Answer-
The total capacitive reactance of the circuit is 442.10 Ω.
Solution-
Three capacitors, a 12 mF, a 20 mF, and a 30 mF, are connected in series to a 60 Hz source.
The effective/equivalent capacitance of the circuit is,
[tex]\Rightarrow \dfrac{1}{C}=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\\\\\Rightarrow \dfrac{1}{C}=\dfrac{5+3+2}{60}=\dfrac{10}{60}=\dfrac{1}{6}\\\\\Rightarrow C =6\ mF[/tex]
When capacitors are joined in series, the total value of capacitance in the circuit is equal to the reciprocal of the sum of the reciprocals of capacitance of all the individual capacitors.
We know that,
[tex]X_c=\dfrac{1}{2\pi fC}[/tex]
Where,
[tex]X_c[/tex] = Capacitive Reactance (in Ω)
f = frequency (in Hz)
C = Capacitance (in F)
Here given,
[tex]X_c[/tex] = ??
f = 60 Hz
C = 6 mF = [tex]6\times 10^{-6}[/tex] F
Putting the values in the formula,
[tex]X_c=\dfrac{1}{2\pi \times 60\times 6\times 10^{-6}}[/tex]
[tex]X_c=\dfrac{10^{6}}{720\pi}[/tex]
[tex]X_c=442.10\ \Omega[/tex]
Therefore, the total capacitive reactance of the circuit is 442.10 Ω.
