# Can someone clarify this for me re:electric circuit

1. Feb 9, 2016

### Seekinfo

Hello
I have never had a good grasp of electricity/circuits but it's never late to learn so
I hope you all can help me so I can help my son.
If I connect a battery to a resistor and form a circuit, I know the "voltage drop"
current and resistance relate by V=IR. Now, my understanding is that in this circuit, the whole source voltage would be dropped across the resistor, so... what pushes the electrons (once past the resistor) back to the low zero potential terminal??

Thanks for your help, I appreciate it

2. Feb 9, 2016

### cnh1995

Ideally speaking, the wires have zero resistance. So, voltage across the wires is zero. But practically, there's some resistance(but very small compared to the resistors), hence, there is a very small voltage across the wires, which is approximated to 0.

3. Feb 9, 2016

### Seekinfo

Thank you sir. That makes sense to me. Thank you

4. Feb 9, 2016

### cnh1995

You are welcome! But I'm not a 'sir'. I'm of your son's age(maybe a couple of years older..).

5. Feb 9, 2016

### Seekinfo

great. I'm glad you helped us out

6. Feb 9, 2016

### sophiecentaur

You are clearly worthy of the title 'sir'. Don't denigrate yourself.
Agewise, I am a 'sir' for most of the people I meet but many of them still call me a dozy old twozzer.

7. Feb 9, 2016

### cnh1995

Thank you! That is very kind of you!

8. Feb 12, 2016

### mpresic

cnh1995 points out the there is little voltage drop across the wires (ideally). This is true. Suppose you take a voltmeter and measure the potential difference from a point right after the battery, to the negative terminal (usually the flat surface without the nub in the battery). The voltmeter measures zero potential difference, so it is natural to ask, "what drives the current back to the negative terminal?".(You call it the zero potential terminal).

But if you take the voltmeter and measure the potential difference from a point right after the positive terminal, and a point right before the resistor, you will also get a potential difference of zero. You could just as easily ask, "what drives the current towards the resistor in the first place?"
The answer is that a voltage difference drives the current. A voltmeter measuring the potential difference across the terminals of the battery might be for instance 1.5 V. As cnh correctly points out the wires are good conductors so they offer almost no voltage drop. The voltage drop across the resistor is approximately the voltage drop across the battery terminals. This potential difference drives current through the resistor.

The chemistry of the battery demonstrates two terminals with unequal affinity for electrons. (My apology to the chemists, the rest may be half-truths, it has been over 40 years since I took chemistry.) The positive terminal may be +1.5 V with respect to the negative terminal at 0 V. This means electrons will want to flow to the positive terminal. (Unlike charges attract). Ultimately, it is this unequal affinity that drives the current in the circuit. Eventually the battery does wear out, and the two terminals approach an equilibrium potential with 0 potential difference, and the battery is disposed of, or recharged or whatever