Circuits questions help( supposedly they are easy.)

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The discussion focuses on solving circuit problems involving resistors in series and parallel, as well as applying Kirchhoff's laws. The user successfully calculated equivalent resistance and total current using the formula I=V/R, while also addressing the impact of voltmeter resistance on circuit measurements. Key insights include the importance of using fixed values for E and r when adjusting rheostats, and the necessity for voltmeters to have high resistance (in the mega ohm range) to avoid significant current draw that could affect readings.

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  • Familiarity with Kirchhoff's laws
  • Basic knowledge of Ohm's law (I=V/R)
  • Concept of equivalent resistance in circuits
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[SOLVED] Circuits questions help(...supposedly they are easy.)

Homework Statement


Here is a picture of the question with the diagrams
http://img153.imageshack.us/img153/9741/questioniy3.jpg


Homework Equations


Resistors in series/parallel,kirchoff's laws.

The Attempt at a Solution


For the part b) I just found the equivalent resistance and then used the fact that [itex]R_1=25R_2[/itex] and then found the total current using [itex]I=\frac{V}{R}[/itex]
and got:

[tex]I=(\frac{26}{26r+25R_2})*E[/tex]


so to find a desired value for I you would need to choose E,r and [itex]R_2[/itex] accordingly?


For part C) [This is where I think I did the question wrong]
p.d. through the voltmeter and [itex]60 \Omega[/itex] resistor =2V
So that the pd through the [itex]90 \Omega[/itex] resistor=6-2=4V

Current through the [itex]90 \Omega[/itex] resistor,I=[itex]\frac{4}{90}=.0444A[/itex]
Current through the [itex]60 \Omega[/itex] resistor,[itex]I_1[/itex]=[itex]\frac{2}{60}=0.033A[/itex]

By Kirchoff's first law:[itex]I=I_1 + I_2[/itex] ((I_2 is the current through the voltmeter))
so that [itex]I_2=I-I_1[/itex]
[tex]\frac{2}{R_v}=\frac{1}{90}[/tex]

and [itex]R_v=180 \Omega[/itex]

cii)

Total resistance=1038.46ohms
total current,I=6/1038.46=0.005777777~.0058A

pd through 900 ohm resistor=0.005777777*900=5.2V

pd through voltmeter and 600ohm resistor=6-5.2=0.8V


Not too sure what this says about voltmeters except that if they have resistance comparable to the other resistances they draw too much current and then some of the pd is lost in it.
 
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Yes. I did not check the math you did, but your results are what I would expect, so I'm guessing the math is good.

So for the first one, the one subtlety that you missed out on was that (assuming an ideal battery) E and r are supposed to be fixed values. You will get the current you want by adjusting the rheostats, a.k.a. trimpots.

The second one you have exactly right. Most of the voltmeters these days have resistances in the mega ohm region, 10^6 ohms. So your voltmeter will give the right result as long as you are a magnitude or two above the resistances you are using.
 

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