High school Kirchoff’s voltage law question.

[Mark]

Hi and thanks for looking at my question, many many years ago I knew this like the back of my hand, however the old memory isn’t what it use to be.

I suppose my issue with the supplied answer is why isn’t current from the 6V battery flowing through the 3 ohm resistor and the 2 ohm resistor. It seems this path has been ignored. In which case I would have 2A - 1.2A across the 2 ohm resistor.

In short they deal with two current paths and I think there are three current paths. I did try putting the circuit into Google AI which gave me a completely different answer.

At the moment I’m looking various videos and sites for an answer and they don’t seem to cover this scenario.

I would appreciate some assistance so I may help my son with his physics exam that’s on the horizon.

IMG_2572.jpeg

[martin manning]

I suppose my issue with the supplied answer is why isn’t current from the 6V battery flowing through the 3 ohm resistor and the 2 ohm resistor. It seems this path has been ignored. In which case I would have 2A - 1.2A across the 2 ohm resistor.

The current I2 is actually opposite to the direction shown, so current from both resistors returns to the 4V negative.

In short they deal with two current paths and I think there are three current paths. I did try putting the circuit into Google AI which gave me a completely different answer.

There are three loops, but one is redundant. Note in my solution below I did not need the the KVL equation from the large loop.

PS Your high-schoolers must be quite advanced over there.

[nworbetan]

I suppose my issue with the supplied answer is why isn’t current from the 6V battery flowing through the 3 ohm resistor and the 2 ohm resistor. It seems this path has been ignored.

My memories of how to apply Kirchoff's voltage law are pretty distant too, maybe someone else can help with getting the + and - signs in the right place to make each loop = 0 like they're supposed to...

I might be able to help with the one thing I quoted though. The junction on the lower left, where the two resistors meet the 4V battery, let's call that junction 0 volts*. The junction on the right side where the two batteries and the 2 ohm resistor meet, that junction is at 4V. So using conventional current flow from the 4V on the right side of the 2 ohm resistor to the 0V on the left side, I2 = 2A.

*We could arbitrarily pick any junction to set at 0V but this one is the lowest voltage point in the circuit, so it makes sense to use this one.

[GAStan]

It didn't make sense to me until I wrote in the voltages. The 3 ohm has 10 volts, the 2 ohm 4 volts.

20250722_140553.jpg

[Mark]

Firstly I want to thank all of you for helping me help my son with his upcoming physics exam, I really appreciate you taking time to help me.

I think my problem is with coming up with the equation, the math is simple enough.

As shown below I see loop 1 and loop 2 opposing each other. I think loop 1 minus loop 2 would give 2 - 1.2 =0.8

If I could further impose upon you could you please explain where I have gotten it wrong.

Thanks again for your help and time.

IMG_2573.jpeg

[martin manning]

Your error is that there is not a single current I in the loops. You have to use the named branch currents and their assumed directions for the KVL equations. Also, as I mentioned above you only need two loop equations.

[Mark]

Thanks Martin. I appreciate it.

[GAStan]

Mark, the Loop 1 in your drawing was throwing me too. Until I wrote in the voltages. If you consider that the 2 ohm resistor has 4 volts across it, then apply Ohm's Law, you can see Loop 1 cannot exist.

[martin manning]

If you consider that the 2 ohm resistor has 4 volts across it, then apply Ohm's Law, you can see Loop 1 cannot exist.

All three loops "exist" in that you can arbitrarily assign current directions (as shown in the problem circuit) and write KVL equations as I did above, but then you have to solve them to get the actual magnitudes and directions. In this case the direction assumed for I2 was found to be incorrect. If I had been a little quicker, I would have recognized that the large loop around the perimeter (my KVL3) gets I1 directly, and then I could have used that and the I2 from my KVL1, plus the KCL for the one true node to get I3. In that case the upper loop (my KVL2) is not used.

[GAStan]

Admittedly "exist" was probably a poorly chosen word.

What I do, to keep things straight in my own mind, is make them as simple as possible.

