# Circuits questions help( supposedly they are easy.)

1. Jan 5, 2008

### rock.freak667

[SOLVED] Circuits questions help(...supposedly they are easy.)

1. The problem statement, all variables and given/known data
Here is a picture of the question with the diagrams
http://img153.imageshack.us/img153/9741/questioniy3.jpg [Broken]

2. Relevant equations
Resistors in series/parallel,kirchoff's laws.

3. The attempt at a solution
For the part b) I just found the equivalent resistance and then used the fact that $R_1=25R_2$ and then found the total current using $I=\frac{V}{R}$
and got:

$$I=(\frac{26}{26r+25R_2})*E$$

so to find a desired value for I you would need to choose E,r and $R_2$ accordingly?

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

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

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

and $R_v=180 \Omega$

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.

Last edited by a moderator: May 3, 2017
2. Jan 6, 2008

### Mindscrape

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.