- #1

B3NR4Y

Gold Member

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## Homework Statement

The ammeter in the Wheatstone bridge of measures zero current when the resistance Rvar of the variable resistor is set to 193Ω .

What is the current IL through the left side of the bridge?

## Homework Equations

V1 + V2 + V3 + V4 + ... + VN = E

Rx = Rvar*R3/R1

## The Attempt at a Solution

To start, I calculated the equivalent resistance of the diamond part in the center, which is [itex] (\frac{1}{70+139} + \frac{1}{210+579} )^{-1} [/itex], where I calculated the R6 to be 579 Ω.

I used this and V1 + V2 + V3 + V4 + ... + VN = E to find the voltage across the whole diamond part (using Kirchoff's Current Rule). The current into the diamond should be equivalent to the one out of it, 0.04 A. Using this knowledge, and the voltage I calculated across the diamond part, and divided it by the equivalent resistance on the left side, which should give me current. I found the current on the right side the same way, and sure enough they added to 0.04 A. I got an answer of 0.02382 A for the left side, which is wrong.

Another thing I tried is considering only the loop through the left side, it's voltages still will sum to zero, so I found the voltage across the left side in a similar manner and divided by the resistance and got 0.0324 A, a different number. But I have one attempt left and don't want to try it. Any EEs here to help?