Using this drawing, if one were to place a VOLT meter across each resistor the voltage measured would be as shown, with points B and C being the positive sides. Using Ohm's Law the current's values and directions can be determined. We know I1 & I2, we just have to solve for I3. Much simpler and, to me, less confusing.

20250723_112917.jpg

Considering the above I do not see how Loop 1 (referencing Mark's latest drawing) has any influence on the overall circuit.

Edit in BOLD

[Mark]

Mark, the Loop 1 in your drawing was throwing me too. Until I wrote in the voltages. If you consider that the 2 ohm resistor has 4 volts across it, then apply Ohm's Law, you can see Loop 1 cannot exist.

Sorry I haven’t responded sooner, unfortunately I came down with a rather nasty cold, and I was largely sleeping. I don’t want you to think that I don’t value people trying to help me. I’m extremely grateful to people assisting me. I even feel bad about having to impose upon others.

I found redrawing the circuit was quite helpful. I think my problem was that I hadn’t established a negative or a positive voltage source.

Thanks again for your assistance. Hopefully, it will have clicked with my son.

IMG_2583.jpeg

[Mark]

If you consider that the 2 ohm resistor has 4 volts across it, then apply Ohm's Law, you can see Loop 1 cannot exist.

All three loops "exist" in that you can arbitrarily assign current directions (as shown in the problem circuit) and write KVL equations as I did above, but then you have to solve them to get the actual magnitudes and directions. In this case the direction assumed for I2 was found to be incorrect. If I had been a little quicker, I would have recognized that the large loop around the perimeter (my KVL3) gets I1 directly, and then I could have used that and the I2 from my KVL1, plus the KCL for the one true node to get I3. In that case the upper loop (my KVL2) is not used.

Martin thanks again for clarifying this circuit for me. I respect your knowledge of electronics and appreciate all your help.

[Mark]

I suppose my issue with the supplied answer is why isn’t current from the 6V battery flowing through the 3 ohm resistor and the 2 ohm resistor. It seems this path has been ignored.

My memories of how to apply Kirchoff's voltage law are pretty distant too, maybe someone else can help with getting the + and - signs in the right place to make each loop = 0 like they're supposed to...

I might be able to help with the one thing I quoted though. The junction on the lower left, where the two resistors meet the 4V battery, let's call that junction 0 volts*. The junction on the right side where the two batteries and the 2 ohm resistor meet, that junction is at 4V. So using conventional current flow from the 4V on the right side of the 2 ohm resistor to the 0V on the left side, I2 = 2A.

*We could arbitrarily pick any junction to set at 0V but this one is the lowest voltage point in the circuit, so it makes sense to use this one.

Thank you for your help and taking the time to help my son and I. I appreciate your time and effort.

[Ten Over]

We are going to use Kirchoff's Loop Rule and Kirchoff's Point Rule to find the various currents.

Kirchoff 1.png

Conventional current is used in this exercise, so current flows from positive to negative. The given directions of the currents dictate the resistor polarities.

Kirchoff 2.png

Point E to F to A to B all have the same current (i1). Point B to C to D to E all have the same current (i3).

We will pick a loop and a direction then add the voltages. The algebraic sum of these voltages must equal zero.

[Ten Over]

Loop A: Point F to point A has a rise of 6V, A to B has a drop of 3 x i1, and B to E has a drop of 2 x i2. "Rise" is positive and "drop" is negative.
+6-(3 x i1)-(2 x i2)=0
We can't solve for one unknown current because another unknown current is also in the equation.

Loop B: Point B to point E has a drop of 2 x i2 and D to C has a drop of 4V.
-(2 x i2)-4=0
-4=2 x i2
-4/2=i2
i2 = -2A

The negative current means that the current is going in the opposite direction to that shown.

Now that we have a value for i2, we can find i1 using Loop A. Point F to point A has a rise of 6V, A to B has a drop of 3 x i3, and B to E has a rise of 4V (the polarity reversed when the direction of i2 reversed).

Kirchoff 3.png

+6-(3 x i1)+4=0
6+4=3 x i1
i1=10/3
i1=3.33A