Does a Voltmeter Measure EMF Accurately in High Internal Resistance Scenarios?

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In a circuit with a cell that has internal resistance and a voltmeter with extremely high resistance, the voltmeter reading approximates the electromotive force (emf) of the cell. This is because the high resistance of the voltmeter results in negligible current, making the voltage drop across the internal resistance effectively zero. Consequently, the terminal voltage equals the emf, which the voltmeter measures. The reasoning is based on the relationship between emf, terminal voltage, and internal resistance. Thus, the conclusion that the voltmeter reading equals the emf is correct.
MBBphys
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Homework Statement


Say we have a circuit where a cell with internal resistance r is connected to a voltmeter with extremely high resistance. This means practically all of the p.d would be across the voltmeter as the ratio of its resistance to the internal resistance is, say, infinitely high.
So the voltmeter reading equals emf of cell?

Is this reasoning correct?
Thanks!

Homework Equations


(V1/V2=R1/R2)

The Attempt at a Solution


(n/a)
 
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MBBphys said:

Homework Statement


Say we have a circuit where a cell with internal resistance r is connected to a voltmeter with extremely high resistance. This means practically all of the p.d would be across the voltmeter as the ratio of its resistance to the internal resistance is, say, infinitely high.
So the voltmeter reading equals emf of cell?

Is this reasoning correct?
Thanks!

I can't tell from what you've written. Which part is the question, which part is the answer, and which part is the reasoning.
 
Mister T said:
I can't tell from what you've written. Which part is the question, which part is the answer, and which part is the reasoning.
Well, I am saying that, for the circuit described, the voltmeter reading will equal approximately the emf of the cell. Is this right? Thank you!
 
What is your reasoning for that answer?
 
Mister T said:
What is your reasoning for that answer?
Well, I thought that if the voltmeter reading was infinitely high:

We know:

emf = terminal voltage - (current * internal resistance)
Hence, if we increase the resistance of the load by putting a voltmeter with infinitely high resistance, current will be zero, so (current * internal resistance) will be zero, so the p.d. across the internal resistance would be zero, so the terminal voltage would equal the emf, and as the voltmeter reads the terminal voltage, the voltmeter reading will equal the emf.

Is that right? Thanks
 
Correct.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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