
#1
Apr2013, 10:08 AM

P: 19

There are a few sentences in by book that are confusing me and I was wondering if someone can clarify what they are trying to say.
"The current in the circuit is given by emf=V=IR. This relationship is saying that as a charge moves through the circuit, the potential increase in the emf is equal to the potential drop V as that charge moves through the resistor" I thought that emf is similar to voltage, what is this "increase in the emf equal to voltage drop" they are talking about? Thanks so much 



#2
Apr2013, 12:24 PM

Sci Advisor
PF Gold
P: 11,392

Start with the Voltage (Potential Difference). It is the equivalent of the Potential Energy of a mass above the surface of the earth (which needed to be raised at some point in the past). If one Coulomb of charge 'falls' through a Volt of Potential, then one Joule of work is done. A battery needs to provide one Joule to raise its potential in order for this to happen. The emf is the term for the 'driving' Volts that the battery provides.
So the PD makes things happen to the Charge. Rate of charge flow is called Current. A resistance of R will allow a current to flow according to the formula I = V/R. That equation can be rearranged to work out any of the three if you are given the other two. Resistance is just the Ratio of V and I and it is best not to try to use 'descriptive' words to 'explain any more about what Resistance 'is'. Kirchoff's Second law puts things this way. It says that the sum of the emfs in a circuit (the total energy supplied by batteries or generators) is equal to the energy that is dissipated by the other components in the circuit so the emf from the battery will be equal to the IR of (the voltage drop across) the resistor. I think that is what your book is getting at. 


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