# Finding Vg with Wheatstone Bridge: R1=40, R2=70, R3=50, R4=20

• mjosephh
In summary: To do this, one needs to use Ohm's law:V=IRHere, V is the potential difference across Rg, I is the current flowing through Rg, and R is the resistance of Rg. Substituting these values into the equation for potential difference, we get:V=60*I*RgTherefore, the potential difference across Rg is 60*I*Rg-I*(60*Rg), or 600V.
mjosephh
1.
I am having trouble finding the Vg for a problem involving a Wheatstone Bridge circuit. If I am labeling the problem correctly, R1=40, R2=70.0 R3=50.0 and R4=20.0.
Voltage - in = 20.0V

I am using the formula: ((R1*R3) - (R2*R4) *V-in) / (R1 + R2) * (R4+R3) to find Vg between points B and D.

R1=40 (resistance between points A and B)
R2=70 (resistance between points B and C)
R4=20 (resistance between points A and D)
R3=50 (resistance between points C and D)
+ -
V-in = 20V

2. I think my problem is that I don't know what to do with the resistance of Vg (between points B and D), which is 60.0 ohms.

3. The answer for Vg in the book is 0.94 V, and I am getting 1.56V. What am I doing wrong?

I attached a graphic of the problem

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• image..jpg
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Can anyone help with this? I have tried everything!

Picture is very unclear. Could you try and repost it?

nevermind about reposting. I see it clearly now. You are trying the find the voltage across that 60 ohm resistor?

I will attach a larger image, but you can try and expand the other one to view it better.
Thanks for looking into this.

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• image.jpg
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Last edited:
Yes I am trying to find that voltage between B and D. I used the above formula (which is given when all resistances are known, but it didn't work out. I am sure that I need to distribute the 60 ohm resistance across the other resistors somehow, but I am not sure how to. I couldn't find anything online to help me. As most problems are geard to find an unknown resistance.

Well, I dislike using Wikipedia, but try this

http://en.wikipedia.org/wiki/Wheatstone_bridge

Follow the equations and rewrite according to the orientation of one's problem as necessary, then follow the steps, but realize that the bridge is unbalanced and Ig is not zero.

The input voltage is given, and one needs to determine the potential difference across Rg, which is related to the potential differences VAB and VAD.

## What is a Wheatstone Bridge and how does it work?

A Wheatstone Bridge is an electrical circuit that is used to measure an unknown resistance by balancing two legs of a bridge circuit. It works by comparing the ratio of the known resistances (R1 and R2) to the unknown resistance (Rx) through the use of a galvanometer.

## How do I calculate the value of the unknown resistance using a Wheatstone Bridge?

The formula for calculating the unknown resistance (Rx) in a Wheatstone Bridge is: Rx = (R2/R1) * R3. In this case, with R1=40, R2=70, and R3=50, the calculation would be: Rx = (70/40) * 50 = 87.5 ohms.

## What do the values of R4 and the galvanometer have to do with finding the unknown resistance?

R4 and the galvanometer are used to balance the bridge circuit and determine the unknown resistance. R4 is adjusted until the galvanometer reads zero, indicating that the bridge is balanced. At this point, the ratio of R1 to R2 is equal to the ratio of R3 to Rx, allowing for the calculation of Rx.

## Can a Wheatstone Bridge be used to measure any type of resistance?

Yes, a Wheatstone Bridge can be used to measure any type of resistance, as long as the known resistances (R1 and R2) are within the range of the unknown resistance (Rx).

## Are there any limitations or sources of error when using a Wheatstone Bridge?

Yes, there are a few limitations and sources of error when using a Wheatstone Bridge. These include the accuracy of the known resistances, the precision of the galvanometer, and the potential for temperature changes to affect the resistances in the circuit. It is important to use high-quality components and carefully control the temperature when using a Wheatstone Bridge for accurate measurements.

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