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What approach do I have to take to find Vs. I don't really want an answer to this question but I want to know how to solve these circuits.

I'll first tell the approach I tried. I took KVL for all the loops, and then I was quite confused with the current directions I assumed. Please help me in figuring out the approach.

  • 2
    \$\begingroup\$ You can't "find" Vs because it is a voltage source. If you meant it to be a developed potential due to the rest of the circuit then you need to draw a different symbol or names the nodes or some other method. \$\endgroup\$
    – Andy aka
    Dec 3, 2023 at 11:38
  • \$\begingroup\$ I'll post the question. this is not homework, this is just strengthening the basics. I'm doing the book called basic engineering circuit analysis by erwin \$\endgroup\$
    – Fredrick
    Dec 3, 2023 at 11:49
  • \$\begingroup\$ I clearly don't know how to calculate the power supplied by 3A current source. Please help me with the basic approach \$\endgroup\$
    – Fredrick
    Dec 3, 2023 at 11:51
  • 1
    \$\begingroup\$ @Fredrick If a current source is supplying 3 A then it must have 4 V across it for 12 W. You should know that 3 A * 4 V = 12 W. It's basic. Do you not see this? \$\endgroup\$ Dec 3, 2023 at 12:02

3 Answers 3


I don't really want an answer to this question but I want to know how to solve these circuits.

  1. List the information that you have. For starters, this is the schematic and this data point:
  • 3A current source produces 12W
  1. See if the information you have allows you to determine some voltage or currents on the schematic. via Ohm's law, KCL, KVL, etc
  • 3A current source produces 12W, P=VI, therefore it has 4V across it (mind the polarity of the voltage, which is given by the mention it is producing power and not consuming).

  • 3A current source is in series with 4 ohm resistor, V=RI, therefore there's 12V across the resistor.

  • 5 ohm resistor is in parallel with 10V source, so there's 2A current through it.

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  1. Add the new information to your list and repeat until you solved the circuit.
  • Two voltages on top are in series, so we get voltage on the 4 ohm resistor
  • Thus we get current in the 4 ohm resistor

enter image description here

  • Now we know currents in 2 wires to the 3 wire node on top of the 10V source, apply KCL to get current through the third wire.
  • 3 components form a loop and 2 voltages are known, so apply KVL to get voltage on the third

enter image description here

  • etc

Basically, write everything on the schematic and every time you spot: known voltages in series, known currents in a branch, nodes with current known in all wires except one, loops with voltage known in all elements except one, etc... then apply the corresponding rule, and every time you do so you get a new piece of information.

  • \$\begingroup\$ But I have another doubt, if the 10 V is across the 5 ohm resistor, does the current value also depend on rest of the circuit as well? Let's say a voltage supply of 8 V is connected across the 4 ohm resistor near to the current source of 3 A in the figure, then according to I = V/R, through the 4 ohm resistor, 2 A current should flow. But you see a 3A current source is connected in series with the 4 ohm resistor and exclusively across the 4 ohm resistor, we have a 8 V supply, will that not affect the current value flowing through the resistor 4 ohm? \$\endgroup\$
    – Fredrick
    Dec 4, 2023 at 15:26
  • \$\begingroup\$ An ideal voltage source (as shown in your schematic) sets the voltage, so the 5V resistor that is in parallel with the 10V source will have 10V across it and thus 2A flowing through it. However, current through the voltage source itself depends on the rest of the circuit. So in your example (8V source across 4 ohm resistor) putting a current source in series with these will not change the current in the resistor, but it will set the current in the voltage source. Basically the ideal voltage source either outputs or absorbs whatever current is necessary to set the voltage. \$\endgroup\$
    – bobflux
    Dec 4, 2023 at 17:02
  • \$\begingroup\$ Why is the direction of 2A current flowing through the 4 ohm resistor parallel to 3A current source and 4 ohm resistor is positive direction, if the Voltage across the 3A current source and the 4 ohm resistor is 4V - 12 V = -8 V (8V on other direction), then is the current direction right? \$\endgroup\$
    – Fredrick
    Dec 4, 2023 at 17:53

We need to find the voltage drops across the current source. V = 12W/3A = 4V. Now we can apply the KVL and find the voltage at all the nodes in the circuit (one node, the left one).

We also see that the voltage at the left node is 10V + 12V - 4V = 18V (KVL) Next, we can calculate all the currents in this node. Thus, we end up with this:

enter image description here

And now with the help of KCL, we can find Ix current and Vs voltage value by using KVL.

  • \$\begingroup\$ Thanks for the answer, But how did come up with the current through the 6 ohm resistor as 3A? \$\endgroup\$
    – Fredrick
    Dec 4, 2023 at 15:48
  • \$\begingroup\$ @Fredrick I used the Ohms law 18V/6Ω = 3A \$\endgroup\$
    – G36
    Dec 4, 2023 at 16:10

I really love the answers you already have (+1 to both) and I really think that what they've written is more than enough to get the job done for you. But I may as well offer the thought process of yet another person. You've got two thrown at the wall to see if either sticks. So here's a third.

