How Do You Apply Kirchhoff's and Ohm's Laws to Analyze Circuit Voltage?

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Discussion Overview

The discussion revolves around applying Kirchhoff's and Ohm's laws to analyze a specific circuit voltage, particularly focusing on determining the voltage at a point in the circuit and ensuring the solution aligns with the principle of power conservation. The context is primarily homework-related, involving theoretical and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to calculate the current and voltage using Kirchhoff's and Ohm's laws, expressing uncertainty about the application of these laws in the given circuit.
  • Another participant questions the sign convention used in the current calculation, suggesting that the negative sign indicates a specific direction of current flow.
  • There is a discussion about the nature of a voltage source in the circuit, with references to it being a Current Controlled Voltage Source (CCVS) and implications for voltage calculations.
  • One participant proposes that if the voltage across a component is -5V, then the current through it would be -5/2A, leading to a calculation of voltage at another point in the circuit.
  • Another participant challenges the assumption that the current through two resistors in series can be treated independently, emphasizing the need to consider total resistance for accurate current calculations.
  • There is a clarification regarding the relationship between current and voltage in dependent sources, with an acknowledgment of the complexity involved in these concepts.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of Kirchhoff's and Ohm's laws, particularly regarding current flow and voltage calculations. There is no consensus on the correct values or methods, and the discussion remains unresolved.

Contextual Notes

Participants highlight potential limitations in their assumptions and calculations, particularly concerning sign conventions and the treatment of dependent sources. There are unresolved mathematical steps related to the total resistance and current in the circuit.

bhsmith
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Homework Statement


1. Use Kirchoffs and Ohm's rule to find the voltage of vknot (to the far right of the picture)
2. show that your solution is consistent with the constraint that the total power supplied in the circuit equals the total power dissipated.

(Picture of Circuit attached)




Homework Equations



I know I need to use V=i(R), but I'm not sure how to apply kirchovs law here.

The Attempt at a Solution



first, i found the current i(s)= -10/6
then the current 3i(s) would be -5A. , so that means that i(0) would be -5A(2ohms)= -10v
Then I'm not sure how to find the voltage from the current voltage and resistance. I'm not sure I did the part with the 2ohm resistor correctly.
 

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first, i found the current i(s)= -10/6

Your sign convention shows that the current flows up through the 6R resistor and into the positive terminal of the 10V source, since the sign is negative on your result. Is this correct?


then the current 3i(s) would be -5A. , so that means that i(0) would be -5A(2ohms)= -10v

The diamond has the + and - polarity, indicating it is a voltage source. That it is proportional to a current just means it is a Current Controlled Voltage Source (CCVS) as opposed to CCCS, VCVS or VCCS. This means that it would be -5V using your sign convention.

If it was -5A then the current in series would mean that i(0) would also be -5A, but since it is not a current source, you have -5V, and two resistances, and need to find the series current to find the voltage Vo.

-5A(2ohms)= -10v would be correct if your -5A assumption was correct, but it is not.
 
Great! That helped a lot.
So, if the voltage is -5v through the diamond then the current would be -5/2A.
since they are in series then the two have the same current leaving the voltage v(knot) to be 15/2V?
is this correct?
And also if I am finding the voltage using 3i(s), and the current has units of Amperes, how do I get voltage from that?
 
So, if the voltage is -5v through the diamond then the current would be -5/2A.

No. The series current is determined by the total resistance.

If the current through the 2R resistor by Ohms Law is -5/2 Amps, then it stands that the current through the 3R load resistor by Ohms Law is -5/3 Amps, but since the two resistors are in series, the current must be in series, and so you are saying that -5/3 A = -5/2 A, which it does not, so something has gone wrong.

The wrong part being using a single resistance instead of the total resistance to find the series current.

since they are in series then the two have the same current ...

This is correct.

... leaving the voltage v(knot) to be 15/2V? is this correct?

If the current was -5/2 A then the voltage would be -5/2 * 3 = -15/2 V. So you would be almost correct (the sign would be wrong) if your initial assumption of -5/2A was correct, which it isn't.

And also if I am finding the voltage using 3i(s), and the current has units of Amperes, how do I get voltage from that?

The internal circuity of a dependent source converts an incoming signal into an outgoing signal, in this case an incoming current into an outgoing voltage. The specifics will be covered by a much later course, but they're complicated and not relevant, so for now you will have to trust the model in that "that's just what happens" (I hate these kinds of explanations myself!) and that the dependent voltage source is 3*i(s) Volts despite i(s) being measured in Amps.
 

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