I have the following definition and explanation of the superposition theorem. I don't understand how the resistance values are calculated. I understand that \$R=V/I\$, so \$I=V/R\$. But how is the equivalent resistance calculated, and how do we know which resistances are effecting the respective branch currents?

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2 Answers 2


If you simply follow along closely in the example in your book, it will make more sense. In your picture they have just done the first half of solving by Superposition by short-circuiting V2.

You understand why they use Ohm's Laws, but to understand what resistance values to use, think about it this way:

For i'1

With V2 shorted, r2 and r3 are now just two resistors in parallel, since both their end terminals are connected. Do you see that? You should now easily recognize the form of the equation for two resistors in parallel: Rparallel = Ra * Rb / (Ra + Rb) and now substituting r2 and r3 for Ra and Rb it becomes more clear.

Then you see that the parallel combination we solved above is in series with r1. This total resistance has V1 dropped over it, hence the equation for i'1 using Ohm's law.

For i'2

This one may be a little harder to see, but basically you are forming a current divider with the resistors. I challenge you to go over that section in your textbook again, or read through this carefully.

Keep at it and you will start to get it.

  • \$\begingroup\$ Thanks for the explanation.I understand the first part for obtaining i'1.I will take a look at current divider rule and get back. \$\endgroup\$
    – elecman
    Commented May 2, 2016 at 3:36
  • \$\begingroup\$ I now understand the second part.Is it because we dont have a voltage source in mesh 2,we need to use the current divider rule? \$\endgroup\$
    – elecman
    Commented May 2, 2016 at 3:43
  • \$\begingroup\$ @elecman No, you could do it with nodal analysis, it is just faster when you recognize it as a current divider. Please don't forget to mark the answer as correct if this answered your question. \$\endgroup\$ Commented May 2, 2016 at 16:20

I think it would help to re-draw the circuit:


simulate this circuit – Schematic created using CircuitLab

That is the second circuit you have on your post. By the theorem of superposition, \$ V2 \$ was set to zero and a subsequent step \$V1\$ will be the one set to zero. But your question is about the equivalent resistance.

In order to know the value of \$i_1\$, you need to know the equivalent resistance of the parallel combination formed by \$R2\$ and \$R3\$, let's call this equivalent resistance \$R_a\$.

Now, the circuit looks like this:


simulate this circuit

And how would you calculate \$i_1\$ at this point? Simply $$i_1=\frac{V_1}{R_1+R_a} $$

Since \$R_a\$ is the parallel equivalent of \$R_2\$ and \$R_3\$ $$ R_a = \frac{1}{\frac{1}{R_2}+\frac{1}{R_3}} = \frac{R_2R_3}{R_2+R_3}$$

If you plug \$R_a\$ into the \$i_1\$ equation, you get:$$i_1=\frac{V_1}{R_1+\frac{R_2R_3}{R_2+R_3}} $$ which is exactly what your text has as the answer for \$i_1\$.

As far as \$i_2\$ is concerned, your text utilizes the current divider equation. Which is a shortcut to finding an unknown current. It would be good for you to read this: https://en.wikipedia.org/wiki/Current_divider

  • \$\begingroup\$ Thanks for your answer.I now understand the current divider rule used to calculate the part 2 of the answer....Is it because we dont have a voltage source in mesh 2,we need to use the current divider rule \$\endgroup\$
    – elecman
    Commented May 2, 2016 at 3:52
  • \$\begingroup\$ You could find I2 with other methods too (e.g node voltage, mesh). The thing is that the current divider provides a faster result in this case , but yes, since there is no voltage source in that mesh, current divider can be used \$\endgroup\$
    – Big6
    Commented May 2, 2016 at 4:09

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