I've recently come across a question where I need to solve using Nodal Analysis:

enter image description here

I'm a bit confused about the part where the arrow V1 points in the opposite direction to the voltage source. I've worked on the question and this is what I've come with so far (all in polar form):

I1 = I2 + I3

I4 = I3 + I5

I1 = (10/_0 - V1) 2

I2 = (10/_30 - V1) j5

I3 = (V1 - V2) (4 + j4)

I4 = (V2) (6 - j8)

I5 = (5/_0 - V2) (5)

And my nodal equations look like this:

At node 1 (Polar Form):

$$ \sqrt{3}*90^\circ = V_1(\frac{3\sqrt{10}}{40}*18.43495^\circ) - V_2(\frac{\sqrt{2}}{8}*-45^\circ)$$

At node 2 (Polar Form):

$$ 1*0^\circ = -V_1(\frac{\sqrt{2}}{8}*-45^\circ) + V_2(0.3867*-6.6666599^\circ)$$

Would this be correct? And does the direction of arrow V1 have any significance?

  • \$\begingroup\$ I see nothing wrong with the arrows. You can declare \$V_1\$ etc. to be wherever you wish. \$\endgroup\$ – Chu Sep 28 '15 at 0:03

What I would do is completely ignore any of the voltage arrows for your analysis. You should be labeling the nodes and have a reference node anyways, which isn't present in that diagram.

For example, here I've removed those arrows, renamed nodes with lowercase letters, and designated a ground node at the bottom. Do all your analysis using Va, Vb, Vc, Vc, Vd, and Ve. Branch currents are assumed to follow the direction of the arrow on the branch (I1, I2, I3, I4, and I5).

enter image description here


Apparently, arrows denote the assumed positive potential as being from arrowhead minus arrowtail, so we can find that:

\begin{align} V_1 = V_b\\ V_2 = V_c \end{align}

| improve this answer | |
  • \$\begingroup\$ Where does it say or imply 'voltage through'? Arrows alongside components denote voltages; that's common practice. \$\endgroup\$ – Chu Sep 28 '15 at 0:05
  • \$\begingroup\$ Interesting, I learned something new today. It's a notation I've never seen or used before. I've exclusively used or seen +/- to denote polarity before (which interestingly the original diagram uses for voltage sources). \$\endgroup\$ – helloworld922 Sep 28 '15 at 0:36

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