Ohms Law Lab

  • #1
OHM's Law Lab

this lab involved a voltmeter and an ammeter...a voltage source was turned on and sent to a resistor, from the resistor there was a current probe and a voltage probe...the idea of the lab was to give 6 readings for a resistor, each reading would exceed by .5v intervals...during each interval, the number in the current probe was recorded, and the number in the voltage probe was recorded...through ohm's law i computed the resistance and graphed the results...now all of that was cake until one of the analysis questions which asks..."Describe the proper placement of the voltmeter in a circuit, why must it be in parallel with the resistance. the 2nd part of the question reads, describe the proper placement of the ammeter in a circuit, why must it be in series with the resistance? any help on the question would be extremely appreciated

if you need me to clarify on anything, just let me know...i appreciate any responses
 

Answers and Replies

  • #2
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When a voltmeter is use, it's necessary to take into consideration its internal resistance. In general, this resistance is enormous in comparison with the emf's internal resistance and the resistance that it's been arrange with.
The ideal voltage under these conditions should be: V=(emf)R/(R+r); but when the voltmeter is used: V=(emf)(RR(v)/(R+R(v))/(r+(RR(v)/(R+R(v)).
After doing some math the relative error is:
E(%)=(1-R(v)(R+r)/(R(v)(R+r)+Rr))x 100.
Where R(v) is the voltmeter's resistance, R is the resistance in parallel, r the emf's resistance. If R(v)-> intinity then E(%)= 0: but if R(v)->0 then E(%)~100%.
The point of having a voltmeter with a big resistance is due to protection for the device and also stoping a big current going through; so when is connected in parallel most part of the current is going to pass through the resistance parallel to it.
 
  • #3
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I hope this helps.
 
  • #4
Integral
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Voltage is a potential difference. Difference is a key word, you can only measure the change in voltage between two point, therefore 2 points are required to connect a voltmeter.

Current is the FLOW of electrons therefore you must channel the flow though the ammeter. Since you need ALL of the current flow through the meter you need to insert the meter at a single point. Note that the circiut must be broken and meter inserted, essentially replaceing a wire in the circiut.

For a Ammeter low resistence is necessary to minimize effects on the circiut being measured.
 
  • #6
i mean what is the emf in this experiment?
 
  • #7
977
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It's the voltage of the source, maybe you know it as epsilon: [tex]\epsilon [/tex].
 
  • #8
the voltmeter's purpose is clear to me now, but what about the ammeter...what is the proper placement of it in a circuit? and why must it be in series with the resistance?
i appreciate ur responses integral and wisky, they were extremely helpful
 
  • #9
Integral
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How else can I explain it? I thought what I posted above should cover it. Read my first post again, think about it.
 
  • #10
copy and paste the part u talk about the 'ammeter'
 
  • #11
Integral
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Current is the FLOW of electrons therefore you must channel the flow though the ammeter. Since you need ALL of the current flow through the meter you need to insert the meter at a single point. Note that the circiut must be broken and meter inserted, essentially replaceing a wire in the circiut.
 
  • #12
oh wait a sec, i mistook u for wisky, now that i reread it i get it, thx
 

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