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I have the following circuit and I'm asked to calculate the equivalent conductance at the ends of the circuit.

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The textbook gives the following solution: \$G_{eq}=\frac{1-\beta}{R}\$.

My question is: why does the generator have its own resistance?

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The equivalent conductance is

$$G_{eq}=\frac{I_i}{V_i}$$ where \$I_i\$ is the input current and \$V_i\$ is the input voltage.

If the generator changes the result of this calculation, then of course it changes the equivalent conductance.

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The dependent current source doesn't have a resistance itself, but it affects the amount of current entering the circuit as a function of the voltage applied to the circuit.

So, the entire circuit has an effective resistance that depends on the behavior of the dependent current source, specifically the value of \$\beta\$. To see this for yourself, apply some unknown voltage \$V_x\$ to the two terminals on the left and figure out how much current will flow into the top terminal. If you divide the applied voltage by the resulting current you will have the effective resistance.

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