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Consider a circuit with only batteries and resistors. Suppose we choose a resistor at random. Then to find the amount of current through this resistor, we can do it by finding the contribution of each individual battery to the current through this resistor.

For example, I choose one battery to remove all the others, find the current which goes through the resistor, note that value, do this for all batteries, and add the quantity. That will be the total current in the original situation itself.

But, I don't understand why exactly this should work. Could someone explain why this works in an intuitive way?

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  • \$\begingroup\$ Real batteries have an internal resistance and other complexities. Do you mean a voltage source? \$\endgroup\$ Commented Jul 26, 2021 at 21:51
  • \$\begingroup\$ Yes, that would be nice @PeterMortensen \$\endgroup\$
    – Babu
    Commented Jul 27, 2021 at 14:13

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You already did describe it in the intuitive way.

As to why it works.

\$V=IR\$ describes a resistor and is linear. Linearity allows superposition by definition. The definition of superposition is almost literally "using the property linearity".

For just a single resistor, if you go \$V=IR\$ it can be split into \$V(I_1+I_2)R\$ which ends up being like \$V = V_1+V_2=I_1 R+I_2 R\$ What else can it be if the equation for the component is linear? You run more "stuff" (current) through the resistor, and what you get out (voltage) goes up by exactly how much extra current you stacked inside it, directly proportionally. No weird scaling. The change that results in the extra stuff you added is independent of what was there previously. That's linearity.

Simplest case is when you have two series voltage sources V1 and V2 rather than just one V=V1+V2. You couldn't do that simple substitution if it wasn't linear.

Superposition doesn't work if you have nonlinear components in your circuit. Ideal resistors, capacitors, and inductors are all linear, except for inductors and capacitors it is the rate of change that is stacked.

Transistors and diodes are not linear and superposition does not work for circuits containing them unless you are using a linearized model which is only applicable over a narrow range of operating conditions (i.e. in small signal analysis here you superimpose a small AC perturbation about DC bias point).

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  • \$\begingroup\$ Perhaps qualify "resistors, capacitors, and inductors are all linear"? All real devices are non-linear if the voltage is high enough (RUD - rapid unscheduled disassembly). \$\endgroup\$ Commented Jul 26, 2021 at 21:53
  • \$\begingroup\$ @PeterMortensen Changed to ideal. \$\endgroup\$
    – DKNguyen
    Commented Jul 26, 2021 at 21:54
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Honestly, it sounds like the explanation you gave yourself is the "intuitive" explanation to the concept of superposition... That is, each source contributes it's own part of the overall effect on each component, and you simply count all of the contributions in order to get your final answer.

I won't try and go into the underlying fundamentals that lead to superposition principle, as they have already been well-explained elsewhere. For instance, Khan Academy has a good explanation of superposition and the underlying mathematical concepts used, linked here. Right before it in their sequence is an explanation of linearity, here, which is the crucial mathematical concept to understanding why superposition works.

On the topic of "why," I can only really say that "math" is what makes it work. You have two choices: take it for what it is, accepting that there are mathematical reasons behind the concept that work; or, you may dig deeper into the mathematics that derive superposition. Hope this helps!

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In physics, and specifically when dealing with waves of different kind, one calculates the "strength" of the wave/field at a certain point with the concept of Super Position.

As an intuitive example, imagine you have 3 lasers, each having P1, P2 and P3 units of power (imaginary units ofc!). imagine you have 3 see-through pieces of materials, and put them behind each other and you shine each laser on them on a test point, calculate how much optical power you receive at each one, and add them all up to get the total power when all three are shined together!

Same in a circuit with batteries and resistors, those resistors are your test points and your lasers are your sources of power into your system, here an electrical circuit. you measure the effect of each one, and adding them will result in the overall effect.

One thing that DKNguyen addressed above is how this would not work in non-linear circuits. if you are mathematically intuitive, that would absolutely make sense because the circuit has a linear relationship. very simple explanation: V=IR, so if you have a "bunch of currents" (very wrong terminology) going through one resistor, you can add them up because the relationship does not have a squared term for current.

Since you only mentioned batteries and not any current sources, I am going to refrain from looking into why current sources are going to be open circuits and voltage sources to be short circuits, but going back to the laser example, it makes sense to model the voltage sources as short circuits, as if they are non-existent.

Hope this gives you an intuitive understanding of this concept. further info can be found here.

P.S: I think an example of 2 water pumps in a pluming setup would have been more intuitive!

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