# Using superposition to find current

I'm given the following problem

I'm a little bit confused about superposition problems when current flows through a voltage source but not the resistor.
1) When I looked at the solution for this problem, I found that when current source was zeroed, to find $R_{eq}$ they combined two bottom resistors ($5\,\Omega$ and $10\,\Omega$ resistors) that were in series and then combined it to a $5\,\Omega$ resistor that was in parallel.

However, I don't understand why two $5\,\Omega$ resistors are not in series but $5\,\Omega$ and $10\,\Omega$ resistors are? Is it because of a voltage source?

2) When a voltage source was zeroed, they used the following to compute $i_s$:
$$-1 * \frac{10}{(10+5)}.$$ So, is this a current divider formula that was used to compute a current through a $5\,\Omega$ resistor? If so, why is it negative? And how do I know that the current will go down before node 2?

Also, for some reason they wrote that the current through the top $5\,\Omega$ resistor is 0 which I don't understand.

• A reason that you might want to decide whether or not resistors are in series is that if they are in series you can just sum their resistances to get the equivalent resistance. In the derivation of the series equivalent resistance formula an assumption is made that the currents through both resistors is the same. With the correct source open circuit (a gap in the circuit where the current source was) the same current does flow though the 5 ohm and 10 ohm resistors but not through the 5 ohm and 5 ohm resistors as there is a current 1is flowing out at the node where they join. Feb 23 '16 at 5:51
• @Farcher I see, thank you ! Just a quick question, why when you short the voltage source, it also shorts that upper 5 ohm resistor? Is it because they are in parallel? Also, if there was a resistor in series with a voltage source and we decide to short the voltage source, would this short the resistor as well?
– Jack
Feb 23 '16 at 5:57
• They voltage source and the top 5 ohm resistor are in parallel. So if the voltage source is shorted (zero resistance) all the current would go though the short. Put another way whatever the current through a short the voltage across it zero. So the voltage across the 5 ohm must also be zero. No voltage across the 5 ohm means no current through the 5 ohm. If you replace the voltage source with a short circuit you keep an series components in place. So voltage source plus resistor in series become just a resistor. Feb 23 '16 at 6:05
• EE.SE has a canonical question on superposition you may want to read. Feb 23 '16 at 6:20
• It will help you to actually re-draw the two versions of the circuit with the two different sources zero'd. For your first question, remember that the ground node is just another node that has no magic about it. Feb 23 '16 at 6:24

1) To look at the behavior due to only the voltage source, you open the current source. At that point the upper $5\,\Omega$ is parallel to the source, and the other $5\,\Omega$ and $10\,\Omega$ end up in series with each other, parallel to the source. Draw a picture, combining the two dots near the $v_2$ reference.
2) To look at only the current source, short the voltage source, replacing it with a wire. That also shorts the top $5\,\Omega$, so it gets no current. $i_s$ is then the current travelling up through the lower $5\,\Omega$ resistor. Yes, it's a current divider, but $i_s$ is opposite the +1 A source.