Voltage vs. EMF: What's the Difference?

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SUMMARY

The discussion clarifies the distinction between electromotive force (emf) and voltage, specifically in the context of batteries. An ideal battery maintains a constant emf, while real batteries experience voltage drops due to internal resistance. The term "ems" is identified as a Swedish equivalent of emf, and it is confirmed that in equations like P=VI, V represents the terminal voltage, which may differ from the emf due to internal resistance. Practical calculations involving current and resistance are demonstrated, emphasizing the importance of understanding internal resistance in circuit design.

PREREQUISITES
  • Understanding of electromotive force (emf) and voltage
  • Basic knowledge of electrical circuits and Ohm's Law
  • Familiarity with internal resistance in batteries
  • Ability to perform calculations involving current, voltage, and resistance
NEXT STEPS
  • Study the concept of internal resistance in batteries and its impact on performance
  • Learn about circuit analysis techniques using Kirchhoff's laws
  • Explore the differences between ideal and non-ideal components in electrical circuits
  • Investigate the applications of electromotive force in various electrical engineering scenarios
USEFUL FOR

Electrical engineering students, hobbyists working with batteries, and professionals designing circuits who need a clear understanding of voltage and emf concepts.

Drizzy
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Homework Statement



Can somebody explain what ems is for a person (me) who has only heard of voltage?

http://imgur.com/Y1JQy7w

Lets say the batteries volt is 10V. Is it always going to be 10? Is the batteries ems 10V?

Homework Equations

The Attempt at a Solution

 
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I think you mean "emf", or electromotive force.

An ideal battery will always present the same emf. Real life batteries depend upon chemical reactions and non-ideal materials that, over time and use, change the properties of the cell. Even if the emf provided by the underlying chemistry remain the same, the emf presented at the batteries terminals will degrade due to factors like internal resistances.
 
okay but am I ever going to use emf? For example in this formula: P=VI Is V the emf or is it the ems? basically is ems always ε or can the ems value be put in an equation that containc V?
 
Drizzy said:
okay but am I ever going to use emf? For example in this formula: P=VI Is V the emf or is it the ems? basically is ems always ε or can the ems value be put in an equation that containc V?
I am not familiar with the term "ems". Can you provide a definition or a source for where you found it?

A good approximate model for a real battery is an ideal cell with a particular emf (otherwise known as its voltage) in series with a small resistance that represents the battery's internal resistance -- the non-ideal manifestation of the real chemistry and materials that make up the battery. The analysis and design of practical circuits might involve this resistance.

This internal resistance may come into play when you need to determine how the voltage that appears at a battery's terminals changes with the electrical load on the battery (current drawn), or how much power is lost inside the battery (causing it to heat up!) for a given load. It also plays a role in charging a (rechargeable) battery, since it can limit the charging current for a given charger voltage.
 
It occurs to me that "ems" may be the same thing as "emf", but coming from a different language. Can you tell me the source language of the document where you found ems used?
 
It is in swedish :P I am wondering if the ems is used in P=UI and when do I need the ems value?
 
Drizzy said:
It is in swedish :P I am wondering if the ems is used in P=UI and when do I need the ems value?
Ah! Swedish. That would explain it then. So I then interpret ems to be the same as emf in English.

In your circuit, if the battery is said to have an ems of 10 Volts, and since no internal resistance is given for that battery then that ems of 10 V will be impressed across the load resistor R (5 Ohms). This results in a current ##I = ems/R = 10/5 = 2 A## to flow, as shown in your figure.
 
okay so let's say the internal resistance is 1 ohm then the volt is U= 10 - (1*2)= 10 - 2 = 8
so the voltage is 8 then?
 
Drizzy said:
okay so let's say the internal resistance is 1 ohm then the volt is U= 10 - (1*2)= 10 - 2 = 8
so the voltage is 8 then?
You would have to calculate the new current given the new total resistance; Adding resistance to the circuit will decrease the current.
 
  • #10
oh right I forgot!

the total resistance is 6 ohms
the voltage is 10 V
The current is 10/6

U = 10 - (1*10/6) = 10 - 10/6 = (60/6) - (10/6) = 50/6 = 8.3333333...

is this correct?
 
  • #11
Drizzy said:
oh right I forgot!

the total resistance is 6 ohms
the voltage is 10 V
The current is 10/6

U = 10 - (1*10/6) = 10 - 10/6 = (60/6) - (10/6) = 50/6 = 8.3333333...

is this correct?
Yes. You could also have found the potential drop across R using the current flowing through it: V = (10/6 A)(5 Ω)
 
  • #12
  • #13
Yes and we also write U=RI not V=RI
 

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