How do I calculate I2 by nodal analysis only?

Question

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?

Answer 1

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.

So in conclusion: Your math was incorrect and your simulator was measuring the current going into your Node A from your picture.

Lovely tip for next time: Make sure your current conventions are all the same. Don't have one current going clockwise and another going counterclockwise. This will confuse you and it's going to mess up your math.

Answer 2

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\$.