Question

Two capacitors A and B are connected in

series across a 100V supply and it is
observed that the p.d. across them 60V and
40V respectively. A capacitor of 2uF
capacitance is now connected in parallel
with A and the p.d. across B rises to 90V.
Calculate the capacitance of A and B in
microfarads.​

Answer 1

`Capacitor A = 0.24\mu F\\Capacitor B = 0.16\mu F`

Step-by-step explanation:

I'll upload my work shortly as an attachment, but here is my process in words:

1. In our first situation we have two capacitors in parallel, which means the charge distribution on both of them is the same. With that, we can find a ratio between the values of Capacitors A and B.
2. In our second situation , we add a capacitor parallel to A (I called it C). Because A and C are in parallel, we know that they must have the same potential difference; which should come to be 10V since 90V of the total 100V is on B. Also, the equivalent charge distribution across A and C must be equal to that of the charge distribution at B, because A&C are in series with B. So I added the charges on A&C and set that equal to the charge on B.
3. Next, I used the ratio from the first situation to substitute Capacitor A out of the equation. This allows us to solve for B's capacitance. (Note: You could have also substituted B for A and solved for A first if you wanted to.)
4. Finally, I used B's capacitance to plug back into the ratio from the first situation to find A's capacitance. And they wanted the answer in micro-Farads, so I went ahead and converted each answer to micro.