# Why does such a lamp and two battery circuit not work?

Assume I have a lamp and I connect one side to the + pole of one battery, and then the other side to the - pole of another battery (say two batteries of 1.5V). The remaining - and + poles are not connected.

So there is a potential difference between across the lamp, but why does it not turn on? It seems to contradict the basic law of electricity.

Note: I am a newbie in electronics/electricty.

• "I am a newbie in electronics/electricty." Kirchoff's Laws might be a good place to start. Mar 4, 2016 at 1:03

## 3 Answers

Why does such circuit not work?

It's not a circuit to begin with since there is no closed path around which charge can flow through the lamp.

Let's try a different route to see this result. To make your 'circuit' an actual circuit, place a resistor between the "remaining - and + poles" as so

simulate this circuit – Schematic created using CircuitLab

Using elementary circuit laws, we can find the potential difference across the lamp to be

$$V_{\mathrm{lamp1}} = (6\:\mathrm{V} + 6\:\mathrm{V})\cdot \frac{100}{R_1 + 100}$$

So, for example, if $R_1 = 0\:\Omega$ then

$$V_{\mathrm{lamp1}} = 12\:\mathrm{V} \cdot \frac{100}{0 + 100} = 12\:\mathrm{V}$$

But, as the resistance of $R_1$ is increased, the potential difference across the lamp must decrease.

For example, if $R_1 = 1\: \mathrm{M\Omega}$, then

$$V_{\mathrm{lamp1}} = 12\:\mathrm{V} \frac{100}{1,000,000 + 100} = 0.012 \: \mathrm{V}$$

Setting $R_1 = \infty$ is equivalent to specifying that the "- and + poles are not connected".

In that case the potential difference across the lamp is

$$V_{\mathrm{lamp1}} = 12\:\mathrm{V} \frac{100}{\infty + 100} = 0 \: \mathrm{V}$$

in contradiction to the claim "So there is a potential difference between across the lamp". In fact, there is no potential difference across the lamp and, as stated earlier, it is for the simple reason that there is no closed path around which charge can flow through the lamp.

• Since the impedance of your R1 can't be infinite (although it is very large), there must be some miniscule charge flowing in the circuit and, therefore, some potential across the lamp. Mar 4, 2016 at 17:33
• @EMFields, I have given your comment precisely the amount of consideration it is due. Mar 4, 2016 at 21:25
• Hmm... I wasn't aware that Argumentum ad hominem was acceptable as a substitute for facts around here, and your logic is weak in that you've chosen to comment on what you claim is unworthy of consideration. ;) Referring to your diagram, though, since BAT1 plus and BAT2 minus are connected through LAMP1, then there must exist a potential difference between BAT1 minus and BAT2 plus through which charge flows. Not very much, mind you, but since the impedance/resistance of the medium connecting BAT1 minus and BAT2 plus isn't infinite, then an infinitesimal charge must flow through LAMP1 Mar 4, 2016 at 21:55
• @EMFields, a statement of fact does not make an "Argumentum ad hominem". I stated a fact and that fact is precisely this: I do not value your comment (original or follow-up). Mar 4, 2016 at 22:21
• You didn't state that, you snidely alluded to it with a prissy little air of arrogance about you. More to the point, you preferred to attack me personally instead of pointing out why you disagree with my comment(s), which is precisely argumentum ad hominem. You've also chosen to escalate the argument because, in my opinion, you can't refute the fact that the impedance between BAT1 minus and BAT2 plus (R1 in your drawing) isn't infinite and, consequently, are trying to kill the messenger instead of merely owning up to your error. If you'd like the last word, take it; I'm done with this... Mar 5, 2016 at 0:19

The potential difference is only between the terminals of each battery.

There is no "global" potential difference.

This is why a circuit must be complete in some manner for current to flow.

Since the resistance of the lamp is vanishingly small compared to the resistance between the unconnected battery terminals, the potential difference across the lamp will be vanishingly small as well, with largely all of the drop occurring across the huge resistance of the medium separating the "unconnected " terminals.

That resistance will be so large that it'll limit the flow of charge through the circuit to a value so low that it couldn't possibly light the lamp.