Solving for the Voltages and Currents in a (simple?) Ciruit

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Discussion Overview

The discussion revolves around solving for voltages and currents in three circuit diagrams, focusing on the application of voltage dividers and superposition. Participants seek clarification on their calculations and approaches to the problems presented in the figures.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that the voltage at point A in figure 1 can be calculated using the voltage divider formula, suggesting a value of 5*(1/21).
  • Another participant corrects this by stating that the voltage at A should be found by subtracting the voltage drop across the top resistor from 5V.
  • Some participants express confusion regarding the application of superposition in figure 2, with one stating that they assumed no current flows through a specific resistor when a source is turned off.
  • A participant shares a measured value of approximately 2.62V for point A, questioning the discrepancy with the calculated value.
  • Another participant suggests a different calculation method for figure 1, emphasizing the need to find equivalent resistances and apply the voltage divider principle correctly.
  • Discussion about the term "probe" arises, with one participant questioning its meaning in the context of figure 2.
  • One participant clarifies that the voltage at A can be calculated using the formula Vout=Vin(R2/(R1+R2)), indicating a different approach to the calculation.
  • Another participant suggests that both upper resistors in figure 2 can be connected together, which may simplify the analysis of the circuit.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating voltages and currents, with no consensus reached on the solutions for the circuits. Disagreements exist regarding the application of the voltage divider and the implications of the floating node at point B.

Contextual Notes

Some calculations depend on assumptions about circuit configurations and the interpretation of terms like "probe." There are unresolved steps in the calculations, particularly regarding the voltage at point A and the treatment of floating nodes.

runawayshoes
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Homework Statement



For the three (simple?) circuits, I'm asked to solve for the voltages at A and B, and also the current I. For figure 1, I assume that is just a simple voltage divider, with the voltage at A to be 5*(1/21) correct?

I'm a little more confused with the circuit in figure 2. What I attempted to use was superposition. So I assumed the right source is turned off, which means that there is no current flowing through the 5k resistor, and the bottom 10k and 1k are then connected in parallel correct. Then I just use voltage divider again, then I do the same thing for the other side and then just sum the voltages. The answer I got was 3.15V. If someone could verify this that would be SO great. Now as far as solving for the current I, I'm completely stumped. I'm used to solving for the current through a component so when it is asking for the current through just the wire I'm confused.

As for figure 2, he states that B is floating until a probe is connected. I'm not sure how that affects the problem. So I just assumed that no current flows through the right 1k and just used voltage divider again. I would really appreciate it if someone can help me with these problems.

Thanks in advance.

circuitsbl_09082008.jpg

 
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runawayshoes said:
For the three (simple?) circuits, I'm asked to solve for the voltages at A and B, and also the current I. For figure 1, I assume that is just a simple voltage divider, with the voltage at A to be 5*(1/21) correct?
One missing step. That gives you only the voltage drop across the (top) 10K resistor. You need to subtract this from 5V to find the voltage at A.

runawayshoes said:
I'm a little more confused with the circuit in figure 2. What I attempted to use was superposition. So I assumed the right source is turned off, which means that there is no current flowing through the 5k resistor, and the bottom 10k and 1k are then connected in parallel correct. Then I just use voltage divider again, then I do the same thing for the other side and then just sum the voltages. The answer I got was 3.15V. If someone could verify this that would be SO great. Now as far as solving for the current I, I'm completely stumped. I'm used to solving for the current through a component so when it is asking for the current through just the wire I'm confused.
You're complicating it unnecessarily. Don't think you have it right, though. One simple way to approach this would be to find the equivalent resistances between 5V and A, and A and ground. Then apply voltage divider principle. Then the potential at A is what you want.

runawayshoes said:
As for figure 2, he states that B is floating until a probe is connected. I'm not sure how that affects the problem. So I just assumed that no current flows through the right 1k and just used voltage divider again. I would really appreciate it if someone can help me with these problems.
What is a "probe" here? And anyway, considering that B is an open circuit node, your approach is correct. I don't know what "probe" means though.
 
Ok, the first one is wrong. The voltage at point A for figure one should be around 2.62V because this is what I actually measured in the lab. I don't get it, this should just be a simple voltage rule right? How come when I apply it I don't get the right voltage using 5*(1/21) ?
 
I already told you what's wrong with your approach for 1. You omitted a final step.
 
For the problem 1, you are doing 5*(1/21), instead of 5*(11/21), which will give 2.619.
Remember the formula is [tex]Vout=Vin(\frac{R2}{R1+R2})[/tex].
 
Last edited:
For figure 2, both the upper resistors are connected to +5V, so you can connect their upper terminals together. The same with the two lower resistors, that are connected to ground.
So, you have two sets of two parallel resistors and can solve the problem with a voltage divider.
 

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