# Basic questions about voltage drop in DC circuit

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.

• 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? Nov 12, 2012 at 16:22

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 $V_2=I R_2$.
• 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 $V_l=\frac{V}{R_1+R_2} (R_1+R_2)=V$.. Nov 12, 2012 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.