Question

In Fig. P24.59, each capacitance C1 is 6.9 mF, and each capacitance C2 is 4.6 mF. (a) Compute the equivalent capacitance of the network between points a and b. (b) Compute the charge on each of the three capacitors nearest a and b when Vab

Answer 1

(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 × $10^{-6}$ F

B) C(eq) = 2.3 × $10^{-6}$ F

Vab = 400 V (Given)

Charge on each of the three capacitors nearest a and b (Q) =?

Q = C(eq) × Vab

   = 2.3 × $10^{-6}$ × 400 = 9.2 × $10^{-4}$ 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.

Learn more about Capacitance here brainly.com/question/13578522

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