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To start, imagine the branch containing the current source as a node, a point currents flow in and out of. This gives us: $$i_3+i_4-i_1-i_2=12mA$$.

We also know by summing all currents flowing in and out of Node1 that: $$-i_1-i_2=12mA$$.

From these 2 equations, we know $$i_3=-i_4$$.

We know $$i_3=\frac{V_2}{6000}$$ and $$i_4=\frac{V_3}{3000}$$. So we can conclude that $$V_2=-2V_3$$. (V2 is the voltage at node 2)

We need another equation invovling V2 and V3 to solve them. If we look at the battery, we see that: $$V_3-V_2=6$$. We solve these 2 equations to get V2=-4V and V3=2V. Now we need to find voltage at node1, V1.

To do that, we look back to the equation summing all currents coming in and out of node1: $$12mA-i_1-i_2=0$$. We plugged in the expressions for i1 and i2 in terms of V. We have:$$0.012-\frac{V_1-V_3}{4000}-\frac{V_1-V_2}{2000}=0$$ Solving it and plugging in the values we got for V2 and V3, we have $$V_1=14V$$.

NOW there is a problem with these values for V. I tried testing them by plugging into the equation for summing currents at node2. $$i_2+i_5-i_3=0$$.

$$i_2=\frac{V_1-V_2}{2000}=9mA$$ $$i_5=\frac{V_3-V_2}{1000}=6mA$$ $$i_6=\frac{V_2}{6000}=-0.667mA$$

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  • \$\begingroup\$ At least your second equation is wrong. i1+i2=12mA \$\endgroup\$ – Chupacabras Sep 24 '17 at 4:27
  • \$\begingroup\$ How, there is conservation of charge at node1 \$\endgroup\$ – most venerable sir Sep 24 '17 at 15:10
  • \$\begingroup\$ The sum of currents that are flowing into the node must be equal to the sum of currents that are flowing out of the node. So, look at the node 1. i1 and i2 are flowing from this node, and 12mA is flowing to this node. That's why i1+i2=12mA, that matches your arrows. \$\endgroup\$ – Chupacabras Sep 24 '17 at 18:29
  • \$\begingroup\$ "imagine the branch containing the current source as a node", I've never seen this technique before. And I think it's wrong, because it brings you to say that \$ I_3 = -I_4 \$ while it's clearly \$ I_3 + I_4 = 12 \, mA \$. Where did you learn this approach? \$\endgroup\$ – joe electro Nov 13 '18 at 14:06
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When you have multiple sources in a linear circuit, whether these sources are independent or controlled (see here for more details), the superposition theorem is often your friend. In this example, I have chosen to determine the voltage at node 2. I will draw two small sketches where a) the 12-mA source is set to 0 A (open circuited) while the 6-V source remains and b) the 12-mA source is back in place while the 6-V source is reduced to 0 V (short circuited). You end up with the below sketch:

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For the upper sketch, the voltage at node 2 (that I call V21) is simply the current source multiplied by the parallel combination of the 6- and 3-\$k\Omega\$ resistances. Then, when the current source is open-circuited, the node voltage across the 6k resistance is the circulating current (\$\frac{6\;V}{6\;k\Omega+3\;k\Omega}\$) multiplied by the 6-\$k\Omega\$ resistance. The node voltage is the sum of these two sub-values. From there you have the voltage at node 3 (you subtract 6 V) and for node 1, a simple KCL gets you there. The below sheet shows the step and the resulting values are confirmed by the quick simulated dc point.

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  • \$\begingroup\$ How is the voltage at node 2=parallel combination of 6k and 3k multiplied by the current source? 6k and 3k are not in paralle because of the 1k. \$\endgroup\$ – most venerable sir Sep 24 '17 at 15:36
  • \$\begingroup\$ Because in superposition, the two sources are alternatively set to 0: the 6-V source being 0 V in the first definition, you replace it by a short circuit across the 1-k resistance which disappears and you see that the 6- and 3-k resistances come in parallel. This is not the voltage at node 2, but the intermediate value \$V_{21}\$ which summed to \$V_{22}\$ (determined when the current source is set to 0 A and disappears from the circuit) will give you the voltage at node 2. Please look back at the small intermediate sketches that I have drawn. \$\endgroup\$ – Verbal Kint Sep 24 '17 at 16:16
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For a start you can forget all about the 1 k resistor until right at the end - its current is ONLY determined by the 6 volt source so that one is easy to find (if you need it and I don't think you do).

Then, try using superposition instead. In case you didn't know, with superposition you analyse each source on its own and produce a small table of currents. To do this you could start with the current source and the voltage source is replaced with a short circuit. If you wanted, you could start with the voltage source but then the current source is open-circuited for that part of the analysis.

It looks fairly simple if you start with the current source because then the 4 k and 2 k resistors become in parallel and also the 3 k and 6 k resistors become in parallel leading to a net impedance across the current source of 1.333 k in series with 2 k. This then tells you the voltage across the current source and from therein you can easily work out voltages and currents for each resistor.

Step and repeat by open circuiting the current source and reinstating the voltage source. Note all currents in both scenarios (sign is important) and finally sum currents for each resistor from both scenarios and you are very close to your answer.

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  • \$\begingroup\$ What is wrong with my calculation? Is i3 =-i4? Is V2=-2V3? \$\endgroup\$ – most venerable sir Sep 24 '17 at 15:45
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    \$\begingroup\$ Your calculation even if it were right is not the method I would choose. That's why I'm not analysing your math because the analysis of someone's math makes it a math question and not an EE question. \$\endgroup\$ – Andy aka Sep 24 '17 at 17:40
  • \$\begingroup\$ If you look at the ground node, you can clearly see \$ I_3 + I_4 = 12 \, mA \$ so no, it's not true that \$ I_3 = -I_4 \$. \$\endgroup\$ – joe electro Nov 13 '18 at 14:08

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