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Reading of a voltmeter and an ammeter

  • Thread starter Jahnavi
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  • #1
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Homework Statement



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Homework Equations




The Attempt at a Solution



Part a)

Voltmeter can be replaced by an open circuit and ammeter can be replaced by a short circuit .By doing so the current flows in the outer loop consisting of both the batteries and both the resisters .The current in the circuit is given by i = (45-15)/(20+10) = 1A .

Potential drop across Voltmeter is U2-iR2 = 45-(1)(20) = 25 V

So reading of ammeter is 1A and that of voltmeter is 25V .

Is that the correct answer for part a) ?

I am having some trouble with the use of "ideal voltmeter" .An ideal voltmeter has infinite resistance .It can be replaced by an open circuit . But if voltmeter has to read a potential difference it must draw some current .If no current flows through the voltmeter ,then it's reading would be simply zero .

Sorry for being a bit confused .
 

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  • #2
phinds
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Is that the correct answer for part a) ?
yes

I am having some trouble with the use of "ideal voltmeter" .An ideal voltmeter has infinite resistance .It can be replaced by an open circuit . But if voltmeter has to read a potential difference it must draw some current .If no current flows through the voltmeter ,then it's reading would be simply zero .
Your analysis is correct, BUT ... ideal volt meters are assumed to draw an infinitesimally small current that has zero effect on the circuit. It's a problem-simplification technique.
 
  • #3
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OK

And in part b) when the two meters are interchanged

No current flows through U1 and R1 , so entire terminal voltage of U1 appears across the voltmeter .It reads 15V .

Current flows in the loop consisting ammeter and U2 and R2 .Current i = 45/20 = 2.25A .

Ammeter reads 2.25A and voltmeter reads 15V .

I hope this part is also correct .
 
  • #4
cnh1995
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  • #5
CWatters
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Just for info... Some specialist Voltmeters have very high input resistance, as much as 10*109 Ohms on some ranges.
 

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