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This is the circuit I am having trouble with

Data given: electromotive force of the ideal voltage source \$E=20V\$, power output of the ideal voltage source \$P_e=3.333W\$, current of the ideal current source \$I_s=4A\$, power output of the ideal current source \$P_i=56/3W\$. The task is to find the resistances \$R_1\$ and \$R_2\$.

So here are the steps that I've taken: \$P_e=E \cdot I\$, therefore, \$I_1=1/6(A)\$, and the resistance at the i.v.s. is therefore 120 ohms. With this we can calculate \$I_2\$, namely it is \$23/6 (A)\$. Then using Kirchhoff's laws, I found that \$R_2 = 1.6 \Omega\$ and \$R_1 = 277 \Omega\$.

I am having a trouble figuring out how deal with these sources, and I am not sure of my answers. When having an i.v.s. do I place the appropriate resistance in parallel with it? What about i.c.s.? In series? I've searched many sites online, but didn't find the explanations I was looking for. We are only allowed to use Kirchhoff laws here. Can you recommend some good, valid sources to learn from?

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If the sources are ideal, you don't have to put any internal resistance on them. Also, I think you are mistaken in assigning a resistance value at the voltage source. It is true that you have a voltage and a current, but the voltage is constant while the current depends on the circuit.

In order to solve the problem, you correctly found the current on the voltage source, and since the text says that it produces 3.333W, it must be the inverse of the arrow.

Therefore, if you apply the KCL at the upper node, you get that \$I_2 = I_S - I_1\$, but since I1 is negative you have to sum the absolute values. I actually don't now how you got the 23/6 A value.

Since you know the voltage on Is, you can calculate the voltage on the two resistances, and since you know all the currents you can obtain the resistance values.

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The basic idea in a problem like this is to find the voltage at each node and the current in each branch. Once you do that, you can calculate the resistances using Ohm's Law. It's kind of like a crossword or Sudoku puzzle where you fill in the gaps based on what you already know. Often, you'll only have voltages and currents, but here you're given power figures as well. You can use \$P = V*I\$ for those.

To start with, you have to define one node as ground (0 V). Node B is on the negative end of all of the supplies, so it's a good choice. You're given the current and power of \$I_S\$, so you can get the voltage across it. That gives you the voltage of node A. Likewise, you're given the voltage and power of \$E\$, so you can get the current through it. From there, you can get the voltage across R1 and calculate its resistance. You'll need to use KCL to figure out R2. I'll leave that to you as an exercise.

Be sure to pay close attention to the signs of the voltages and currents!

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  • \$\begingroup\$ So I have managed to come up with these solutions R1=120 ohms, and R2=76.46 ohms. \$\endgroup\$ – Shemafied Dec 15 '14 at 18:11
  • \$\begingroup\$ Neither of those is correct. R2 is going to be very small. What did you get for the VA and VR1? \$\endgroup\$ – Adam Haun Dec 15 '14 at 18:55
  • \$\begingroup\$ I got Va=4.66 Volts, and Vr1=20V(since it's in the same branch as E). But I kinda could've guessed I was wrong all along. Help me out. \$\endgroup\$ – Shemafied Dec 15 '14 at 19:01
  • \$\begingroup\$ Your value for VA is correct, but VR1 is not 20V. Remember, voltage is only shared in parallel, not in series. 20V is the voltage across E. R1 has its own voltage drop. To get VR1, you have to go from node A through the voltage source. This gives VR1 = VA - 20V. The answer is a negative voltage, which is expected since the current is flowing from ground (node B) towards E. \$\endgroup\$ – Adam Haun Dec 15 '14 at 19:30
  • \$\begingroup\$ Achh, now I see. It's even more obvious if I do a loop from B to B. \$\endgroup\$ – Shemafied Dec 15 '14 at 20:51

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