- #1

nagyn

- 26

- 0

## Homework Statement

The ammeter in the Wheatstone bridge of (Figure 1) measures zero current when the resistance Rvar of the variable resistor is set to 189 Ω .

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

## Homework Equations

Junction rule:

ILeft + IRight = I4

I4 + I2 = I1

Loop rule:

ILeft(70) - IRight(210) = 0

R6 = (189)(210)/(70) = 567

## The Attempt at a Solution

Using the loop rule, ILeft = 3IRight, and using the junction rule ILeft = (3/4)I4. So I figured in order to determine ILeft I needed to find I4, and the only way to find that was to find I1 and I2.

The entire equivalent resistance of the diamond: the left and right bridges have in-series resistors, so on the left: 189+70 = 259. On the right, 210+567 = 777. The bridges themselves are in parallel, so 1/Rdiamond = (1/777) + (1/259).

Rdiamond = 194.25

Using the same approach for the rightmost branch of the entire circuit, 1/Req = (1/Rdiamond) + (1/350)

Req = 124.92

So the total resistance is Req + 50 + 150 = 324.92

I1 = V/Rtotal = 15/324.92 = 0.0462

Then this might be where I'm messing up: I tried using the loop rule again to find the ratio between I2 and I4 like I did for ILeft and IRight.

I4(194.25) - I2(350) = 0

I4 = 1.8*I2

I1 = 1.555*I4

So I4 = 0.0462/1.555 = 0.0297

And now ILeft = (3/4)I4 = (3/4)(0.0297) = 0.0223 A

This is not the correct answer. Like I said above I think I might be misunderstanding how to determine I2 and I4, but I'm not sure what the right way to approach it is. I hope my work is fairly easy to follow