Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I understand all the concepts of what voltage is using all the analogies but some things related to the drop of voltage across a circuit confuses me.

  1. If I had a short circuit and attached a voltmeter I would get a potential difference reading of 0 volts. How is current then going through the wire if it is 'X' volts at any point in the wire?

  2. Let's say I have a DC 9V battery with a load in the middle. I read somewhere (probably mistaken here) that the voltage drop in this situation must be 9V or rather, that the sum of load resistances must be equal to voltage of the source. I mean, there is a variable amount of resistance to each load so at the base of the load I might have 9V and at the end of the load I might have 5V with a 4V potential difference.

share|cite|improve this question
Are you confused if there is a current flowing at all in the first case? I'm confused by your question. And your second question '9V battery with a load in the middle' - i.e. a circuit with only a load resistance and a battery? Or do I have the wrong image in my head? – Kitchi Nov 12 '12 at 16:22
up vote 0 down vote accepted

for the first question, you normally don't have a "real" short circuit, but a very low load (say a milliohm). In this case there is a very low (non measurable with normal instruments) difference of potential and this will cause the current to flow through your load. The amount of current flowing in the load will depend on the internal resistance of the generator. If the internal resistance is $R_i$ and the load resistance is $R_l$, with a open circuit potential difference of $V$, the current will be $I=\frac{V}{R_i + R_l}$. If you had a proper zero resistance load (a real short circuit), the difference of potential across the load would be exactly zero but, being the resistance also zero, you would not need a potential difference to support a current (think of electrons flowing frictionless in your load, so you don't need to provide energy to allow the electrons to win the friction, therefore current without difference of potential). In this case the current would be $I=\frac{V}{R_i}$. This should help understanding the second question as well. If your load is constituted by two resistors in series (say of resistances $R_1$ and $R_2$), the total current in the circuit would be $I=\frac{V}{R_i + R_1 + R_2}$, the difference of potential across the load would be $V_l=I (R_1+R_2)$ and the drops across the resistors would be $V_1=I R_1$ and $V_2=I R_2$.

share|cite|improve this answer
Thanks for the answer. In terms of the second part, I get that the difference of potential across the load would involve the sum of the resistances but don't people mention that the total voltage drop across all loads must equal the voltage of the source? If I have a 9V source with only one load causing a voltage drop of 4V, how is the above statement accurate? – Dohk Nov 12 '12 at 20:03
Yes, the difference of potential across the load is equal to the sum of the resistances (times the current, don't forget :-) IF you forget the internal resistance of the generator i.e if you put $R_i=0$ in the formulas above. In this case you will see that the current in the load will become $\tilde{I}=\frac{V}{R_1+R_2}$ and the final difference of potential across the load will become $V_l=\frac{V}{R_1+R_2} (R_1+R_2)=V$.. – Marco Aita Nov 12 '12 at 20:39

1) The steady-state solution of a short circuit is that all points on the circuit have the same electric potential, yes, but usually what happens is that one has two rails separated by a potential $V$ and a third wire is added that connects the two rails. There is a brief time when current suddenly flows through this wire, heating it up based on whatever small resistance it really does have, and causing all the mayhem we usually associate with shorts.

But that's all very quick, and again, as I said, the steady-state that is quickly settled into is one where the potential is (nearly) the same everywhere.

2) An ideal battery forces the potential difference between its terminals to be some fixed value. Varying the resistance changes the current but not the potential difference. For something where the potential difference changes, look into circuits made with ideal current sources--ones whose currents are modeled as having a fixed value.

share|cite|improve this answer

protected by Qmechanic Jan 5 '14 at 12:08

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.