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Measuring emf confusion

Hi so I'm being told different things by different sources.

This voltmeter is recording the potential difference across the entire circuit yes? The work done moving the charge through the entire thing.

So in this circuit with the switch open, common sense and my revision guide tell me that the potential difference recorded by the voltmeter will be 0, because the current isn't flowing. Also I know this to be true because of the entire thing about doing a required practical on extrapolating data to find the emf as it can not be found directly.

However in this website:
http://spmphysics.onlinetuition.com.my/2013/08/measuring-emf-and-internal-resistance.html
and in past papers it tells me that when the switch is open the reading on the voltmeter is in fact equal to the emf. Yet when it's closed instead of reading what would've been the lost volts within the battery it begins reading the terminal p.d. instead?

Can someone please tell me what's going on here because it makes no sense.

Reply 1

phys981

With the switch open, it measures the emf of the cell.

With the switch closed, it measures the terminal potential difference (tpd), in other words the voltage delivered to the external circuit.

If you have any opportunity to do so, set it up in a lab to prove it to yourself.

(edited 6 years ago)

Reply 2

Radioactivedecay

Original post by Retsek

Hi so I'm being told different things by different sources.

Pd can be induced without any current flowing. However, because no current is flowing in this instance, no pd is dropped to the battery's internal resistnace, so the pd recorded is the emf.

Reply 3

DrPhysicsWatford

A voltmeter can do two things, it can measure the energy (per charge) transferred to a component as current flows, and measure the energy (per charge) transferred to the passing current by a cell, or would be transferred if a current was flowing.

The difficulty comes from the role of the internal resistance. When no current flows, there is no work done passing through it. When current flows then work is done and energy is lost and some of the emf from the cell is lost to this internal resistance.

Reply 4

Retsek

Original post by phys981

Original post by Radioactivedecay

Original post by DrPhysicsWatford

Okay so what's the point in the practical with this circuit where we decrease the current (by increasing the resistance) to find the y-axis intercept which is the the emf value. If we can just record the emf value directly?

Reply 5

Radioactivedecay

Original post by Retsek

I believe this experiment is for finding the internal resistance of the cell, not it's emf no?

Reply 6

Retsek

Original post by Radioactivedecay

I believe this experiment is for finding the internal resistance of the cell, not it's emf no?

You right, my guide tells me it's for finding emf an internal resistance, but it puts emf first as if it's the more important part and my brain completely forgot about the internal resistance. My teacher didn't explain it very well either, just gave us the sheet and told us to follow the instructions. :/

Thanks for your help.

Reply 7

phys981

The experiment (if it's the one I think you mean) tells you the internal resistance from the gradient and the EMF from the intercept on the voltage axis.

Original post by Retsek

Original post by Radioactivedecay

I believe this experiment is for finding the internal resistance of the cell, not it's emf no?

(edited 6 years ago)

Reply 8

Eimmanuel

Study Forum Helper

As a side note, we can also use potentiometer circuit to determine the emf of an unknown source and internal resistance of the unknown source. A potentiometer is a device used for comparing potential differences.

Sometimes, people would say potentiometer is a better way of finding the emf of the unknown source.
https://www.youtube.com/watch?v=JJB85Mortt0