Question

In the figure, each
capacitance C1 is 6.9
UF, and each
capacitance C2 is 4.6
uF. a) Calculate the
equivalent
capacitance of the
network between
points a and b. b)
Determine the charge
on each of the three
capacitors closest to
a and b when Vab =
420 V.​

Answer 1

Respuesta:

Explicación:

1) 3 on the right add up to 6.9/3 = 2.3

in parallel with C2, this is 4.6+2.3 = 6.9

again we have 3 x 6.9 in series, so equivalent is 2.3

in parallel with next C2 this is again 2.3

again we have 3 x 6.9 in series, so equivalent is 2.3

ceq = 2.3uF

2) from a/b we have 3 6.9uf in series, so the voltage is

equal on each, equal to 440/3 or 147 volts on each.

Q = CV or 6.9uf x 147 = 1012uC on C1/C2

C2 has same voltage, so Q = 4.6 x 147 = 676 uC

3) We have 147 volts on the leftmost C2. That divides equally into 3 parts, so 49 volts appear across right C2

and 1/3 of that or 16 volts appears across the far right C1