# How do I calculate I2 by nodal analysis only?

I am trying to do a nodal analysis of this circuit. and I need to do this by nodal analysis only.

I need to find $i_1$, $i_2$ and $i_3$.

I have found that

$i_1 = \frac{V_B - V_A}{R_1} = \frac{40 - (-15)}{10} = \frac{55}{10} = 5.5 A$

$i_3 = \frac{V_A}{R_2 + R_3} = \frac{-15}{5 + 25} = \frac{-15}{30} = -0.5 A$

Both these value are correct, according to the simulator.

But this is the problem:

$i_2 = i_1 + i_3 = 5.5 + (-0.5) = 5 A$

But the simulator says -6A.

What is the correct way to approach that?

• ??? please explain better. – Duck Sep 14 '18 at 16:30
• sorry about that. It was a typo. Simulator shows -6A. SEE HERE – Duck Sep 14 '18 at 16:34
• My simulator says -6A as well (using LTSpice). However, I believe it's measuring the current going into Node A, hence why it's -6A. – KingDuken Sep 14 '18 at 16:51
• i3=0.5A........ – Chu Sep 14 '18 at 17:21
• @KingDuken, i3=0.5A with the current direction the OP has specified. – Chu Sep 14 '18 at 19:55

I checked with my simulator as well with LTSpice and I got the same result of -6A. I believe your simulator is trying to read the current that is going into your assigned Node A from your picture, which is why the current is negative. Current going into a node is always negative.

As far as your math goes, it's not entirely correct. Because the direction of $i_1$ and $i_2$ is going in a clockwise direction, your sign for $i_3$ is actually going to be negative because it's traveling counterclockwise.

Hence $i_2=i_1-i_3=5.5A-(-0.5A)=6A$, which makes sense because you have $i_2$ leaving the node and thus being positive. However, if you had $i_3$ going in the clockwise direction, it would be $i_2=i_1+i_3=5.5A+0.5A=6A$...

And I know what you're thinking...

Even if $i_3$ was going clockwise, how would I get a negative current reading?

If you let $i_3$ go into the clockwise direction, it would actually be $\displaystyle i_3=\frac{15}{30}$ because the current is going into the positive terminal of the voltage source.

The voltage at Node A is $\small-15V$, hence $\small i_1=\frac{55}{10}=5.5A$, and $\small i_3=\frac{15}{30}=0.5A$. Currents at Node A: $\small i_1+ i_3=i_2=5.5+0.5=6A$.