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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\$ – user287001 Jun 14 at 9:00
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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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Well, let's solve this mathematically. We have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_7=\text{I}_1+\text{I}_3\\ \\ \text{I}_\text{i}=\text{I}_5+\text{I}_7\\ \\ \text{I}_9=\text{I}_1+\text{I}_2\\ \\ \text{I}_2=\text{I}_3+\text{I}_8\\ \\ \text{I}_4=\text{I}_\text{i}+\text{I}_8\\ \\ \text{I}_5=\text{I}_4+\text{I}_{10}\\ \\ 0=\text{I}_6+\text{I}_9+\text{k}\cdot\text{I}_3\\ \\ \text{I}_{10}=\text{I}_6+\text{k}\cdot\text{I}_3 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_\text{i}}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_2}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_1-\text{V}_2}{\text{R}_6} \end{cases}\tag2 $$

Now, I wrote and executed the following Mathematica code that solves your problem:

In[1]:=FullSimplify[
 Solve[{I7 == I1 + I3, Ii == I5 + I7, I9 == I1 + I2, I2 == I3 + I8, 
   I4 == Ii + I8, I5 == I4 + I10, 0 == I6 + I9 + k*I3, 
   I10 == I6 + k*I3, I1 == V1/R1, I2 == (V1 - Vi)/R2, I3 == Vi/R3, 
   I4 == (Vi - V2)/R4, I5 == V2/R5, I6 == (V1 - V2)/R6}, {Ii, I1, I2, 
   I3, I4, I5, I6, I7, I8, I9, I10, V1, V2}]]

Out[1]={{Ii -> ((R3 R4 R5 + R2 (R3 + R4) R5 + R3 (R4 + R5) R6 + 
        R2 (R3 + R4 + R5 - k R5) R6 + 
        R1 (R4 (R3 + R5) + 
           R2 (R3 + R4 + R5) + (R3 + R4 + k R4 + 
              R5) R6)) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  I1 -> ((R3 (R2 + R4) R5 + (-k R2 + R3) (R4 + 
           R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  I2 -> -(((R3 R4 (R1 + R5) + (k R1 + R3) (R4 + 
             R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
          R2 R4 R5 + (R1 + R2) (R4 + R5) R6))), I3 -> Vi/R3, 
  I4 -> ((R2 R3 (R1 + R5) + (R1 + R2) (R3 - 
           k R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  I5 -> ((R1 R3 (R2 + R4) + (R1 + R2) (R3 + 
           k R4) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  I6 -> -(((-R1 R3 R4 + R2 R3 R5 + 
          k (R1 R2 R4 + R2 R4 R5 + 
             R1 (R2 + R4) R5)) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + 
          R1 R4 R5 + R2 R4 R5 + (R1 + R2) (R4 + R5) R6))), 
  I7 -> ((R1 R2 R4 + R1 R2 R5 + R2 R3 R5 + R1 R4 R5 + R2 R4 R5 + 
        R3 R4 R5 + (R1 + R2 - k R2 + R3) (R4 + 
           R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  I8 -> -(((R1 (R4 (R3 + R5) + 
             R2 (R4 + R5) + (1 + k) (R4 + R5) R6) + (R2 + R3) (R5 R6 +
              R4 (R5 + R6))) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 +
           R2 R4 R5 + (R1 + R2) (R4 + R5) R6))), 
  I9 -> -(((R3 (R1 R4 - R2 R5) + 
          k (R1 + R2) (R4 + R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + 
          R1 R4 R5 + R2 R4 R5 + (R1 + R2) (R4 + R5) R6))), 
  I10 -> ((R3 (R1 R4 - R2 R5) + 
        k (R1 + R2) (R4 + R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + 
        R1 R4 R5 + R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  V1 -> (R1 (R3 (R2 + R4) R5 + (-k R2 + R3) (R4 + 
           R5) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6)), 
  V2 -> (R5 (R1 R3 (R2 + R4) + (R1 + R2) (R3 + 
           k R4) R6) Vi)/(R3 (R1 R2 R4 + R1 R2 R5 + R1 R4 R5 + 
        R2 R4 R5 + (R1 + R2) (R4 + R5) R6))}}

So, for \$\text{V}_1\$ we get:

$$\text{V}_1=\frac{\text{R}_1\text{V}_\text{i}(\text{R}_6(\text{R}_4+\text{R}_5)(\text{R}_3-\text{k}\text{R}_2)+\text{R}_3\text{R}_5(\text{R}_2+\text{R}_4))}{\text{R}_3(\text{R}_6(\text{R}_1+\text{R}_2)(\text{R}_4+\text{R}_5)+\text{R}_1\text{R}_2\text{R}_4+\text{R}_1\text{R}_2\text{R}_5+\text{R}_1\text{R}_4\text{R}_5+\text{R}_2\text{R}_4\text{R}_5)}\tag3$$

| improve this answer | |
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  • \$\begingroup\$ Thanks for the downvote, and especially thanks for the explanation why! \$\endgroup\$ – Jan Jun 14 at 19:20
  • \$\begingroup\$ I appreciate the answer Jan, but it doesn’t really answer my question. The question asked how you would arrive at one of the 6 choices of a node voltage equation for node A, seen under the circuit. PS: I am not the one who downvoted you. \$\endgroup\$ – Carl Jun 14 at 20:31

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