# capacitors in series, determine voltage.

#### rcmango

A 3.36uF and a 5.47uF capacitor are connected in series across a 25.0-V battery. A 7.43uF capacitor is then connected in parallel across the 3.36uF capacitor. Determine the voltage across the 7.43uF capacitor.

I struggled with this problem and tried to use an equilent capacitance formula: Q = CV

1/C = 1/3.36 + 1/5.47

#### clombard1973

Hello

A 3.36uF and a 5.47uF capacitor are connected in series across a 25.0-V battery. A 7.43uF capacitor is then connected in parallel across the 3.36uF capacitor. Determine the voltage across the 7.43uF capacitor.

I struggled with this problem and tried to use an equilent capacitance formula: Q = CV

1/C = 1/3.36 + 1/5.47

I would start by combining the two parallel capacitors getting 10.79uF. Capacitors in parallel sum like resistors in series. So you have a basic circuit with a DC supply of 25V and two capacitors which are 5.47uF and 10.79uF.

Now it is correct to use Q=VC, but in this case the charge on each capacitor must be the same. This is because when charge is displaced from one capacitor, the other being in series also displaces the same amount of charge. In other words, for every electron that is moved from one capacitor, one is also moved from the next one in series back to the supply.

So let the 10.79uF capacitor combination be C2 and the 5.47uF capacitor be C1. The voltage across the C2 combination is V2 and across C1 is V1. The 7.43uF and 3.36uF are in parallel and so will have the same voltage drop across them. Therefore knowing the charge displaced by one is the same as that displaced on the other you can write:

V1C1 = V2C2

V1 = (C2/C1)V2 = (10.79uF/5.47uF)V2

You know also that V1 + V2 = 25, so substituting you get:

(10.79uF/5.47uF)V2 + V2 = 25

V2[1 + (10.79uF/5.47uF)] = 25

So,

V2 = 25/[1 + (10.79uF/5.47uF)]

V2 = 8.41 V

Therefore V1 is 25 - 8.41 = 16.59 V

Since the 7.43uF has V2 across it the answer is 8.41 V.

The combination of 7.43uF and 3.36uF yielding 10.79uF could actually store more charge than it is doing. It has a greater "capacity" than the 5.47uF. The 5.47uF is smaller though and so it only allows what charge is displaced from it to also be displaced from the 10.79uF. Less charge on a greater capacitance means less voltage is dropped. (Q=VC, so V = Q/C)

Many Smiles,
Craig

rcmango

#### rcmango

Thankyou for your great deal of help with this one, I was really stumped for this problem because of the parallel and series circuitry combined. Thanks for your help its much, much more clear now!