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The question attached in the image : the question

I have found the I in the left loop I=1.19A then I found the vo=-11.9

Here two questions in my mind :

1)does that mean that the voltage across the top resistor equalls to the vo or it is the same with positive since the polarity is the oopisite ?

2)how can I determine the polarity of across the current source?

Thanks a lot.

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Your first question is a bit unclear. The marking for \$V_O\$ means that there is a voltage named "\$V_0\$" that is the voltage across the resistor, and it is assumed to have a higher voltage on the left side of the resistor.

For the second question, I assume that you mean the polarity of the voltage across the current source. Whenever you are looking for a voltage, it is a good bet that KVL will help.

You don't show your work and I won't say anything about your answers, but my guess is that you will need to write a couple of independent equations then solve them for \$V_O\$.

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  • \$\begingroup\$ I mean if he asked me to find the voltage across the resistor (at the top) it will be -vo ? So if vo= -11.9 v=-vo=11.9 @ElliotAlderson \$\endgroup\$ – Mhd Ghd Sep 25 '19 at 14:50
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In general, set up your KVL and KCL equations with the assumptions that quantities have a "higher" voltage on the "+" end of the resistor or voltage source and that currents are positive in the direction of the arrow. By being consistent with this, your units should always make sense (positive voltages across resistors, etc) as long as you did your systems of equations correctly.

As an aside, once you pick a KCL loop equation convention (writing resistors as drops and sources as rises or vice versa) stick to it slavishly and you shouldn't have any problems regardless of which method you pick.

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Because you get a negative \$V_O\$ value which means that the voltage at the right side of the top resistor is higher by 11.9 volts than the left side.

This also means that in "reality" you have this situation:

enter image description here

And you can find the voltage across the current source do the KVL.

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  • \$\begingroup\$ I mean if he asked me to find the voltage across the resistor (at the top) it will be -vo ? So if vo= -11.9 v=-vo=11.9 \$\endgroup\$ – Mhd Ghd Sep 25 '19 at 14:55
  • \$\begingroup\$ All you need to do is to find the Vo voltage and power absorbed by the VCCS. \$\endgroup\$ – G36 Sep 25 '19 at 15:00
  • \$\begingroup\$ I know but what if he asked for the voltage across that resistor ? \$\endgroup\$ – Mhd Ghd Sep 25 '19 at 15:03
  • \$\begingroup\$ But if you know the Vo voltage you also know the voltage across that resistor. \$\endgroup\$ – G36 Sep 25 '19 at 15:05
  • \$\begingroup\$ My question is if the old vo = -11.9 the v across the resistor is it v=-vo=11.9 ? \$\endgroup\$ – Mhd Ghd Sep 25 '19 at 15:19

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