Current through left bridge of a Wheatstone Bridge

In summary, 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.
  • #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 .

Mazur1e.ch31.p78.jpg


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
 
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  • #2
Your work looks fine to me, and the answer looks good. Perhaps they're wanting it expressed as mA rather than A? Or perhaps they're quibbling about significant figures?
 
  • #3
gneill said:
Your work looks fine to me, and the answer looks good. Perhaps they're wanting it expressed as mA rather than A? Or perhaps they're quibbling about significant figures?

I dropped the third sig fig and the answer was accepted. Which seems strange to me, but I guess you were right. Thank you anyways!
 

Related to Current through left bridge of a Wheatstone Bridge

1. What is the purpose of the left bridge in a Wheatstone Bridge?

The left bridge in a Wheatstone Bridge is designed to provide a reference current for comparison with the unknown current in the right bridge. By using this comparison, the unknown current can be accurately measured.

2. How does current flow through the left bridge of a Wheatstone Bridge?

The current through the left bridge of a Wheatstone Bridge flows from the positive terminal of the power source through the left resistor, through the galvanometer, and back to the negative terminal of the power source. This creates a known current for comparison with the unknown current in the right bridge.

3. What is the relationship between the currents in the left and right bridges of a Wheatstone Bridge?

The relationship between the currents in the left and right bridges of a Wheatstone Bridge is that they are equal when the bridge is balanced. This means that the ratio of the left bridge resistors is equal to the ratio of the right bridge resistors, and the unknown resistance is equal to the known resistance in the right bridge.

4. How does changing the current through the left bridge affect the Wheatstone Bridge circuit?

Changing the current through the left bridge of a Wheatstone Bridge will affect the balance of the circuit. If the current is increased, the galvanometer will deflect further, indicating that the right bridge is unbalanced. If the current is decreased, the galvanometer will deflect less, indicating that the right bridge is closer to being balanced.

5. What factors can affect the current through the left bridge of a Wheatstone Bridge?

The current through the left bridge of a Wheatstone Bridge can be affected by factors such as the voltage of the power source, the resistance of the left bridge resistor, and the resistance of the galvanometer. Any changes in these factors can alter the balance of the circuit and the accuracy of the current measurement.

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