# Hardest circuit diagram I have provided an example of the type of problem I am having issues with. In the given diagram, the resistances and batteries are identical and the batteries do not have internal resistances(for the sake of the argument). How can we compare the current over these resistances?

(I could solve it by using Kirchoff's laws granted I had little to no conceptual understanding of the nature of current and electric fields. Kirchoff's laws simply grant no understanding. They are overly simplified) But please, if you can form a logical explanation as to how you have come to compare the currents from the idea of Voltage, you will have provided an answer to this question.

• Unless you can make the circuit diagrams simpler, eg by combining parallel and/or series resistors and/or cells, you will need to use Kirchhoff's laws. Redrawing a circuit diagram often helps. – Farcher Oct 30 '18 at 10:08
• What is your difficulties? Have you heard of Kirchhoff's laws? – K_inverse Oct 30 '18 at 10:08
• my difficulties are caused because you simply cannot draw the circuit in any other form which makes the question less difficult. Breaking up the circuit into loops does not work well for every circuit. @K_inverse – ten1o Oct 30 '18 at 10:13
• @Farcher I could solve it by using Kirchoff's laws granted I had little to no conceptual understanding of the nature of current and electric fields. Kirchoff's laws simply grant no understanding. They are overly simplified . – ten1o Oct 30 '18 at 10:15
• I would like an explanation from those who use the idea of voltage across arms (kirchoffs laws) to save this problem as to what connection they have formed to transfer the idea of voltage to in this case, current. – ten1o Oct 30 '18 at 10:19

"But please, if you can form a logical explanation as to how you have come to compare the currents from the idea of Voltage, you will have provided an answer to this question."

I do not understand what you say here, but here is my answer. You have use the fact that Ohm's law is linear, and we have only linear elements in our circuit, so you can use principle of superposition.

That is, assume only one battery is switched on, and calculate the resulting currents through the resistors (tedious, but easy). Now do this for every one of the four batteries present. After you have done this, sum up all the contributing currents through each resistor.

As an example, let's assume we have switched on the topmost battery V only. You easily see that the currents are

through K: zero

through L: V/R (flowing from right to left)

through M: V/R (flowing from right to left)

Now do this kind of analysis for the remaining three batteries.

EDIT: Okay, I carry out the remaining analysis to be done.

The net contributions are:

current through K: 0 + (-V/R) + 0 + 0 = - V/R

current through L: (V/R) + (-V/R) + 0 + 0 = 0 (easy to see, since all batteries have the same potential and L connects to cathodes on both ends).

current through M: (V/R) + (-V/R) + (-V/R) + (-V/R) = -2V/R (positive being from right to left)

the circuit diagram you have shared with us is easy enough that you can read the current just off the diagram, but the kind of analysis I have used using linear superposition should help when there is a more complex one.

• Is superposition in this case any easier than doing loop analysis? I'm not sure it is. – Bob D Oct 31 '18 at 22:14
• @BobD, it may not be, but I feel that sometimes this kind of approach is better when you have many more loops in the circuit, so I wanted to demonstrate what may be a useful point of view to the OP. – wcc Nov 1 '18 at 10:04
• understand. The more ways to approach it the better – Bob D Nov 1 '18 at 11:07

The circuit is actually quite simple and you can quickly determine the currents by a judicial selection of the loop currents. I have redrawn the diagram below showing the loop currents. You say all the batteries and resistors are identical. Lets call the battery voltage V.

Consider loop $$I_1$$. It only contains resistor M. What’s more, the loop currents $$I_2$$ and $$I_3$$ do not contribute any voltage to resistor M.

Then for loop 1:

$$+V+V-V+V-I_{1}M=0$$

$$I_{1}=\frac{2V}{M}$$

From this we can quickly determine the voltage between nodes (a) and (b). Moving from terminal a to terminal b through the branch containing resistor M we have:

$$+V-M\frac {2V}{M}+V$$ Which gives us zero voltage between a and b which means zero voltage across L.

Since there is no voltage across resistor L, $$I_3$$ must be zero, because any non zero value of $$I_3$$ would produce a voltage across resistor L. Since $$I_3$$ is zero, the only current in K is due to the single voltage V, giving,

$$I_{2}= \frac {V}{K}$$

Note that you could have solved the loop 2 and 3 equations which would have given you $$I_{3}=0$$. The idea here is you can avoid that extra work by simply looking at the results of the loop 1 equation.

It is not always necessary to grind through simultaneous equations to determine voltages and currents. In this case we were able to quickly determine the currents by simple inspection of the loop currents.

Hope this helps • You'd need to explain why there is no voltage across L... – ZeroTheHero Oct 30 '18 at 13:12
• @ZeroTheHero Going from a to b through resistor M, you have a voltage drop across M of 2V and voltage rises of +V+V = 2V, for a total voltage between a and b of zero. If you are uncomfortable with that, solve 2 equations and 2 unknowns for loops 2 and 3. You will find that I3 is zero. If I3 is zero the voltage across L must be zero. – Bob D Oct 30 '18 at 13:23
• I'm not uncomfortable with this just pointing out you could improve you answer by expanding on this point. – ZeroTheHero Oct 30 '18 at 14:00
• @ZeroTheHero Got it. I will edit the answer to hopefully improve it for you. – Bob D Oct 30 '18 at 14:03

I had little to no conceptual understanding of the nature of current and electric fields. Kirchoff's laws simply grant no understanding. They are overly simplified