# Measure voltage when battery is connected to lead without resistance

When we use voltmeter without resistance,(parallel connection) I thought it might have potential difference between two points, because of this equation V=Ed I thought There should be small difference 'd' between two points. So i connected what i said and I was embarrased to see "0" on my voltmeter. So i wonder why this happen although V=Ed says there might be potential difference between two points. I'm not a English user, so I'm sorry for my bad English. But I hope someone could help me get out of this chaos.

• Some more questions, Q1. IS it possible a flow of electricity without an electric field? Q2. Moving of particle with electric charge in electric field makes potential difference, isn't it? Then, let's suppose no resistance in wire. IF Q1 is true, wire with a flow of electricity makes an electric field, and in the electric field an electron moves. Then potential difference must exist. But when measured, there is no potential difference. – Yelim Aug 20 '17 at 6:37

simulate this circuit – Schematic created using CircuitLab

Figure 1. Battery equevalent Circuit with load resistor.

• Every battery has some internal resistance. This is shown in Figure 1.
• You have created a voltage divider by placing your "resistor" across the battery terminals.
• With the values I have shown there will be about 1% of battery voltage on your test wire. For a 1.5 V cell that will be 15 mV.
• If you measure at both ends of that wire you might see that much voltage - but it depends on the resistance of the wire.

If you can find a long length of resistance wire and repeat your experiment you won't be so embarrassed. If you don't have suitable wire then make a chain of resistors and measure various points along it.

• Thank you for answering the question:) But I can not ensure that the reason why the two points are at the same potential.And Why electric field doesn't exist between that two points? And to one more, if there potential difference exist in lead, then electric field is formed by voltage? – Yelim Aug 19 '17 at 13:08
• Add a diagram of your measurement wiring (into your question) and I will try to answer. – Transistor Aug 19 '17 at 13:14
• I added it. I know it's logically wrong(my battery was getting very hot) – Yelim Aug 19 '17 at 13:28
• According to your drawing, you have a wire directly connecting the positive and negative terminals of the battery - that is a short circuit, and will draw a very high current from the battery. As the wire has extremely low resistance, no voltage will be developed between any two points on it. – Peter Bennett Aug 19 '17 at 15:52
• @Yelim: See the update. – Transistor Aug 19 '17 at 17:30

You might read about battery ESR being some value but remember that is only for a specific range of battery currents. Batteries are highly non linear devices with highly time dependent parameters especially at higher discharge rates.

I have done a lot of work with batteries and "ESR" can be very different with state of charge, temperature and load current.

• You mean it depends on situation I set up. Thank you for answering:) – Yelim Aug 20 '17 at 4:11
• Then, will it be different if I use another power supplyer? I mean, Is it possible "0" voltage on wire in ideal world? (Not real world) – Yelim Aug 20 '17 at 4:15