How Is Current Calculated in a Wheatstone Bridge Setup?

• B3NR4Y
In summary: No, it is not series with the two resistors...it is only in series with the resistor of 150 ohm...the resistor of 50 ohm is in parallel with the diamond...In summary, the conversation discusses a problem involving a Wheatstone bridge and a variable resistor with a resistance of 193Ω. The question is asked about the current through the left side of the bridge, and the conversation delves into various attempts at solving the problem. The expert summarizer provides a brief overview of the key points discussed, including the use of Kirchhoff's laws and knowledge about series
B3NR4Y
Gold Member

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 $(\frac{1}{70+139} + \frac{1}{210+579} )^{-1}$, 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?

Is the variable resistance 193 ohm or 139 ohm?

How did you get 0.04 A for the current through the diamond?

ehild said:
Is the variable resistance 193 ohm or 139 ohm?

How did you get 0.04 A for the current through the diamond?
It is 193 ohms, to get 0.04 A for the current through the diamond I calculated the equivalent resistance of the diamond, then added it to the two other resistors that form a loop with the diamond (50 and 150), then I divided the Emf of the battery by this resistance and got my answer.

B3NR4Y said:
It is 193 ohms, to get 0.04 A for the current through the diamond I calculated the equivalent resistance of the diamond, then added it to the two other resistors that form a loop with the diamond (50 and 150), then I divided the Emf of the battery by this resistance and got my answer.

The diamond is not connected in series with the other two resistors to the left.

ehild said:
The diamond is not connected in series with the other two resistors to the left.
Then it's connected in parallel?

B3NR4Y said:
Then it's connected in parallel?
No. Not parallel either.

The diamond is in parallel with the 350 Ω resistor.

SammyS said:
No. Not parallel either.

The diamond is in parallel with the 350 Ω resistor.
I understand that, but the only two things we have learned about are parallel and series...

You can solve the problem with series and parallel resistors, but you need to know what series connection and parallel connection means. What do you know about the current flowing through series resistors? What do you know about the voltage across parallel resistors?
You calculated the resultant resistance of the diamond. You can replace it with the resultant.

ehild said:
You can solve the problem with series and parallel resistors, but you need to know what series connection and parallel connection means. What do you know about the current flowing through series resistors? What do you know about the voltage across parallel resistors?
Current through series resistors is the same, voltage through parallel resistors are the same.

B3NR4Y said:
Current through series resistors is the same, voltage through parallel resistors are the same.
Splendid. Is the current through the diamond equal to I1? Is the voltage across the 350 ohm resistor equal to the voltage between b and d?

ehild said:
Splendid. Is the current through the diamond equal to I1? Is the voltage across the 350 ohm resistor equal to the voltage between b and d?
I-1 is not the current through the diamond, because there's a junction there. And the voltage across the 350-ohm resistor is equal to the voltage across b and d... I think. Circuits confuse me.

So how are the diamond (Rd) and the 350 ohm resistor connected?

ehild said:
So how are the diamond (Rd) and the 350 ohm resistor connected?

View attachment 82176
The diamond and the 350 ohm resistor are connected in parallel, and therefore should have the same voltage across them? And from there I can find the current in the left side of the diamond because I know the voltage across it?

B3NR4Y said:
The diamond and the 350 ohm resistor are connected in parallel, and therefore should have the same voltage across them? And from there I can find the current in the left side of the diamond because I know the voltage across it?
How do you know the voltage between b and d?

ehild said:
How do you know the voltage between b and d?
Because voltages in parallel resistors are the same, and I can find the voltage through the 350-ohm resistor fairly simply, which is in parallel with the diamond.

B3NR4Y said:
Because voltages in parallel resistors are the same, and I can find the voltage through the 350-ohm resistor fairly simply, which is in parallel with the diamond.
Yes, how do you find the voltage across the 350 ohm resistor? Show it, please.

ehild said:
Yes, how do you find the voltage across the 350 ohm resistor? Show it, please.
Firstly, the 350 ohm resistor is in series with the two resistors in the left, if you ignore the diamond, so sum the resistances and get 550 ohms. Now divide the emf of the battery by this resistance to get the current through this loop, 0.027-A. Now, by kirchhoffs law, emf = V-1 + V-2 + V-3, or E-I(R-1 + R-3) = R-2 (where R-2 is the 350 ohm and R-1 is the 50, R-3 being the 150). This is equivalent to the voltage across the 350 ohm resistor. 9.5454 volts

B3NR4Y said:
Firstly, the 350 ohm resistor is in series with the two resistors in the left,
No, it is not series with the two resistors on the left. Check the currents. Is I1 the same as I2?

Resistors are connected in series, if nothing else is connected to their common point. They must look as pearls of a necklace.

Determine the parallel resultant of the 350 ohm with the diamond. That resultant is in series with the 150 ohm and 50 resistors.

ehild said:
No, it is not series with the two resistors on the left. Check the currents. Is I1 the same as I2?

