# How can the charge of capacitors in parallel + series exceed the total charge?

I'm solving this question:

From this video

We can simplify the circuit to be a single capacitor with capacitance of 50 micro farads.

Using Q = CV, we can determine the charge on each capacitor in series is 4800 micro-coloumbs.

So far so good.

The confusing part:

The 1st capacitor has a charge of 4800 micro coloumbs, as expected

The 2nd "capacitor" is actually a parallel circuit. We calculate the charge of C2 (70 microfarads) using (7/10 * 4800) = 3360 microfarads.

This means the charge of the other path (C3 + C4 in series) is 4800 - 3360 = 1440.

Still, so good.

However, since C3 & C4 are in series, they both must have the same charge. So they both have 1440 micro-coloumbs!

This means the total charge on the circuit composed of (C2,C3,C4) ... adds up to 3360 + (1440 * 2) = 6240, which is greater 4800.

My understanding of what is going is: C3 + C4 should be treated as 1 big capacitor, so the 1440 charge is on both of them at the same time rather than 2 instances of 1440 micro-coulombs on each individual capacitor.

Does this make sense?

• If your car was travelling at 50 mph, because you have 4 wheels would you conclude that the total speed is 200 mph? Nov 2 at 9:16
• Answer of GT Electronics is correct. You need to treat parallel capacitors as series resistors and series capacitors as parallel resistors mathematically (sum vs inverted sum of inverses)
– Ilya
Nov 2 at 13:47