# Why is the polarity of the $3 \ \text{V}$ source necessary to calculate the current flowing through the $8 \ \Omega$ resistor?

I'm currently studying the textbook Fundamentals of Electric Circuits, 7th edition, by Charles Alexander and Matthew Sadiku. Chapter 1.7 Problem Solving gives the following example:

Example 1.10

Solve for the current flowing through the $$\8 \ \Omega\$$ resistor in Fig. 1.19.

The authors begin by stating that a problem is that we do not know the polarity of the $$\3 \ \text{V}\$$ source, but I don't understand why the polarity of the $$\3 \ \text{V}\$$ source is even necessary to calculate the current flowing through the $$\8 \ \Omega\$$ resistor. Couldn't we just calculate the current flowing through the $$\8 \ \Omega\$$ resistor by using the other half of the circuit, where we have the $$\5 \ \text{V}\$$ source? Why is the polarity of the $$\3 \ \text{V}\$$ source necessary to calculate the current flowing through the $$\8 \ \Omega\$$ resistor?

You absolutely need to know the polarity of the 3V source. The 3V source creates a potential that wants to make a current. This current has to go somewhere and that 8ohm resistor is a pretty good place for it to go.

See these two images. If you flip the 3V source it will drive current the other way.

In the 2nd case: It doesn't mean that current is going through both ways at the same time. They will somewhat cancel out in the 8ohm resistor and flow in one direction, but less. (I didn't do the calculations, so I don't know which way is the final flow) Anyway, there will be more current flowing through the 2 ohm and the 4ohm resistor as now the two supplies add together.

See this last image.

• In your illustrations, you're showing the physical direction of the current (+ to -), but the conventional direction of the current (- to +), as used in electronics, would go in the opposite direction, right? Commented Nov 25, 2021 at 10:42
• @ThePointer current flows identically. You are now mixing that with flow of electrons which is opposite to current. Commented Nov 25, 2021 at 10:46
• @ThePointer. Yes elections flow in the opposite direction to the current. Its best to just always go with the + to - current flow when talking about things at this scale. Many electrical engineers wish they could go back in time and fix the current flow direction: xkcd.com/567 Commented Nov 25, 2021 at 10:49
• Oops, that was a typo: it should be "physical direction of the current (- to +)" and "conventional direction of the current (+ to -)." Yes, it seems I misinterpreted your illustration. Commented Nov 25, 2021 at 11:02
• I've not heard anyone say "physical current" outside of school textbooks. You don't need to say "physical" and "conventional". Current is always (+to-), Just like Voltage for that matter. Then use "Election flow" for the cases when its really necessary do describe what happens at the atomic scale. (i.e.: physics of junctions and batteries, e.t.c.) Commented Nov 25, 2021 at 13:19

Because if you measure current without unknown supply connected, then connecting the unknown supply will either make the current through resistor either smaller or larger depending on which way the unknown supply is added.

So no, you can't just use the 5V supply to calculate current when there is another supply connected.