Resistors are connected in series, if nothing else is connected to their common point. They must look as pearls of a necklace.

If they're not in series, and they're not in parallel, the only thing I can think of is 0 V across the 350 ohm resistor, and I don't think that makes sense.

B3NR4Y said:
If they're not in series, and they're not in parallel, the only thing I can think of is 0 V across the 350 ohm resistor, and I don't think that makes sense.
Resistors in a circuit are not all parallel or all series. Why do you think that there is 0 V across the 350 Ω resistor? Does current flow through it? If the current is not zero, is the voltage across it zero?

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ehild said:
Resistors in a circuit are not all parallel or all series. Why do you think that there is 0 V across the 350 Ω resistor? Does current flow through it? If the current is not zero, is the voltage across it zero?
If the current is nonzero, voltage can't be zero by Ohm's law.

What throws me off is calculating the currents when junctions happen, I suppose

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Calculate the resultant resistances, and replace the circuit between the points b and d by the resultant. What is the resistance of the diamond? What is the resultant if it is connected parallel with 350 ohm? Show the numerical values, please.

B3NR4Y said:
If the current is nonzero, voltage can't be zero by Ohm's law.

What throws me off is calculating the currents when junctions happen, I suppose
What does Kirchhoff's Nodal Law state? The algebraic sum of the currents at a node must be zero. Charges can not accumulate at the node. The net current flowing in must flow out.

@ B3NR4Y,

Don't forget, you do know how to get the equivalent resistance for resistors in parallel. (That formula actually comes from knowing that the voltage is the same for both resistors.)

The "Diamond" is in parallel with the 350 Ω resistor, so what is that equivalent resistance?

ehild said:
Calculate the resultant resistances, and replace the circuit between the points b and d by the resultant. What is the resistance of the diamond? What is the resultant if it is connected parallel with 350 ohm? Show the numerical values, please.
I will do this when I come back home, in about two hours, thank you so much for your help. I feel like I kind of understand what's going on now.

SammyS said:
@ B3NR4Y,

Don't forget, you do know how to get the equivalent resistance for resistors in parallel. (That formula actually comes from knowing that the voltage is the same for both resistors.)

The "Diamond" is in parallel with the 350 Ω resistor, so what is that equivalent resistance?
The total resistance of the "diamond" I calculated to be 579-ohms. So $\frac{1}{R_{eq}} = \frac{1}{R_{diamond}} + \frac{1}{R_{350}}$

B3NR4Y said:
I will do this when I come back home, in about two hours, thank you so much for your help. I feel like I kind of understand what's going on now.The total resistance of the "diamond" I calculated to be 579-ohms. So $\frac{1}{R_{eq}} = \frac{1}{R_{diamond}} + \frac{1}{R_{350}}$
That can't be right for the "diamond". It was R6 that was 579Ω.

In the Original Post for this thread you wrote.
B3NR4Y said:
To start, I calculated the equivalent resistance of the diamond part in the center, which is (170+139+1210+579)−1 ##(\frac{1}{70+139} + \frac{1}{210+579} )^{-1}## , where I calculated the R6 to be 579 Ω.
Of course, there's a typo because later you told ehild that Rvar = 193 Ω.

So ##\displaystyle\ R_\text{diamond}=\left(\frac{1}{70+193} + \frac{1}{210+579} \right)^{-1}\ .\ ## That actually turns out to be 0.75(70+193) .

1. What is a Wheatstone Bridge?

A Wheatstone Bridge is a circuit used to measure electrical resistance. It consists of four resistors connected in a diamond shape with a voltage source and a galvanometer. It was invented by Samuel Hunter Christie in 1833 and later popularized by Sir Charles Wheatstone.

2. How does a Wheatstone Bridge work?

A Wheatstone Bridge works by comparing two unknown resistances to two known resistances. The voltage source applies a known voltage to the two known resistors, creating a voltage drop across each. The unknown resistances are then adjusted until the voltage across them is equal, indicated by no current flow through the galvanometer. This allows for the calculation of the unknown resistances.

3. What is the significance of the "null point" in a Wheatstone Bridge?

The "null point" refers to the point at which there is no current flow through the galvanometer, indicating that the two unknown resistances are equal to each other. This is significant because it allows for accurate measurement of the unknown resistances.

4. How is a Wheatstone Bridge used in practical applications?

Wheatstone Bridges are commonly used in electronic circuits to measure resistance, such as in strain gauges, temperature sensors, and pressure sensors. They can also be used in medical devices to measure blood pressure and in geophysical instruments to measure soil moisture.

5. What factors can affect the accuracy of a Wheatstone Bridge measurement?

The accuracy of a Wheatstone Bridge measurement can be affected by factors such as temperature, wire resistance, and the quality of the resistors used. To ensure accurate measurements, these factors should be carefully controlled and calibrated before use.

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