# Wheatstone bridge?

Discussion in 'Homework Help' started by Bigcountry, Jan 14, 2010.

1. ### Bigcountry Thread Starter Active Member

Jul 4, 2008
71
0
I really stuck on this one here is what I am suppose to do to the first attached document

1.Determine V1 and V2 mathematically using the equations below.
2.Construct the Figure 1-5 below using Multisim
3.Connect voltmeters to V1 and V2 and simulate the circuit, what are your readings
4.Print and send in your Multisim circuit along with your solution to question 1 and 3.
5.Adjust R10 until V1 = V2, what is percentage of R10?
6. Verify you results in step 5 mathematically

Then I did it on the the software ( multisim) when I did the the problems I could up with V1=3.125 V and V2= 3.77 V but I can not seem to adjust the potentiometer to get the V2 of 3.77 V. The only way I got it to make 3.125 V was to make R10 0%.

I think I am overlooking something very simple I just can't see what it is on the page. thanks for any help in advance..

File size:
256.2 KB
Views:
178
File size:
136 KB
Views:
71
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
For V1=V2
$\frac{R2}{R1}=\frac{R11}{R10}$
R10=(1KΩ*330Ω)/2.2K=150Ω

3. ### Bigcountry Thread Starter Active Member

Jul 4, 2008
71
0
thanks for the mathematically part but I stumped on this percentage part. I think it is 30% but I am really confused.

4. ### Jojo_B New Member

Nov 17, 2009
17
0
Never used MultiSim before, but it seems that the resistance of the pot is determined by the percentage. So according to your diagram...if the pot is tuned to 50% of the "key" 500 Ohms...then it's actual value is 250 Ohms.

So to answer the percentage question, whatever value you figure out R10 (pot) to be, take that value and divide by 500 to get your percentage.

edit: or take the number 1 minus your percentage, not sure which way it goes.