Capacitors (7 of 11) in Series, Calculating the Charge Stored

How to calculate the charge on series capacitors.

Capacitors in series will store the same amount of charge on their plates regardless of their capacitance's. This is because the charge stored by a plate of any one capacitor must have come from the plate of its adjacent capacitor. For capacitors in series the voltage of the source will be divided across each of the individual capacitors. The voltage drop across each capacitor will be different and is dependent upon the values of the individual capacitances. For capacitors in series, the inverse of the total capacitance is equal to the sum of the inverses of each of the individual capacitances.

Capacitors in parallel will have the same voltage drop. This is because each capacitor is connected directly to the voltage supply. Capacitors in parallel will store different amounts of charge on their plates. The amount of charge stored is dependent upon the capacitance of the capacitor and the voltage of the source. For capacitors in parallel, the total capacitance is equal to the sum of all the individual capacitances.

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[michaeljones]

Thanks for this video! None of my textbooks explained why the charge on series capacitors is equal so I'm very glad I found this video.

[DEV.estaAqui_]

i just got a new way of viewing how charges are assigned on capacitor plates 😀 thank u soo much sir

[maryanthonettemoreno]

Hi i got failed in my physics subject this sem. But my prof gave me another change to take exam. This video has extremely helpful for me. I hope I passed my exam so i cant get a failed mark.

[starrysky]

Summary: *in series, the equivalent charge is equal to individual charges on capacitors, and all capacitors have same charge irrespective of their capacitances. Qeq=Q1=Q2=Q3.... *Calculate the equivalent capacitance by 1/Ceq=1/C1+1/C2+1/C3 ... *Multiply with potential difference (Q=C*V) and you've got the charge!

[a-yagomarianstefanie]

How about where the capacitance is 4 x 10^3, do I still have to solve for it (4000 F) or it remains like 4 x 10^3, and how will I multiply it from the voltage?
For example, Voltage=12 Capacitance=4 x 10^3
4 x 10^3 F x 6 V
4000 F x 12 V = 48000
C
OR
48 x 10^3 C?

[salvatorehayes]

So Say I have 6 Capacitors in series charging at 120amps. Does that mean 120amps goes through each capacitor? Or does 20amps go through each capacitor. It's 120amps correct?

[dondavir]

I like the way he says "capacitors".

[pranamshrivastava]

does the value of v(current) effects the capacitor on moving in cpacitance

[Muck-qyoo]

What about putting already charged caps in series?

[imahinasyon]

Thanks, and I could really live with that explanation but I just want to know the "Logic" behind it.
If we say that "Capacitance" is the measure of how much charge a capacitor can store, then a 5 Farad capacitor will store less than a 12 capacitor, and logically since they're connected together both of their capacitance should combine, like combining two barrels, a small one and a large one, lets assume that the small one has a capacity of 10 liters/Farads and the bigger one has 20 liters/Farads, so combined they can have 30 liters/Farads (I just placed the Farads just to make the analogy between water molecules and charged particles because both can be quantified). The equation is easy to understand but I just don't see any logic to it and I would love to be enlightened.

[Festus]

You never explained the "logic"as to how each capacitor has x-charge which is also equal to the total charge. This makes no sense. Why shouldn't they add up to total charge?

[owillypete]

Use a script, please. You repeat yourself many times and it makes it difficult to follow