[Guest]

A standard AA torch cell (it is technically not a "battery") has a voltage of 1.5 V. I don't know, even after Googling, what its internal resistance is, but assume for this question that it is 1 Ohm.

Twelve such cells are connected in series so as to form a closed loop. If one connects a voltmeter across any three successive cells, what reading would the meter show?

[bonanova]

0.

Arrange the cells on a clock face.

We are asked the voltage difference between Noon and 3:00.

By symmetry the same voltage appears between 3:00 and 6:00,

between 6:00 and 9:00 and between 9:00 and Noon.

Call that voltage V.

By Kirchoff's law the sum of the voltages around a closed loop is zero.

4V = 0.

V = 0.

The 12 cells in series produce a potential of 18 v.

The resistance around the loop is 12 Ohms.

The current in the loop is thus 18v/12 Ohms = 1.5 amps.

That current flowing through a 1-Ohm resistance produces a voltage drop of 1.5 v.

The sum of each cell's potential and voltage drop is thus 0.

The potential at all 12 points in the loop is the same.

You need new batteries.

Nice puzzle.

[Guest]

Believe it or not, as long as the internal resistances are approximately the same in each battery, in the closed circuit like this, the potential difference across any number of the batteries will be 0 volts.

However, if the circuit isn't closed, the potential difference will be the sum of the voltages of the number of batteries you cross.

(By the way, the typical internal resistance of a 1.5 volt battery is 150 to 300 milliohms...decreasing as temperature increases, and generally rising during discharge).

[Guest]

Well, you two demolished that one pretty smartish. Let's hope my next one is slightly more difficult!