# Total energy in a system of capacitors

#### davidbenari

Can someone provide for me a proof that the total energy of my equivalent capacitor is equal to the sum of the energy in all capacitors? Or, if this is untrue, tell me why?

Is this applicable to current and resistors too?

Thanks a lot.

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#### phinds

Gold Member
How do you define energy in a capacitance? Can you do the math yourself to compare total energy on a simple set of series parallel capacitors and then do the same for the equivalent capacitance?

Since resistors don't store energy, I don't know what you might have in mind for "Is this applicable to current and resistors too?"

#### davidbenari

My problem was proving it for series capacitors. Parallel capacitors seem more straightforward, for two reasons, one is that they add $Ceq=C_1+C_2...+C_n$ the other is that they all have the same $V$.

Series capacitors don't always have the same $V$ and they add differently. So I can't do it at the moment. :(

What i meant with resistors was the power they consume. Is the power an equivalent resistor consumes equal to the sum of all the powers on all resistors?

#### phinds

Gold Member
What i meant with resistors was the power they consume. Is the power an equivalent resistor consumes equal to the sum of all the powers on all resistors?
So I ask for that as well, can you figure the power consumed in a resistive circuit and then do the same for the equivalent resistance?

What is it that prevents you from doing it for series capacitors?

#### willem2

My problem was proving it for series capacitors. Parallel capacitors seem more straightforward, for two reasons, one is that they add $Ceq=C_1+C_2...+C_n$ the other is that they all have the same $V$.

Series capacitors don't always have the same $V$ and they add differently. So I can't do it at the moment. :(

If both series capacitors start out uncharged, they will always have the same charge, because the current through them is always the same.
In this case their energy is:

$$\frac {1}{2} \frac {Q^2} {C_1} + \frac {1}{2} \frac {Q^2} {C_2}$$

The equivalent capacitance is

$$\frac {C_1 C_2} {C_1+C_2}$$

and the energy of the equivalent capacitance is

$$\frac {1}{2}\frac {Q^2} {C_{eq}} = \frac {1}{2} \frac {Q^2 (C_1 + C_2) } {C_1 C_2} =$$ wich is equal to the sum of the stored energy of C1 and C2 calculated above.

#### phinds

Gold Member

If both series capacitors start out uncharged, they will always have the same charge, because the current through them is always the same.
In this case their energy is:

$$\frac {1}{2} \frac {Q^2} {C_1} + \frac {1}{2} \frac {Q^2} {C_2}$$

The equivalent capacitance is

$$\frac {C_1 C_2} {C_1+C_2}$$

and the energy of the equivalent capacitance is

$$\frac {1}{2}\frac {Q^2} {C_{eq}} = \frac {1}{2} \frac {Q^2 (C_1 + C_2) } {C_1 C_2} =$$ wich is equal to the sum of the stored energy of C1 and C2 calculated above.
I would have appreciated it if you would have waited for him to answer my question before you spoon fed him the answer. My goal was to help him figure it out on his own.