I am trying to solve the following circuit:

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I believe the answer I'm getting for \$i_b\$ is wrong because I put it into LTSpice and I'm getting that \$i_b = -3.63636\$

This is my LTSpice diagram:

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I found \$i_b = 1\$mA by doing a loop voltage analysis on the left loop; for the voltage drop across the \$200\Omega\$ resistor I assumed that it would be \$i_b + 29i_b\$, which works out to be a nice number and in fact all of the numbers are nice in this case--usually when the numbers are nice, you know you're doing it right.

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At this point, I'm not sure if I incorrectly modeled this in LTSpice, or if I incorrectly assumed which way the current was flowing.

Instead of giving me the answer directly, I would just like to know how to determine whether the current at a node is entering or leaving a branch.

  • 1
    \$\begingroup\$ possible duplicate of Node Analysis - current calculation \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 16 '15 at 16:04
  • \$\begingroup\$ General answer without looking at your example: Leaving and entering is simply a matter of sign convention. If you know the relative polarity of the sources you can deduce current polarity. If not then plugging in what you know in a consistent manner will produce consistent results. The problems usually come from an inconsistent application of this basic concept. \$\endgroup\$ – Russell McMahon Jun 16 '15 at 16:07
  • \$\begingroup\$ @IgnacioVazquez-Abrams The answer given in that question states that you can just "guess the current" and it will work out positive or negative, but in this case if I guess the opposite then I will get a voltage drop of 200*(28ib) at the 200 ohm resistor--vs 200*(30ib). This will change the answer... (I say this unconfidently because I don't know what my mistakes are). \$\endgroup\$ – Klik Jun 16 '15 at 16:08
  • \$\begingroup\$ @RussellMcMahon Since I have a CCCS, won't the answer change, as I described in my previous comment? \$\endgroup\$ – Klik Jun 16 '15 at 16:09
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    \$\begingroup\$ It looks like your Spice simulation is wrong because I1 is an independent current source, not a CCCS. \$\endgroup\$ – Null Jun 16 '15 at 16:48

Your paper analysis is correct, but your LTspice simulation is incorrect. I get the same (incorrect) result as you if I use a gain of \$+29\$ for the F device (your \$I_1\$). But the gain should be \$-29\$ since \$i_b\$ flows from the negative to positive terminal of \$V_{\text{ib}}\$. Changing the gain gives you the correct result.


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F device attributes:

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If I change the gain to \$+29\$ the result is:

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Note that the simulation result is \$v_y = v_{y1} - v_{y2} \approx 98\$V when using a gain of \$+29\$, which is clearly wrong.

The two simulations highlight the importance of maintaining consistency in the direction of currents. The problem statement defines \$i_b\$ and \$29i_b\$ as both flowing toward the middle "T" node. LTspice defines \$i_b\$ as flowing away from it since it defines the current through \$V_{\text{ib}}\$ as flowing from positive to negative terminal. That means you also have to define the CCCS \$29i_b\$ as flowing away from the middle "T" node. In the incorrect simulation (with gain of \$+29\$), \$29i_b\$ is still flowing toward the "T" node while \$i_b\$ is flowing away from it. The correct simulation defines them both as flowing away from the "T" node. Alternatively, you could just switch the direction of the "F" device and use a positive current gain -- it would then also be defined as flowing away from the "T" node.

  • \$\begingroup\$ Thank you very much for going to the length to find the mistake! I spent a lot of time trying to figure out what I was doing wrong. At least I learned something from this. \$\endgroup\$ – Klik Jun 16 '15 at 22:09
  • \$\begingroup\$ @Klik Happy to help. \$\endgroup\$ – Null Jun 17 '15 at 1:45
  • \$\begingroup\$ I don't know if it's too late, but LTspice's currents are distributed in a grid, and their directions are chosen to start from top left, going right and downwards. This can be easily berified by pressing Alt and hovering the mouse over the wires, in any topology. It's just a matter of arbitrary choice for the programming part, the same as current through elements that are fixed at entering into one pin, and exiting at the other: a fixed reference from the solver's perspective. \$\endgroup\$ – a concerned citizen Sep 1 '16 at 17:59

While doing these type of problems don't worry about current directions. First covert \$i_b\$ in terms of the principle node, then apply nodal analysis thereafter and find all the currents in the branch. Then you will know the direction of currents.


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