Arrangement of circuit affecting resistance measured

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Homework Statement


Please refer to the images.

First image-question
Second image-my interpretation/working.
Third image-the answer given.


Homework Equations



V=IR

The Attempt at a Solution


In the second image

--------------------------------------------
I really don't understand what does the answer mean.
I assumed the first half of the answer is referring to circuit A, but still it doesn't seem to make sense.

Someone please explain to me why my working is wrong, thanks.
 

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Answers and Replies

  • #2
CWatters
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I don't follow the book answer either. Is there any other info in part (a) not shown in the image? eg anything that allows us to assume the voltmeter is ideal or at least can be assumed to be ideal.

Do you have two answers for Q (b)(I)(1) ?

Student A will calculate a value of about 3.87 Ohms if he assumes both meters are ideal. I don't think there is enough info for him to calculate any other answer unless it's in part (a).

Question (b)(iii). There are two possibilities....

If the volt meter is ideal then both students will measure roughly the same current. Student B will calculate a value of 6/1.5 = 4 Ohms.

If the voltmeter is not ideal then Student B will measure a lower current than Student A. This will lead him to calculate an even higher value for the resistance.

So regardless of the quality of the volt meter (ideal or not) student B will calculate a higher value for the resistance of the lamp.

I don't see how the statement "Potential Difference measured by B is lower than expected" can ever be true. He's measuring the highest voltage in the circuit (6V) so how can a higher voltage ever be "expected"?
 
  • #3
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Is there any other info in part (a) not shown in the image? eg anything that allows us to assume the voltmeter is ideal or at least can be assumed to be ideal.
Please see image attached.

Do you have two answers for Q (b)(I)(1) ?
Yes. 1. Ammeter has finite resistance. 2. Internal resistance of the battery.
 

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