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So imagine that I have a circuit that only contains a resistor in parallel with a dependent source of current. The resistor has a resistance of $$3 k\Omega$$ and a current I. The dependent source of current has a current of 2I. Now I want to know the Norton resistance of this circuit. The correct answer is $$1 k\Omega$$

However I'm not getting there.

My attempt

Because $$R_N = \frac{U_0}{I_{sc}}$$

I attempted to calculate both. However I'm getting to a voltage of 0V. I might be thinking incorrectly. What I'm thinking is that well the resistance and the source have the same voltage which is equal to the voltage I want to obtain. So $$U= RI $$

Considering my node

$$ I + 2I = 0 $$ so I = 0 and the U = 0. I might be assuming something wrong, if someone could please clarify me. Thanks!

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    \$\begingroup\$ If the current source produces 2I and, half that current flows through a 3 kohm resistor, then the other half of the current flows through another 3 kohm resistor. \$\endgroup\$ – Andy aka Nov 9 '16 at 19:51
  • \$\begingroup\$ I can't make sense of a circuit that only "contains a resistor in parallel with a dependent source of current" and where the dependent current source is \$2\cdot I\$. Assuming the observed current is the current in the resistor, the only two solutions are \$I=0\$ and \$I=\infty\$, so far as I can tell off-hand. \$\endgroup\$ – jonk Nov 9 '16 at 20:53
  • \$\begingroup\$ Please add a schematic of your circuit to make your question clear. For example, you haven't said whether the dependent source and the controlling current are oriented in the same or opposite directions. \$\endgroup\$ – The Photon Nov 9 '16 at 21:29
  • \$\begingroup\$ BTW 0/0 is undefined and hence compatible with "correct" 1kohm result. This is only not the way to get the result. \$\endgroup\$ – carloc Nov 9 '16 at 22:07

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