Capacitor of 6 microf is conected in parallel with the combination of 4 microf capacitor and a 3 microf capacitor that are in series.Find the net...

Since the capacitors 3 microF and 4 microF are in series
will be replaced by a capacitor that has the capacitance
Cs.

1/Cs = 1/4 + 1/3

1/Cs =
(3+4)/4*3

1/Cs = 7/12

Cs =
12/7

Now, the capacitor that has the capacitance Cs = 12/7
and the capacitor of 6 microF are in parallel.

The
capacitors will be replaced by the capacitor that has the capacitance
Cp.

Cp = 12/7 + 6

Cp =
(12+7*6)/7

Cp = (12+42)/7

Cp =
54/7 microF

The net capacitance is Cp = 54/7
microF.

Since the capacitor of 6microF and
the capacitor of 12/7 microF are in parallel, the potential difference of 20 V across 6
microF is the same across 12/7 microF, too.

Now, we'll
determine the charge through 12/7 microF capacitor:

charge
= voltage*capacitance

charge = 20V*(12/7)
microF

charge = 240/7
microC

The charge on the capacitors of 4 and 3 microF will
be the same, since they are in
series.

Potential difference across 3 microF
capacitor is:

240/7*3 = 240/21
= 80/7 = 11.4 V