Circuits & EMF: Understanding Internal Resistance

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Electromotive force (emf) is defined as the potential difference across a source when no current flows, indicating that load resistance is infinitely high. Internal resistance, found within a battery, opposes current flow and affects circuit performance. When current flows, the total emf equals the sum of the potential differences across both the load resistor and the internal resistance. Understanding these concepts is crucial for solving circuit problems effectively. The discussion concludes with encouragement to practice with exam questions for further clarity.
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Hello

Iam stuck on a topic (have been for a while !) about the electro motive force around a circuit (emf). Iam also confused about how internal resistance plays a role in this.
I do vaguely remember a teacher telling me that max current is at max internal resistance (or something around those lines).
I don't do advanced electronics and circuits so I don't think you'll hav to launch into long,advanced explanations :).
Sorry if this is a bit vague. If you're nto sure ill try and reword my problems :).thanks in advance
 
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Hi there Lil ze and welcome to PF,

As your inquiry is a little vague I don't really know how deep to go but I'll start with the basics.

Electromotive Force

The electromotive force (emf) of a source is defined as the potential difference across a source when no current flows. This means that the load resistance (not the internal resistance) is infinitly high (or at least very very high for a good approximation). It is also said the the emf is work done per unit charge.

Internal Resistance (r)

Inside a chemical cell (battery), the chemicals present a resistance to the flow of current, this is known as the internal resistance.

There are usually three things to remember when dealing with an internal resistance problem;

  • When no current flows in a circuit then emf = pd across source
  • When a current flows in a circuit there is a p.d. across the load resistor(R) (V_{R} = IR) and a p.d. across the internal resistance(r) (V_{r} = Ir).
  • From Kichhoff's second law - the current in any part of the circuit is the same, thus emf = sum of the p.d.'s - emf = IR + Ir
 
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thanks a lot ! that seems to hav cleared up some of my problems, ill do some exam Qs and see how it goes :D

thanks hootenanny
 
Lil ze said:
thanks a lot ! that seems to hav cleared up some of my problems, ill do some exam Qs and see how it goes :D

thanks hootenanny

My pleasure :smile: Don't hesitate to come back if you get stuck.
 
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