Answer :
The equivalent capacitance of the network between points a and b is approximately 2.057 μF.
Let's calculate the equivalent capacitance again, considering the correct series capacitance rule:
Total capacitance in series is given by,
[tex]\frac{1}{C_{eqv}} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +...[/tex]
Total capacitance in parallel is given by,
[tex]C_p = C_1 + C_2 + C_3 +...[/tex]
Given capacitance values:
C1 = C5 = 8.4 μF
C2 = C3 = C4 = 5.0 μF
In this given question,
Capacitors C3 and C4 are in series.
[tex]C_{series} = 1 / (1/C3 + 1/C4)[/tex]
[tex]C_{series} = 1 / (1/5.0 + 1/5.0 )[/tex] = 2.5 μF
The series combination ([tex]C_{series}[/tex]) is in parallel with C2.
[tex]C_{parallel} = C_{series} + C2[/tex]
[tex]C_{parallel} = 2.5 + 5.0[/tex] = 7.5 μF
Step 3: The parallel combination ([tex]C_{parallel}[/tex]) is in series with capacitors C1 and C5.
[tex]C_{series_2} = 1 / (1/C_{parallel} + 1/C1 + 1/C5)[/tex]
[tex]C_{series_2} = 1 / (1/7.5 + 1/8.4 + 1/8.4 )[/tex] ≈ 2.057 μF
The equivalent capacitance of the network is approximately 2.057 μF.
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The question is-
In (Figure 1- attached), C1 = C5 = 8.4 MF and C2 =C3 = C4 = 5.0 MF. The applied potential is Vab = 220 V. Part A What is the equivalent capacitance of the network between points a and b?
