Can someone explain how is the current source on the left side below is transformed into a voltage source?:

enter image description here

I thought a current source is shown with resistor in parallel. I'm not used to see the conductance G with a current source. How can the circuit on the right side can be derived from the one on left side step by step?


I tried to plug 1/R1 to G1 and this is what I reached:

enter image description here

Where am I wrong?


1 Answer 1


If you substitute G with 1/R, then the current source becomes a voltage source of \$I_1 \cdot R\$ as per ohms law and the voltage source is in series with a resistor of value R as per Thevenin's theorum.

  • \$\begingroup\$ Please see my edit I substituted G1 with 1/R1 but I obtain different result. I know the result seems odd because of the units but when I follow Thevenin Norton I reach that conclusion. I hope you can help with the correct derivation. \$\endgroup\$
    – user1999
    Sep 27, 2018 at 14:17
  • \$\begingroup\$ @newage2000 Your expression \$i_1*(\frac{1}{R_1})\$ should be \$i_1 / \frac{1}{R_1}\$ - think of the dimensions in your equations. \$\endgroup\$
    – awjlogan
    Sep 27, 2018 at 14:21
  • \$\begingroup\$ Yes but G1 is 1/R1 so circuit becomes I1 is in parallel with (1/R1) Ohm. This look weird but it happens like that when I plug in. And then follow the Thevein V1 = I1*(1/R1). I really dont get it.. \$\endgroup\$
    – user1999
    Sep 27, 2018 at 14:28
  • \$\begingroup\$ @newage2000 the resistor shown as having a value of G1 in the left circuit is measleading because G1 is a conductance value and it will have a resistance of R (= 1/G1). That does not mean the resistor in parallel with the current source has a resistance of 1/R. \$\endgroup\$
    – Andy aka
    Sep 27, 2018 at 16:03
  • \$\begingroup\$ @Andyaka I understand we should see it as a Resistor component with a G1 conductance value \$\endgroup\$
    – user1999
    Sep 27, 2018 at 17:11

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