Voltage sources in series

[hobz]

When I connect two 1.5 V batteries in series, I get a potential difference of 3 V across the terminals.

|+|----
|1|    |
|-|    |
|     3 V
|+|    |
|2|    |
|-|----

(Showing battery 1 and 2 in series).

 

Can anyone explain why the voltage is doubled when voltage sources are connected in series?

Why, for instance, is current not flowing from + of battery 2 to - of battery 1 (conventional current flow)?

[big314mp]

Lets say we have an electron that starts at the positive end of battery 1. It gains 1.5V of energy (since the volt is energy/unit of electrons, this is a semi valid comparison) as it travels across battery 1. It travels across the wire between batteries 1 and 2. It gets to the positive terminal of battery 2, travels across battery 2, and picks up another 1.5V of energy. It now has gained 3V from where it started back on the positive side of battery 1. What this means is that there is a total of 3V across both batteries. The voltage across each individual battery is still 1.5V, but they "stack" on each other to produce higher voltages.

 

You can also see that current does flow from the negative end of 1 to the positive end of 2. Otherwise the circuit would be broken.

[Fuzzwood]

And the reason that 2 batteries connected in parallel give twice as much Amp, is that there are twice as much elektrons pumped into the circuit at a given time limit.

[hobz]

Good explanations, thanks!

 

I have not made my question clear enough though. The 3 V "wire", was actually not meant to be a wire, but a visual reference to the 3 V.

 

My question is: Why does current not flow (or does it?) from the +/- terminals when the first battery is put ontop of the second battery, without the wire.

[npts2020]

You need to have a circuit. An anode touching a diode is only half of the circuit.

[hobz]

I guess you're viewing the battery as a diode. But couldn't I then just flip the diode around?

 

I am basically viewing the battery as a charged (to 1.5 V) capacitor, and the series connection as two capacitors in series.

 

By connecting one capacitor to the other, redistribution of charge will occur, such that the plates that are connected obtain equal potential.

While this happens, current must flow in the capacitors until the charges are distributed to equilibrium. Is this not true for batteries also?

[big314mp]

Ah, now we have an interesting question

 

You have to remember that voltage is a relative thing. Something is +1.5V relative to something else. Let's say that the negative end of batt 2 is at 0V (just as a point of reference). This means that the positive end of batt 2 is at +1.5V relative to the negative end of batt 2, as this is the voltage of the cell. The positive end of batt 2 and the negative end of batt 1 are both at +1.5V, as they are in equilibrium with each other. The positive end of batt 1 is at +1.5V relative to the negative end of batt 1, which is at +1.5V relative to our reference at the negative end of batt 2. This means that the positive end of batt 1 is at a grand total of +3V relative to the reference at the negative end of batt 2.

[hobz]

I follow. But right when you connect the batteries a current must flow as a reaction to the equilibrium being created?

[big314mp]

Well, a current would flow if the two were at different potentials relative to some outside source.

 

Say we measure the negative end of batt 1 and find that it is +5V relative to my sink. Say that the positive end of batt 2 is measured at +3V to my sink. When the two are connected, a small current would flow that would balance them to roughly +4V relative to my sink.

 

The current would be very small. The amount of current is also related to the capacitance of the structure.

[hobz]

I agree.

 

The charge on the area of the terminals will probably be the only current stored in the system, and not that much charge is needed to create such small differences in voltage.

 

Thanks for replies!