How Does Potential Drop Equal Supplied Voltage in a Resistance-Free Circuit?

[danago]

Hey. I've been told that the total voltage supplied to an electrical series circuit will equal the sum of the potential drop across the circuit. I am a bit confused.

Lets say i have a 12V battery in a circuit, so each coulomb of charge obtains 12J of potential energy. What if there are no electrical devices? How can the potential drop equal the supplied potential difference?

 

[ultimateguy]

I'm not quite sure, but if there were no electrical devices in the circuit, wouldn't that mean that there is zero potential drop?

 

[StatMechGuy]

Well, yes. But then you don't really have a circuit since you need some sort of resistor to make your circuit non-trivial.

 

[berkeman]

A power source give you a voltage increase, and current flowing through loads gives you voltage drops. When you have a battery sitting there open circuit, there is no external current flow, so all you have is the battery voltage. When you connect it to an external circuit (like say two resistors in series), the sum of the voltage drops will equal the applied potential. So if you add up the voltage drops going around a complete loop, you get a negative voltage drop (voltage increase) at any sources, and positive drops at loads. The sum around the loop will equal zero.

 

[TheloniousMONK]

A power source give you a voltage increase, and current flowing through loads gives you voltage drops. When you have a battery sitting there open circuit, there is no external current flow, so all you have is the battery voltage. When you connect it to an external circuit (like say two resistors in series), the sum of the voltage drops will equal the applied potential. So if you add up the voltage drops going around a complete loop, you get a negative voltage drop (voltage increase) at any sources, and positive drops at loads. The sum around the loop will equal zero.

Yes, if you need further explanation look up Kirchoff's Voltage Law.

 

[danago]

Im still not understanding why though. If i use a 12V battery, it means there will be a total of 12V potential drop around the circuit, right? Let's say i hooked up each terminal of a battery with a resistance free wire. How does this law still apply?

 

[TheloniousMONK]

I do not know a lot of E&M theory as I am a humble EE, but I think if you look at it in terms of Ohm's Law it might make more sense mathematically.

Ohm's Law is I = V/R. Since R = 0 across the resistance-free wire, I = infinity. Thus, voltage across the circuit can mathematically be anything, even 12 V.

Someone else may come along and correct me.

 

[berkeman]

Im still not understanding why though. If i use a 12V battery, it means there will be a total of 12V potential drop around the circuit, right? Let's say i hooked up each terminal of a battery with a resistance free wire. How does this law still apply?

There's no such thing as a resistance free wire (except for a superconductor, but that's another subject).

Let's say you hook up an 18AWG wire across the battery, and the wire's resistance is a few milliOhms. A real battery (as with any real power source) has an internal resistance associated with its particular battery chemistry. So you model the battery as an ideal voltage source (zero output resistance) in series with the battery's output resistance. The short circuit draws a large current from the battery, and the source resistance of the battery limits the short circuit current.

Don't get caught up in the trap of thinking of real world things like batteries and wires in their simplified ideal model terms. Instead, look at real world things and understand how to model them in terms of combinations of ideal elements. That's what you do with SPICE simulations of real circuits, for example.