Does the voltage across a battery include its internal resistance?

AI Thread Summary
The discussion centers on whether the voltage across a battery includes its internal resistance, questioning if the equation should be 2 = E - iR or simply 2 = E. It is clarified that the internal resistance (r) affects the terminal voltage, meaning the actual voltage depends on the current drawn from the battery. The conversation also emphasizes the difference in measuring voltage and current, noting that voltmeters are connected in parallel and should have high resistance, while ammeters are connected in series with low resistance. Additionally, the concept of internal resistance is described as a useful construct for understanding battery performance and energy loss. The discussion concludes that the ideal battery model used for analysis is a simplified version of the more complex real battery.
laser
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Homework Statement
See description
Relevant Equations
V=iR
Screenshot_4.png


I am uncertain if this represents situation (a) or situation (b).

Screenshot_6.png


i.e. is 2 = E-iR or is 2 = E? Or is the question ambiguous?

Thanks!
 
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laser said:
Homework Statement: See description
Relevant Equations: V=iR

View attachment 344072

I am uncertain if this represents situation (a) or situation (b).

View attachment 344073

i.e. is 2 = E-iR or is 2 = E? Or is the question ambiguous?

Thanks!
Is the r in your diagrams a representation of the internal resistance? If so, it is (b).
The actual voltage on the terminals depends on the current, so it would not make sense to call it a 2V battery if it meant (a).
 
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haruspex said:
Is the r in your diagrams a representation of the internal resistance?
Yes
 
Just to add on to that: Battery voltage (or any voltage you measure really) is measured with a voltmeter connected to the terminals. The voltmeter has a large resistance as to not interfere too much with the measurement - an ideal voltmeter would have infinite resistance implying zero current through it.

Amperemeters are the opposite, they are connected in series in the circuit which you measure and need as low resistance as possible not to interfere with the circuit. An ideal amperemeter would have zero resistance.

Edit: Fixed typo.
 
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laser said:
Homework Statement: See description
Relevant Equations: V=iR

is 2 = E-iR or is 2 = E? Or is the question ambiguous?
How would you envisage measuring the value ξ in diagram b? You'd have to put the probes 'somewhere' in amongst the electrolyte and electrodes. emf is a value of potential difference when no current.is being taken. Internal resistance is just a mental construct which gives a value to the amount of energy dissipated inside the battery for each unit of charge that passes through. We don't know where the loss occurs and nor do we find that the value of r is the same for all currents. r is a handy value to tell us things like how 'good' a battery is and what sort of load it will support.
 
Orodruin said:
Amperemeters are the opposite, they are connected in parallel in the circuit
We measure current in series, voltage in parallel.
 
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DaveE said:
We measure current in series, voltage in parallel.
Yeah, I think that was just a typo in his reply, based on his "opposite" comment (which is correct). I've sent Oro a PM to clarify. :smile:
 
Yes. That’s a typo clearly. Fixing.
 
The "ideal" battery we usually choose to use for analysis is the Thevenin equivalent of the "real" battery which is in fact more complicated internally..
 
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