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Circuit design

The circuit has 2 voltage sources one of which is 0.5UR2, one current source, resistor elements R1 & R2, the load is RT 10 ohms.

Having the following circuit, I have used the Kirchhoff's circuit laws to find the current I1, which is equal to (1-05*UR2)/35.

Solution

I am stuck here. How should I proceed?

After finding out that R2 is not a linear element and that UR2 = 2I1*3 ohms, the solution I got to is the following:

The final solution

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  • \$\begingroup\$ The circuit symbols used are not entirely compliant with the standard ones and this can be misleading. \$\endgroup\$
    – Franc
    Commented Mar 16 at 14:23
  • \$\begingroup\$ @Franc I undestand you but I am unable to change the design. We're using a diffirent standard. If you have particular questions which symbol represents what I could answer them. \$\endgroup\$
    – Chum Chum
    Commented Mar 16 at 14:56
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    \$\begingroup\$ Why do you say that R2 is a non-linear element? \$\endgroup\$
    – G36
    Commented Mar 16 at 15:03
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    \$\begingroup\$ No, R2 is just an ordinary linear resistor. The polarity markings on R2 resistor refers to the "positive" voltage generated by a voltage controlled voltage source 0.5xUR2. And this dependent voltage source is controlled by the voltage drop across R2 resistor. \$\endgroup\$
    – G36
    Commented Mar 16 at 20:28
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    \$\begingroup\$ Did you manage to solve it using my tip? $$U_{R_2} = 2\:I_1 \:3\Omega$$ $$I_1 = \frac{1V - 0.5U_{R_2}}{35\Omega} = \frac{1V - 0.5( 2\:I_1 \:3\Omega)}{35\Omega} $$ Because you are very close to get the solution. \$\endgroup\$
    – G36
    Commented Mar 17 at 8:13

2 Answers 2

Reset to default
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Topological study of the network. From this analysis, obtain the branch voltages, currents and powers using the node voltage method:

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Instead, if we set Vg2=-0.5VR2, the currents and voltages take on different values, what will be the right solution?: enter image description here

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  • \$\begingroup\$ How can this provide any help to the TS? Is this a solution or what? \$\endgroup\$
    – G36
    Commented Mar 17 at 10:06
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    \$\begingroup\$ @G36 What I carried out is the study of the network based on the rules of electrical engineering which is based on the topology of electrical networks. It serves to give an idea of how to do a serious study of the network and that ultimately all electrical network simulators use. Maybe you are not interested but it may be of interest to the requesting user. \$\endgroup\$
    – Franc
    Commented Mar 17 at 10:17
  • \$\begingroup\$ @G36 Since this is a work based on my experience in this regard, it may not be free from errors. \$\endgroup\$
    – Franc
    Commented Mar 17 at 10:24
  • \$\begingroup\$ OK, but your answer seems to be wrong. If I do for example a quick nodal analysis I'm getting that the voltage at node 2 is 0.868421V. \$\endgroup\$
    – G36
    Commented Mar 17 at 10:26
  • \$\begingroup\$ It would be very helpful to know what causes the error. \$\endgroup\$
    – Franc
    Commented Mar 17 at 10:29
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Editing my answer because I see two solutions depending on how the non-standard voltage source symbol is interpreted for voltage polarity.

To G36, I believe that current flowing out of positive terminal is in most cases negative, but not necessarily defined as negative.

two solutions

Both solutions in spice:

solution 1 solution 1

solution 2 solution 2

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  • 1
    \$\begingroup\$ But this is not the correct solution. \$\endgroup\$
    – G36
    Commented Mar 17 at 7:57
  • \$\begingroup\$ I1 can't equal 1/38. Then KVL equation won't hold true. \$\endgroup\$
    – Reynaldo Gonzalez
    Commented Mar 17 at 19:21
  • \$\begingroup\$ Why do you think that KVL will not hold true? \$\endgroup\$
    – G36
    Commented Mar 17 at 19:34
  • \$\begingroup\$ made a rounding error, it does hold true, my bad \$\endgroup\$
    – Reynaldo Gonzalez
    Commented Mar 17 at 19:43
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    \$\begingroup\$ Also, in your CCCS (F1) in LTspice should have V1 -2. Because the current that is flowing out from the positive terminal of a V1 source is negative by definition. \$\endgroup\$
    – G36
    Commented Mar 17 at 19:47

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