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

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SUMMARY

The discussion clarifies that the direction of current does not affect the application of Ohm's Law sign convention when analyzing a circuit. Regardless of whether one traverses the circuit clockwise or counterclockwise, the calculated current remains consistent. Using the standard loop law, both traversal methods yield a current of -0.2 A, indicating that the actual current flows in the opposite direction to the initial assumption. The negative sign signifies that the current through the 4 Ω resistor flows from right to left, confirming the importance of understanding potential differences in circuit analysis.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with circuit analysis techniques
  • Knowledge of voltage and current sign conventions
  • Ability to apply Kirchhoff's loop law
NEXT STEPS
  • Study the application of Kirchhoff's Voltage Law in complex circuits
  • Learn about the implications of negative current values in circuit analysis
  • Explore the effects of resistor configurations on current flow
  • Investigate potential difference and its role in determining current direction
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Electrical engineering students, circuit designers, and anyone involved in circuit analysis and troubleshooting will benefit from this discussion.

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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