Finding current in an electric circuit

  • Thread starter ImpCat
  • Start date
  • #1
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Homework Statement


A voltmeter of resistance 50kΩ is used to measure the electric potential difference in a circuit, as shown. The cell has an electromotive force (emf) of 5.0 V and negligible internal resistance.

upload_2016-12-1_8-38-49.png


Homework Equations


I=V/R
V=IR

The Attempt at a Solution


I=5/150000 I=3.33E-5
V=50000*3.33E-5=1.7 V

But the answer is 1 V. How?
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Answers and Replies

  • #2
Doc Al
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You did not account for the internal resistance of the voltmeter.
 
  • #3
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You did not account for the internal resistance of the voltmeter.

How do you account for the internal resistance of a voltmeter though? It's not provided. Plus isn't voltmeters purposefully created with high resistance so as to create as little change in the circuit as possible?
 
  • #4
NascentOxygen
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How do you account for the internal resistance of a voltmeter though? It's not provided. Plus isn't voltmeters purposefully created with high resistance so as to create as little change in the circuit as possible?
You're told it is 50kΩ. (That's separate to the resistor of the same value in the circuit.)

There may be reasons it can't always designed to be near "infinity".
 
  • #5
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You're told it is 50kΩ.

There may be reasons it can't always designed to be "infinity".

Ohhhh I see, I thought the 50kΩ is the resistance of the resistor! Maybe I should've read the question more clearly. :woot:
 
  • #6
NascentOxygen
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See my amended reply.
 

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