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2000 Q9


3 replies to this topic

#1 jambo david

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Posted 15 May 2006 - 02:19 PM

Right there's a ciruit with two resistors in parallel with an ammeter on one branch and a switch(S) on the other and then (not in parallel now) there's another resistor with a voltmeter over it.

THe question is

"in the following circuit, the supply has negligible internal resistance

switch S is now closed.

Which row in the table shows the effect on the ammeter and voltmeter readings?"

Sorry i can't use a diagram and can't explain it very well, what happens to the voltmeter and ammeter readings and why?

#2 SncZ

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Posted 15 May 2006 - 02:25 PM

if i have drawn it correctly.. which i probably havent ... i would say the ammeter reading will decrease as current splits up in parallel ( i.e is shared) and then for the voltage.. V= IR since I has decreased so will V

is that right unsure.gif ?

#3 AM4R

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Posted 15 May 2006 - 03:45 PM

The answers say the ammeter reading decreases and the voltmeter increases. I dont understand that blink.gif
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#4 Michael

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Posted 15 May 2006 - 05:50 PM

I assume closing the switch adds the other resistor in parallel. In effect this would decrease the resistance. Therefore the overall value of I=V/R would effectively increase. The voltmeter reading would decrease as V=IR would be smaller. I dont understand how he answer is the other way around, any way you could post the diagram?





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