Question

Find the total charge stored in the network of capacitors connected
between A and B as shown in figure.​

Answer 1

Answer:

Total charge simply goes to be 27 µC.

Explanation:

The effective capacitance between A and B is 9 μF.

Since, Q = C × V

So, total charge coming out of the source will be, 9 × 3 μC = 27 μC

So these 27 μC will be distributed among the capacitors.

Since, Q ∝ C , so, 27 μC will be distributed among the capacitors in proportional to their capacitance i.e., 6:12 :: 1:2 ratio.

So, 6μF capacitor gets $\frac{1}{3}$ × 27 μC = 9 μC

and, 12 μF capacitor gets $\frac{2}{3}$ × 27 μC = 18 μC