Solve Wheatstone Bridge Circuit: Voltage B-D

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In summary, the circuit in the attachment is a Wheatstone bridge and the task is to find the voltage between points B and D. The circuit contains two sets of parallel resistors and one odd resistor with a total resistance of 102.5ohm. Using the formula i = V/R, the total current is found to be 0.195A. However, none of the resistors are in parallel or series with each other, so the total resistance may not be correct. The approach to solving this problem depends on what circuit analysis methods have been taught, such as Kirchoff's rules, which is commonly taught in first year physics classes. So far, the class has covered KCL, KVL, Thevenin's
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rave7
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The circuit in the drawing in known as a Wheatstone bridge circuit. Find the voltage between points B and D. The circuit drawing is inside the attachment.

How should i approach this circuit? i can only conclude that there 2 are sets of parallel resistors. 60ohm being the odd one left. so total resistance i found = 102.5ohm

I use i = V/R to get my total current. which is 0.195A. what should i do from here. use current divider?
 

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  • #2
Hi rave7,

rave7 said:
The circuit in the drawing in known as a Wheatstone bridge circuit. Find the voltage between points B and D. The circuit drawing is inside the attachment.

How should i approach this circuit? i can only conclude that there 2 are sets of parallel resistors. 60ohm being the odd one left. so total resistance i found = 102.5ohm

I use i = V/R to get my total current. which is 0.195A. what should i do from here. use current divider?

None of these resistors are in parallel or series with each other, so I don't believe your total resistance is correct.

As far as what approach to use, it depends on what kind of circuit analysis approaches they have taught you or are going to teach you in your class. You can use the Kirchoff's rules approach to solve this problem (with a bit of fun algebra) and I think most first year physics classes teach Kirchoff. Is your class only covering equivalent resistances and Kirchoff's rules as a way of analyzing cirvuits?
 
  • #3
Yes, so far we have only cover KCL, KVL, thevenin's theorem and superposition.
 

FAQ: Solve Wheatstone Bridge Circuit: Voltage B-D

1. What is a Wheatstone Bridge Circuit?

A Wheatstone Bridge Circuit is a type of electrical circuit used to measure an unknown resistance. It consists of four resistors arranged in a diamond shape, with two known resistors on one side and the unknown resistor on the other side. When a voltage is applied to the circuit, the unknown resistance can be calculated using Ohm's law.

2. How does a Wheatstone Bridge Circuit work?

The Wheatstone Bridge circuit works by balancing the voltage across the two sides of the diamond shape. When the voltage between points B and D is equal, the bridge is said to be balanced. This is because the resistors on each side create a voltage drop, and the unknown resistor can be calculated based on the ratio of the known resistors.

3. How do I solve a Wheatstone Bridge Circuit?

To solve a Wheatstone Bridge Circuit, you will need to follow these steps:

  1. Calculate the total resistance of the known resistors on one side of the bridge.
  2. Calculate the total resistance of the known resistors on the other side of the bridge.
  3. Set up an equation using Ohm's law to find the unknown resistance.
  4. Solve for the unknown resistance.

4. What is the purpose of a Wheatstone Bridge Circuit?

The purpose of a Wheatstone Bridge Circuit is to measure an unknown resistance. This can be useful in various scientific and engineering applications, such as determining the resistance of a material or component, or in the calibration of instruments.

5. Are there any limitations to using a Wheatstone Bridge Circuit?

Yes, there are a few limitations to consider when using a Wheatstone Bridge Circuit:

  • The resistors must be in the same temperature and in the same environment to get accurate results.
  • The voltage source must be stable and not fluctuate.
  • The resistance of the unknown resistor must fall within the range of the known resistors for accurate measurement.

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