  1. Usually, the very first thing I do is figure out which node to call ground. And this may mean (not often) that if there is already a ground called out I might change it to simplify my analysis. You get to call any one node ground. So use that power well! In this case, I'd do the same as most everyone else would but also just as the schematic almost already suggests:

    enter image description here

  2. The next thing I do is label parts (if not labeled) and label nodes (if not labeled.) It's always better to get into the practice of working with symbols than specific values unless your only goal is a quick answer on a timed test. Otherwise, label everything under the sun and use the labels.

    enter image description here

  3. The next thing I do is to remove stuff that is getting in the way. And now you know that \$R_4\$ is a complete waste of time worrying over. It's an entirely separate circuit and for now (but not later) should be dead to you except to note that \$I_{R_4}=2\:\text{A}\$.

    enter image description here

  4. It's usually only now that I go back to see what information was provided on the problem. I've cleaned up a few things and managed to completely eliminate one part from my view. One less thing to be confused about, I say! But this is the point where I start to apply what's been added.

  5. The current source is providing \$12\:\text{W}\$ of power. So \$V_3=V_2+\frac{12\:\text{W}}{I_1=3\:\text{A}}=V_2+4\:\text{V}\$!

  6. \$\mid I_{R_5}\!\!\mid=I_1\$ so the voltage drop across \$R_5\$ must be \$\mid V_{R_5}\!\!\mid=I_1\cdot R_5=12\:\text{V}\$, with the (+) side being the one pointed at by the current arrow. it then follows that \$V_1=V_3-12\:\text{V}\$ and \$\therefore V_3=22\:\text{V}\$.

  7. From #5 (\$V_3=V_2+4\:\text{V}\$) and #6 (\$V_3=22\:\text{V}\$), find \$V_2=V_3-4\:\text{V}=18\:\text{V}\$.

  8. Redraw schematic with this new information and add the following now-known currents: \$I_{R_2}=\frac{V_2-V_1}{R_2}=2\:\text{A}\$ and \$I_{R_3}=\frac{V_2-0\:\text{V}}{R_3}=3\:\text{A}\$. The sum of the currents leaving node \$V_2\$ is \$I_1+I_{R_2}+I_{R_3}=8\:\text{A}\$. So add that current, as well, onto the schematic:

    enter image description here

  9. The (+) side of \$R_1\$ is where the tip of the current arrow points into (not out of), so this means the (+) side is on the left of \$R_1\$. Therefore, \$V_s=V_2+R_1\cdot 8\:\text{A}=42\:\text{V}\$.

  10. The last question is about the power in/out of \$V_1\$. Here, you want the current into/out-of \$V_1\$. There is \$I_1\$ entering the \$V_1\$ node, plus \$I_{R_2}\$ also entering the \$V_1\$ node. But \$I_{R_4}\$ exiting the \$V_1\$ node (#3 above.)

    So the resulting magnitude is \$3\:\text{A}+2\:\text{A}-2\:\text{A}=3\:\text{A}\$.

    But this magnitude is going into \$V_1\$ and not leaving out of it. So it is absorbed power. \$V_1\$ is being charged, so to speak.

    This means we use a minus sign to represent its power. So the result is that the power for \$V_1\$ is \$10\:\text{V}\cdot -3\:\text{A}=-30\:\text{W}\$.

Those are the micro-steps I'd mentally take if forced to explain things, in detail. However, what really happened for me was that I could quickly see that there were just \$8\:\text{V}\$ between \$V_1\$ and \$V_2\$ and since I knew that \$V_1=10\:\text{V}\$ I then immediately knew that \$V_2=18\:\text{V}\$. From there it was just a matter of seeing the currents, adding them up, and applying the net result at \$V_2\$ as a current in \$R_1\$ to see that \$V_s=42\:\text{V}\$. And then a similar mental sum to get the power into/out-of \$V_1\$.

It wasn't my intent to provide specific values. I was just trying to do what you asked, show you a process I'd follow. You already had the results, anyway. So including them above is just illustrative only. The main thrust of the above is the approach. Not the quantities, so much. They are just there to help clarify anything I failed to write more about, is all.

  • \$\begingroup\$ Hi @periblepsis, Thank you, But in the second step, you just put V3 at a point, but V3 referenced to which point. I got confused. If there is a usual standard reference please help me understand. \$\endgroup\$
    – Fredrick
    Dec 4, 2023 at 17:31
  • \$\begingroup\$ @Fredrick That point is another node. It's useful to add so that I can refer to it by name when taking. That's all. \$\endgroup\$ Dec 4, 2023 at 18:11

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