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I recently had an exam with the following multiple choice problem:

Consider the circuit below:

enter image description here

The node voltage equation for node A is which of the following?:

  1. \$ \frac{V_A-kV_a}{R_1}+ \frac{V_A-V_3}{R_2} + \frac{V_A-V_B}{R_6} = 0 \$
  1. \$ \frac{V_A+kV_A}{R_1} + \frac{V_A-V_3}{R_2} + \frac{V_A-V_B}{R_6} = 0 \$
  1. \$ \frac{V_A-kV_A}{R_1} + \frac{V_A+V_3}{R_2} + \frac{V_A+V_B}{R_6} = 0 \$
  1. \$ \frac{V_A+kV_A}{R_1} + \frac{V_A+V_3}{R_2} + \frac{V_A+V_B}{R_6} = 0 \$
  1. \$ \frac{2V_A}{R_1} + \frac{V_A+V_3}{R_2} + \frac{V_A+V_B}{R_6} = 0\$
  1. \$ \frac{V_A-V_3}{R_2} + \frac{V_A-V_B}{R_6} = 0\$

My attempt

This is how I would make the node voltage equation for node A:

\$ \frac{V_A}{R_1} + \frac{V_A-V_3}{R_2} + \frac{V_A-V_B}{R_6} + ki_1 = 0 \$

But as you can see, it doesn't match any of the options above and I can't figure out how I'm supposed to reach on of those solutions.

Can somebody help me out?

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  • \$\begingroup\$ fix i1 to i2 in your eq. \$\endgroup\$
    – user136077
    Commented Jun 14, 2020 at 9:00

1 Answer 1

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A tricky case.

V3=(I1-I2)R3. Va=-(I1)R1. With these you can fade the controlled current source from your equation.

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