Answer-
The total capacitive reactance of the circuit is 7.96 Ω.
Solution-
Two capacitors, a 20 mF and a 30 mF, are connected in parallel to a 400 Hz source.
The effective/equivalent capacitance of the circuit is,
[tex]C=20+30=50\ mF[/tex]
As, when capacitors are connected together in parallel, the equivalent capacitance in the circuit is equal to the sum of capacitance of all the individual capacitor.
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 = 400 Hz
C = 50 mF = [tex]50\times 10^{-6}[/tex] F
Putting the values in the formula,
[tex]X_c=\dfrac{1}{2\pi \times 400\times 50\times 10^{-6}}[/tex]
[tex]X_c=\dfrac{10^{6}}{40000\pi}[/tex]
[tex]X_c=7.96\ \Omega[/tex]
Therefore, the total capacitive reactance of the circuit is 7.96 Ω.
The total capacities reactance of the circuit is 7.96 Ω.
Two capacitors, a 20 mF and a 30 mF, are connected in parallel to a 400 Hz source.
The effective/equivalent capacitance of the circuit is,
[tex]C=30+20=50mF[/tex]
when capacitors are connected together in parallel, the equivalent capacitance in the circuit is equal to the sum of capacitance of all the individual capacitor.
What is the formula for capacitive Reactance (in Ω)
[tex]x_c=\frac{1}{2\pi f C}[/tex]
Where,
[tex]X_c[/tex] = Capacitive Reactance (in Ω)
f = frequency (in Hz)
C = Capacitance (in F)
We have to find,
[tex]X_c= ??[/tex]
f = 400 Hz
C = 50 mF = F
Therefore,by using the given value in formula we get,
[tex]X_c=\frac{1}{2\pi *400*50*10^6}[/tex]
[tex]X_c=\frac{10^6}{4000\pi }[/tex]
[tex]X_c=7.6 \Ohm[/tex]
Therefore, the total capacitive reactance of the circuit is 7.96 Ω.
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