## Capacitors in parallel question

If i take a 2microF capacitor charged to a p.d. of 50V, and a 3microF capacitor charged to 100V, and connect them in parallel with their positive plates connected,

a) what is the equivalent capacitance of this combintation?
b) what is the p.d. across each capacitor after they are connected?
c) what is the total energy stored by the capacitors?
d) why is there a loss of stored energy when the capacitors are connected (as opposed to when they are separate)?

Part a is pretty simple (i think!). you just add the capacitance of each capacitor to get 5microF, right?

Part b confuses me. I know that when capacitors are in parallel, the voltage is the same on each one, but I don't know how to find that voltage! Is it just 50V + 100V = 150V? Or is it the average of the two = 75V? Or something else?

For part C I guess i need to use the equation Energy stored = (1/2)CV^2, but for that I need to know the p.d. across the capacitors!

Part d: is it because energy is lost in the wires that connect the capacitors?

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 I'm a little rusty so someone please correct me if I'm wrong, but: A) Yes, 5mF. B) When you close the circuit the 100V capacitor will discharge into the 50V capacitor until the voltages are equalized. C) Add the two together, so: (0.000002 x 50) + (0.000003 x 100) = 0.0004 Joules. This means that the end result is a charge of (0.0004 / 0.000005) = 80 Volts. D) There is a loss of energy from the 100V capacitor as it discharges into the 50V capacitor. Assuming both capacitors are rated at least 80V, the total energy will only drop slightly as a result of heat generated during this transfer of energy.
 Pergatory, You almost got it. But even if there were no heat losses during the transfer the final configuration would be storing less energy than the original one. 5uF at 80 volts is less energy than 2uF at 50V plus 3uF at 100V.

## Capacitors in parallel question

Assuming no heat loss, we begin with 0.0004 Joules, and end with 0.0004 Joules, right? What energy is lost?