Capacitance in Series: Find Charge, Voltage, Energy

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To find the charge, voltage, and energy stored in two capacitors in series with capacitances of 1.6 nanofarads and 5 nanofarads connected to a 24-volt supply, first calculate the total capacitance using the formula 1/Ct = 1/C1 + 1/C2. The total charge can be determined with Q = Ct * V, where V is the supply voltage. Each capacitor stores the same charge, allowing the voltage across each to be calculated using V = Q / C. The individual voltages should sum to the total supply voltage of 24 volts. This method confirms the calculations are accurate and consistent with the expected results.
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1. ok, there are 2 capacitors in series in a DC network, with A having a capacitance of 1.6 nanofarads and B having 5 nanofarads capacitance, but how do i find the charge, voltage and energy stored in each if the supply voltage is 24 volts. if anyone could help within the day that would be great. thanks! :biggrin:
 
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just to say that I am asking u guys just to check the figures because I've got a set of answers here.
 
what is your approach on the problem? Where are you struck up? Could you show the steps you followed and relevant equations used.
 
Here is an approach to solving this type of problem.

First, get the total capacitance of the two capacitors in series.
This is given by 1/Ct = 1/C1 + 1/C2

Now use Q = Ct * V to get the charge which charged the capacitors. V is the supply voltage.

The charge on each of the capacitors is the same as the total charge.

So, rearrange the formula now to get V = Q / C. (Q in Coulombs. C in Farads)

You know each of the capacitor values and you know the charge, so you can work out the voltages across each of the capacitors.
Notice that the voltages must add up to the supply voltage.
 
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yes just checked my answers with that method and it's worked out allright. thanks for that quick response!
 

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