Circuit theory basic problem

Question

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.

Answer 1

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

2. 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.

3. 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

• 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

• 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.

Answer 2

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:

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

Answer 3

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:

   

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.

   

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}\$.

   

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:

   

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.