Calculate the resistance to add in parallel to balance this bridge circuit

In summary, to balance a bridge circuit, you need to calculate the resistance that should be added in parallel to achieve equal voltage across the bridge. This involves using the principles of circuit analysis, specifically applying Kirchhoff's laws and the formula for equivalent resistance in parallel circuits, to determine the correct resistance value that will ensure that the bridge is balanced.
  • #1
Saibot
12
6
Homework Statement
Find resistance R if the voltmeter reading is zero in the circuit below.
Relevant Equations
V = IRtotal, 1/R = 1/R1 + 1/R2
The reading on the voltmeter is 0.
circuit.png


I figure the potential drop through line AB must be the same as the potential drop through AC (since voltmeter is zero). As such, the 10 Ohm resistor has the same as the effective resistance of the parallel branch.

1/10 = 1/8 + 1/R
1/R = -0.025
R = -40 Ohms

This is clearly wrong. Help please!
 
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  • #2
Saibot said:
Homework Statement: Find resistance R if the voltmeter reading is zero in the circuit below.
Relevant Equations: V = IRtotal, 1/R = 1/R1 + 1/R2

The reading on the voltmeter is 0.
View attachment 337192

I figure the potential drop through line AB must be the same as the potential drop through AC (since voltmeter is zero). As such, the 10 Ohm resistor has the same as the effective resistance of the parallel branch.

1/10 = 1/8 + 1/R
1/R = -0.025
R = -40 Ohms

This is clearly wrong. Help please!
I don't know if there is a trick to this, it is true that the voltage drop from AB is the same as AC. There must be different currents the branches though. So they don't have the same effective resistance as far as I can tell.

I'm having some trouble finding enough equations, ( I'm getting 2 equations in 3 unknows), but maybe I'm missing it.
 
  • #3
If the voltmeter reads zero, it means that the potential difference between C and B is zero. If you replace the voltmeter with a resistor, there will be no current through it. It follows that whatever current goes through the 2 Ω resistor must go through the parallel combination. Likewise, whatever current gos through the 5 Ω resistor must go through the 10 Ω resistor. What does this suggest to you? Write some equations down involving the two currents that I mentioned.
 
  • #4
kuruman said:
If the voltmeter reads zero, it means that the potential difference between C and B is zero. If you replace the voltmeter with a resistor, there will be no current through it. It follows that whatever current goes through the 2 Ω resistor must go through the parallel combination. Likewise, whatever current gos through the 5 Ω resistor must go through the 10 Ω resistor. What does this suggest to you? Write some equations down involving the two currents that I mentioned.
Is the total voltage source supposed to be known, and just not shown in the diagram? I really can't find that 3rd independent equation. It seems to me that something is missing...the entire circuit could very well be at zero volts (no current flowing in either branch), and the voltmeter would be none the wiser.
 
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  • #5
erobz said:
Is the total voltage source supposed to be known, and just not shown in the diagram? I really can't find that 3rd independent equation. It seems to me that something is missing...the entire circuit could very well be at zero volts (no current flowing in either branch), and the voltmeter would be none the wiser.
You do have to assume that a nonzero p.d. is supplied by the accumulators.
 
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  • #6
kuruman said:
If the voltmeter reads zero, it means that the potential difference between C and B is zero. If you replace the voltmeter with a resistor, there will be no current through it. It follows that whatever current goes through the 2 Ω resistor must go through the parallel combination. Likewise, whatever current gos through the 5 Ω resistor must go through the 10 Ω resistor. What does this suggest to you? Write some equations down involving the two currents that I mentioned.
Got it, thank you. The current through the top is larger than the current through the bottom. Man, it seems so obvious now.
 
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  • #7
Thanks for your input everyone!
 

FAQ: Calculate the resistance to add in parallel to balance this bridge circuit

What is a bridge circuit?

A bridge circuit is an electrical circuit used to measure unknown resistances by balancing two legs of a bridge circuit. The most common type is the Wheatstone bridge, which consists of four resistors in a diamond shape with a voltage source applied across one diagonal and a galvanometer across the other.

How do I determine if a bridge circuit is balanced?

A bridge circuit is balanced when the voltage difference between the two midpoints (where the galvanometer is connected) is zero. This means that the ratio of the resistances in one leg equals the ratio of the resistances in the other leg.

What is the formula to calculate the resistance needed to balance a bridge circuit?

The formula to balance a Wheatstone bridge is \( \frac{R1}{R2} = \frac{R3}{Rx} \), where \(R1\) and \(R2\) are the known resistances, \(R3\) is the known resistance in the other leg, and \(Rx\) is the unknown resistance. To find the resistance to add in parallel, you need to adjust these values to satisfy the balance condition.

How do I add a resistor in parallel to balance the bridge circuit?

To add a resistor in parallel to balance the bridge circuit, you need to first determine the equivalent resistance needed to achieve balance. Then, use the formula for parallel resistors \( \frac{1}{R_{eq}} = \frac{1}{R_{existing}} + \frac{1}{R_{added}} \) to find the value of the resistor to add.

Can you provide an example of calculating the resistance to add in parallel?

Sure! Suppose you have a bridge circuit with \( R1 = 100 \Omega \), \( R2 = 200 \Omega \), \( R3 = 150 \Omega \), and you want to balance it by adding a resistor in parallel with \( R3 \). First, find the required \( Rx \) using the balance condition: \( Rx = \frac{R3 \cdot R2}{R1} = \frac{150 \cdot 200}{100} = 300 \Omega \). Now, calculate the resistance to add in parallel using \( \frac{1}{300} = \frac{1}{150} + \frac{1}{R_{added}} \), which gives \( R_{added} = 300 \Omega \).

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