6 2~ \mathrm { A } \\ \\ & I _{ C 2}=\frac{ E _{ C 2}}{ X _{ C 2}}\\ \\& I_{C 2}=\frac{120}{35.367}\\ \\& I_{C2}=3.393 ~A\\ \\& I _{ C 3}=\frac{ E _{ C 3}}{ X _{ C 3}} \\ \\& I_{C 3}=\frac{120}{132.62}\\ \\& I _{ C3}=0.905~ A &\end{aligned}
The amount of reactive power for each capacitor can now be computed using Ohm’s law.
\begin{aligned}& \text { VARs }_{ C 1}= E _{ C1}\times I _{ C 1} \\& \text { VARs }_{ C 1}=120 \times2.262 \\& \text { VARs }_{ C 1}=271.442 \\& \text {VARs}_{ C 2}= E _{ C 2}\times I_{C2}\\&\text{VARs}_{C2}=120\times3.393\\&\text{VARs}_{C2}=407.159\\& \text {VARs }_{ C 3}= E _{ C 3}\times I _{ C 3}\\&\operatorname{VARs}_{ C 3}=120\times 0.905 \\& \operatorname{VARs}_{C3}=108.573\end{aligned}
To make a quick check of the circuit values, add the VARs for all the capacitors and see if they equal the total circuit VARs.
\begin{aligned}& \text { VARs }_{ CT }=\text {VARs }_{ C 1}+\text { VARs }_{ C 2}+\text { VARs }_{ C3} \\ \\& \text { VARs }_{ CT}=271.442+407.159+108.573 \\ \\& \text { VARs }_{ CT }=787.174\end{aligned}
The slight difference in answers is caused by rounding off values. The circuit with all values is shown in Figure 15-48.
