Simple potentiometer circuit help

  • Thread starter brad sue
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  • #1
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Hi ,please Can I have some suggestions to do this exercice:

A simple potentiometer circuit used to measure unknown voltages accurately is shown below.
Here Vs is the known source voltage, Vxis the unknown voltage, and the resistor is a variable one from which the values R1 and R2 can be read from the position of the pointer. these resistances are varied until the current in the ammeter is zero. Show that the unknown voltage then has the value Vx=Vs*R2/(R1+R2).


I can not create a picture.
Thank you

B
 

Answers and Replies

  • #2
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How can we give you correct suggestion without seeing your picture?
 
  • #3
Gokul43201
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brad sue said:
Hi ,please Can I have some suggestions to do this exercice:
Brad, please read the guidelines for posting in the coursework forum.
 
  • #4
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Here's my solution to your problem

[tex]V_S=I(R_1+R_2)[/tex]

[tex]V_X=IR_2[/tex](because the current through ammeter is [tex]0[/tex]).

Then the answer is obvious[tex]V_X=\frac{V_SR_2}{R_1+R_2}[/tex]
 
  • #5
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phucnv87 said:
Here's my solution to your problem

[tex]V_S=I(R_1+R_2)[/tex]

[tex]V_X=IR_2[/tex](because the current through ammeter is [tex]0[/tex]).

Then the answer is obvious[tex]V_X=\frac{V_SR_2}{R_1+R_2}[/tex]


Thank you, but please can you give me more precision about Vx=I*R2?
if there is current in the ammeter, this means that Vx should equal to 0 since U=R*I right???
 
  • #6
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Using the first and the second Kirchhoff's law you will easily find out what I mean and remember to use [tex]I_A=0[/tex] in your equations.
 
  • #7
lightgrav
Homework Helper
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Brad Sue, your current i2 = 0 , once the resistors have been properly adjusted.
The I in "IR_2" ... (in post#5) is your i1 ... R1 and R2 have same current
since none of it goes thru the ammeter.
 

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