Respuesta :
(a) Equivalent capacitance of network between points a and b is 2.3μF.
(b) Charge on each of the three capacitors nearest a and b is 920 μC.
A) Let's consider the right side that is farthest from the point 'a'.
Three C1 are connected in series
C= 1/C1 + 1/C1 +1/C1 = C1/3
C = 6.9/3 = 2.3μF
Now this C is connected in parallel with C2
So, C= 2.3 + 4.6 = 6.9μF
Again we get three capacitors of 6.9μF each, connected in series.
C = C1/3 = 2.3μF
It again combines with 4.6μF in parallel
C = 4.6 + 2.3 =6.9μF
Now, this final reduction is the same as that of the first.
In the end, we have 3 capacitors of 6.9μF each connected in series.
Equivalent Capacitance C(eq) = C1/3 = 6.9/3 = 2.3μF
= 2.3 × [tex]10^{-6}[/tex] F
B) C(eq) = 2.3 × [tex]10^{-6}[/tex] F
Vab = 400 V (Given)
Charge on each of the three capacitors nearest a and b (Q) =?
Q = C(eq) × Vab
= 2.3 × [tex]10^{-6}[/tex] × 400 = 9.2 × [tex]10^{-4}[/tex] C = 920μC
Hence, the equivalent capacitance of the network between points a and b is 2.3μF.
And Charge on each of the three capacitors nearest a and b is 920 μC.
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