Transfrom scheme into eq. scheme with one EMF and one resistor.

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    Emf Resistor
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The new circuit is pic.1

What do I need to do with this two branches?OH!It's not R2.It's R3
 
Attachments
  • Circuit schematic with a photodiode-style symbol drawn as a box with two arrows pointing inward, resistor R2 in a vertical branch, resistor R1 in a lower series branch, and two terminals on the right.
    scheme3.JPG
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One of the wires is redundant. You can't have less resistance than no resistance!
Remove one of them and solve for the net resistance across a-b.
 
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Rn=(R1*R3)/(R3+R1)

And new circuit is
 
Attachments
  • Left: two resistors R1 and R3 connected in parallel between terminals a and b. Arrow points right to a single resistor Rn between terminals a and b.
    tt.JPG
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Okay. Now put this Thevenin resistance in series with the Thevenin voltage source that you calculated before. That should be your final circuit, the Thevenin Equivalent for the original.
 


The final circuit is...
E=U=1.6 V
Rn=0.875 Ohms
 
Attachments
  • Circuit schematic with a resistor labeled Rn in series with an EMF source labeled E, connected to two terminals labeled а and б on the right.
    final.JPG
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Yes. Be sure that you pay attention to the polarity of the voltage source. Remember that you found a negative value for E, making point a negative with respect to b.
 


Remember that you found a negative value for E, making point a negative with respect to b.

I need to change the direction of E?
But E>0 because I3>0 & R3>0.

One of the tasks is done(I don't sure in this tense form)(another task is node potential method(only 2 nodes in the circuit so I try to make on my own(with rare success)).Thanks for your help.