# Solve for the current in the following circuit using superposition

I understood why $i_1$ is equal to $V_1/(R_1\parallel R_2)$. But, I can't understand why $i_2$ became right that. Would anyone explain why $i_2$ is not $-V_2/(R_2\parallel R_3)$ (since I got this answer in my calculation)? Or, can you just describe why $i_2$ is $-V_2/R_2$?

• Is i2 the current through V2 or R2? I can't figure it from your sketch and notes. Aug 31, 2016 at 6:39

The superposition theorem just says to replace remaining voltage sources with shorts and current sources with opens and evaluate. Then just sum everything up. I don't think any of your results are correct, yours or the ones you think may be right. But perhaps I just can't read your problem with understanding. I can read the circuit, though. Just to recall, here it is:

simulate this circuit – Schematic created using CircuitLab

Let's focus on $V_1$, shorting out $V_2$ to produce this for step 1:

simulate this circuit

We can now lay out these currents:

$I_{R_1} = \frac{V_1}{R_1}$

$I_{R_2} = \frac{V_1}{R_2}$

$I_{R_3} = 0$

Note that $R_3$ is shorted out by the replaced $V_2$ and so its current must be zero.

Now let's focus on $V_2$, shorting out $V_1$ to produce this for step 2:

simulate this circuit

We can now lay out these currents:

$I_{R_1} = 0$

$I_{R_2} = -\frac{V_2}{R_2}$

$I_{R_3} = \frac{V_2}{R_3}$

Note that $R_1$ is shorted out by the replaced $V_1$ and so its current must be zero. Also notice that the direction of the current in $R_2$ is the opposite to the earlier direction. So here we use an opposing sign. Just be consistent about this.

Now we can simply take the two above cases and sum them up together as though they are happening simultaneously:

$I_{R_1} = \frac{V_1}{R_1} + 0=\frac{V_1}{R_1}$

$I_{R_2} = \frac{V_1}{R_2}-\frac{V_2}{R_2}= \frac{V_1-V_2}{R_2}$

$I_{R_3} = 0 + \frac{V_2}{R_3}=\frac{V_2}{R_3}$

The current in $V_1$ will be the sum of the two returning currents from $R_1$ and $R_2$ or else it will be the sum of the currents through $R_1$, $V_2$, and $R_3$, depending on which way you'd rather look. Either way, it has to be the same.

Just by way of making sense of things, it should be clear that since $V_1$ is directly across $R_1$ that the final summed current through $R_1$ must simply be $\frac{V_1}{R_1}$. And it is. Good. Similarly, it should be clear that since $V_2$ is directly across $R_3$ that the final summed current through $R_3$ must simply be $\frac{V_2}{R_3}$. And it is. Also good.

Back to business. Easiest way to get $I_{V_1}$ is to sum the returning currents in $R_1$ and $R_2$:

$I_{V_1}= I_{R_1}+I_{R_2}= \frac{V_1}{R_1}+\frac{V_1-V_2}{R_2}$

I don't recall seeing that answer in the stuff you provided. Maybe I just didn't see it.

Superposition tells us the current $i$ is given by $i_1+i_2$, where $i_1$ is the current in that same branch as $i$ when $v_2=0$ and $i_2$ is the current in that same branch as $i$ when $v_1=0$. The key is understanding that $i_1$ and $i_2$ are the current in the same branch. So, for $i_2$, we can easily see that, when $v_1=0$, $R_1$ is short-circuited and hence $i_2$ is just the current that comes from the branch with $R_2$. This current, keeping track of the polarity of $v_2$ and remembering that $v_1=0$, is $i_2=-v_2/R_2$.
I think you are considering $i_2$ as the current through $v_2$ when $v_1=0$. That's not what superposition tells us.
From Ohm's law, Ic = V2/R2, but Ic is in the opposite direction of i2, therefore, in the above circuit, i2=-Ic=-V2/R2. Now, you can sum up i1+i2 to find out i.