Question 21.5: Three capacitors having values of 50 μF, 75 μF, and 20 μF ar......

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

NOTE: Refer to the Pure Capacitive Circuits Formula section of Appendix D.

X_{C1}=\frac{1}{377×0.000050\ F}

X_{C1}=53.05\ Ω

X_{C2}=\frac{1}{377×0.000075\ F}

X_{C2}=35.367\ Ω

X_{C3}=\frac{1}{377×0.000020\ F}

X_{C3}=132.626\ Ω

The total capacitive reactance can be found in a manner similar to finding the resistance of parallel resistors:

X_{CT}=\frac {1}{\frac{1}{X_{C1}}+\frac{1}{X_{C2}}+\frac{1}{X_{C3}}}

X_{CT}=\frac {1}{\frac{1}{53.05\ Ω}+\frac{1}{35.367\ Ω}+\frac{1}{132.626\ Ω}}

X_{CT}=\frac {1}{0.05467\frac{1}{Ω}}

X_{CT}=18.292\ Ω

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

E_T=\sqrt{\text{VARs}_{CT}×X_{CT}}

E_T=\sqrt{787.08\ \text{VARs}×18.292\ Ω}

E_T=119.989\ V

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 amount of total current for the circuit and the amount of current in each branch can be found using Ohm’s law:

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

I_{CT}=\frac{119.989\ V}{18.292\ Ω}

I_{CT}=6.56\ A

I_{C1}=\frac{E_{C1}}{X_{C1}}

I_{C1}=\frac{119.989\ V}{53.05\ Ω}

I_{C1}=2.262\ A

I_{C2}=\frac{E_{C2}}{X_{C2}}

I_{C2}=\frac{119.989\ V}{35.367\ Ω}

I_{C2}=3.393\ A

I_{C3}=\frac{E_{C3}}{X_{C3}}

I_{C3}=\frac{119.989\ V}{132.626\ Ω}

I_{C3}=0.905\ A

The amount of reactive power for each capacitor can now be computed using Ohm’s law:

\text{VARs}_{C1}=E_{C1}×I_{C1}

\text{VARs}_{C1}=119.989\ V×2.262\ A

\text{VARs}_{C1}=271.415

\text{VARs}_{C2}=E_{C2}×I_{C2}

\text{VARs}_{C2}=119.989\ V×3.393\ A

\text{VARs}_{C2}=407.123

\text{VARs}_{C3}=E_{C3}×I_{C3}

\text{VARs}_{C3}=119.987\ V×0.905\ A

\text{VARs}_{C3}=108.590

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:

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

\text{VARs}_{CT}=271.415+407.123+108.590

\text{VARs}_{CT}=787.128

The slight difference in answers is caused by the rounding off of values. The circuit with all values is shown in Figure 21–15.