Refer to the figure for the actual equivalent circuit of this set-up. This assuming that all 3 batteries have equal emfs and internal resistances.
Obviously current will flow even if there's is no load connected because of their voltage difference.
By KCL (or this one actually is more of nodal analysis),
(1.5+1.5-Va)/(R+R)=(Va-1.5)/R
(3-Va)/2R=(Va-1.5)/R
Va or Vab=2V (ideal value considering the above assumptions, in actual 1.5V
If you connect voltage sources with different values in parallel their voltage would always equalize and stabilize into a certain value, current will flow in reverse to the source which ever has the lower potential which could actually damage it. Now consider batteries with equal voltage in parallel (without load), since their voltages are equal there will be no current flowing, current would only flow if a load is connected. Now in actual a battery can only supply a limited amount of current, if there is more of them their current supplying capacity would add up (referring to parallel connection). This also answers your question why set-ups are usually symmetrical.
Hope I've enlighten you.
It would be better to say that adding batteries in parallel increases the maximum current possible.
In physics problems, we often use "ideal" voltage sources. Adding two ideal voltage sources in parallel doesn't change anything. Either one can always supply the necessary voltage, so two together produce no change.
In a real circuit, one battery may not be able to supply enough current to hold the voltage where it should be. In that case, adding multiple cells in parallel may be enough.
You would never connect 1 1.5V cell in parallel with a 3.0V battery of cells. That creates a circuit with no work to do. The 3.0V battery would drive current backward through the 1.5V cell, causing damage. Parallel batteries would always be the same voltage.
The circuit in your picture is consuming power from the 3.0V side and pushing reverse current through the 1.5V side. This will eventually run down the 3.0V side and may damage the 1.5V cell.
It is never a good idea to connect batteries in parallel as shown. You should connet two pairs of batteries in series and those pairs in parallel but through a switch which connects both pairs to the load at the same time. In this case the available current would be twice that of one cell and the voltage would be twice that of one cell.
when shut off neither pair would be powering anything.
I = E/R. Increasing the voltage will inrease the current in a circuit. Ohm's Law.
Two 1.5Vs in parallel is still 1.5V. It willl increase the amount of current that is available but it won't increase it if the circuit is the same as the series one, it will decrease if as the voltage remains at 1.5V.
So there are two 1.5v cells connected in series. I understand that connecting cells in series will result in no change in the current, but an increase in voltage equal to the voltage of one cell multiplied by the number of cells in series. So in this case 3v.
My confusion lies when you connect a single 1.5v cell in parallel to the above series group. I understand that connecting cells in parallel results in an increase in current, but not in voltage.
So does this mean the voltage measured will be 3v?
Also, as current developed doesn't increase when adding batteries in series with each other, does adding just one battery in parallel with the series really double the developed current across the flying leads? This seems to be what the rules suggest, but if so why in the real world do we only normally see symmetrical layouts in series parallel connections - 2 lots of 2 cells in series connected in parallel for example?
Thanks!
http://imageshack.com/a/img163/2107/u2ly.jpg
