Here is what I think, you guys tell me what you think.
Lets us think about this in the physics perspective.
- First law of thermodynamics
The law of conservation of energy:the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed (wikipedia)
- electric potential and electric potential energy
I'll go straight to the formulas
electric potential or V =-W/q = U/q is the work done per unit of charge or is the amount of electric potential energy per unit of charge
Since we already know the electric potential of the batteries
electric potential energy or U = qV, this will be the potential energy carried by an electron leaving the negative terminal
Now lets put three batteries in series say, A, B, C. So that the (-) of A connect the (+) of B, the (-) of B connects the (+) of C and (- )of C connects a resistor which connects the (+) of A. All of 1.5 V
When an electron with a potential energy of U pass through the resistor, part of its potential energy is transferred to the resistor, lets say the resistor dissipate the transferred energy as heat, the potential energy of the electron after passing the resistor must reduce (First law of thermodynamics).
But the question is, what is the value of the potential energy U of the electron passing the resistor?
Here my theory(so to speak)
A to B
When electrons leaves from the negative terminal of A to enter to the positive terminal of B, they have potential energy U of qV. This potential energy must be transferred to the battery B.
Now the potential energy in B is no longer U = qV, but U = qV + Ua
where Ua is the potential energy transferred by eletrons entering B.
Since Ua = qV, then
the potential energy in B increase to U = qV + qV = 2(qV) = q(2V)
B to C
When electrons leaves from the negative terminal of B to enter to the positive terminal of C, they have potential energy U of q(2V). This potential energy must be transferred to the battery C.
Now the potential energy in C is no longer U = qV, but U = qV + Ub
where Ub is the potential energy transferred by eletrons entering C.
Since Ub = q(2V), then
the potential energy in C increase to U = qV + q(2V) =q(3V)
C to A
When electrons leaves from the negative terminal of C to enter to the positive terminal of A, they have potential energy U of q(3V).
This is the potential energy carried by the electrons flowing through the circuit.
As V = U / q, you se that the voltage between C and A is
V = q(3V)/q = 3V = 3*1.5 ... the same as 1.5+1.5+1.5
Tell me what you think guys, I'm trying to understand this stuff too ... cheers