Does the Direction of Current Affect Ohm's Law Sign Convention?

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Discussion Overview

The discussion revolves around the implications of current direction on the application of Ohm's Law and the sign convention in circuit analysis. Participants explore whether the direction of current affects the outcomes when traversing a circuit in different directions, focusing on the interpretation of negative current values and their relation to voltage sources and resistors.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the negative sign in current indicates that the actual current flows in the opposite direction to that shown in the diagram.
  • Others argue that the answer does not depend on the direction of traversal through the circuit.
  • A participant clarifies that the actual direction of the current and its numerical value remain consistent regardless of how one chooses to traverse the circuit.
  • Another participant points out that traversing the circuit in either direction while maintaining the current direction as indicated will yield a negative current value, which signifies the actual flow direction.
  • One participant asserts that it is possible to assume an initial current direction and still arrive at a positive current value regardless of the traversal direction.

Areas of Agreement / Disagreement

Participants express differing views on whether the direction of traversal affects the interpretation of current direction and sign. There is no consensus on this issue, as some argue for the independence of traversal direction while others maintain that it does influence the sign convention.

Contextual Notes

Participants reference specific circuit elements and their relationships, such as voltage sources and resistors, but do not resolve the underlying assumptions regarding sign conventions and current direction.

Neon32
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Here the voltage is negative so the current is negative. does it matter if I traverse the circuit through the other direction?
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The negative sign means that the current is actually in the opposite direction that you have shown in the diagram.
 
The answer does not depend on how you traverse the circuit.
 
Sure it does. Why is your sign negative?
 
I think I did not make my meaning clear. The actual direction of the current (irrespective of what you initially choose in the diagram) and the numerical value of the current do not depend on how you decide to traverse the circuit.
If we go round the circuit as shown in the diagram, keeping the direction of current also as indicated in the diagram, and use standard loop law:

-5V - I (6Ω) + 3V - I (4Ω) = 0, giving a current of - 0.2 A, as Neon32 got. As Kuruman points out, the negative sign simply means that the actual current is opposite to what was shown in the diagram. so the actual current is going counterclockwise round the circuit

On the other hand, keeping the current as it is in the diagram, we can traverse the circuit in the anticlockwise direction, and get:

+5V + I (6Ω) - 3V + I (4Ω) = 0, giving a current again of - 0.2 A, with again the same interpretation of the negative sign. This is what I mean when I said the answer does not depend on the sense of traversing the circuit.
 
You're right. I misunderstood what "traversing" meant. Indeed, if you traverse the circuit clockwise or counterclockwise, as long as you assume that the current through the 4 Ω resistor is to the right as indicated by the arrow you drew, then you will get a negative value for the current. The negative sign means that the current through the 4 Ω resistor flows to the left. You can see why this is because the right side of that resistor is connected to the positive terminal of a 5 V battery while the left side is connected to the positive terminal of a 3 V battery. Thus, the right side is at higher potential than the left and current flows right to left regardless oh how you draw the arrow or traverse the circuit.
 
Absolutely. And nothing prevents you from assuming, to start with, that the current is from right to left in the 4 Ω resistor (anticlockwise round the loop), and then, whichever way you traverse the loop, you will get a positive current, as you should.
 

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