The question asks to determine V_x and I_y.

I tried to find it by using KVL in to lower LHS mesh and KCL in the middle node. But, my final value is a contradictory statement(such as I_x equal all real numbers). Any help would be appreciated.

Using KCL in the central node, the current that passes through the 10 ohm resistor is -5-0.1V_x+y(where y is the current that goes through V_x/20 ohms resistor). Then I used KVL in lower LHS mesh, so -50-10(-5-0.1V_x+y)-V_x=0 As you can see V_x cancel and 10y=0 is left. However V_x=20y.

  • \$\begingroup\$ I haven't analyzed the circuit, but "I_x equals all real numbers" is not actually a contradictory statement. A contradictory statement (like "X=0 and X=1 simultaneously") means either the circuit is non-physical or you made a calculation error. Your result could mean that the circuit is non-physical, but it could just mean that you haven't used enough information from the circuit. \$\endgroup\$ – Justin Sep 24 '13 at 13:37
  • \$\begingroup\$ Its implications in my equation(which may be wrong) do form a contradiction. I_x-I_x+10x=0 -->x=0(where x is the current at V_x) which is not possible since there is a current there. \$\endgroup\$ – user29568 Sep 24 '13 at 13:40
  • \$\begingroup\$ I think your solution may be correct. I solved it using another method (KCL at the center node and the top node), and I also calculate that the current through the bottom 20 ohm resistor is 0. What makes you think it must be non-zero? \$\endgroup\$ – Justin Sep 24 '13 at 14:08
  • \$\begingroup\$ I have the answer given to be Vx = 70 V, Iy = 3 A. Maybe there is something wrong with the question. Random Question: Is it possible for a larger current to pass through an independent current source. For example say there is a 5 A source, the current coming from behind it MUST always be 5 A? \$\endgroup\$ – user29568 Sep 24 '13 at 14:12
  • \$\begingroup\$ I agree something is wrong with the answer Vx = 70V (though Iy = 3A is correct). An ideal current source (independent or dependent) forces the current to equal the specified value -- no more, no less. Same for an ideal voltage source. \$\endgroup\$ – Justin Sep 24 '13 at 14:26

KCL at the center node gives:

$$ \frac{-V_x-50}{10} + 5 + 0.1V_x + \frac{-V_x}{20} = 0 $$

Which is only an equation of \$V_x\$ and can be immediately solved to get \$V_x = 0\$.

KCL at the top node (with voltage \$V_T\$):

$$ \frac{V_T - 50}{20} - 5 + I_Y = 0 $$

Ohm's law on the 25 ohm resistor gives \$ I_Y = \frac{V_T - 5I_Y}{25} \implies V_T = 30I_Y\$. Using this in the preceeding equation gives:

$$ \frac{30I_Y - 50}{20} - 5 + I_Y = 0 \implies I_Y = 3 \text{A}$$

So, along with your own answer and Alfred Centauri's answer, we can conclude that your book's answer of \$V_X = 70 V\$ is incorrect.


This problem is, I think, a good case for using superposition with dependent sources.

Zero all sources (including dependent sources) except for the 50V DC source.

Then, by voltage division and Ohm's Law:

\$V_x = -50V \dfrac{20}{10 + 20} = -33.33V\$

\$I_y = \dfrac{50}{20 + 25} = 1.111A \$

Now, zero that source and enable the 5A current source.

Then, by inspection and current division:

\$V_x = 5A \cdot 20\Omega || 10 \Omega = 33.33V\$

\$I_y = 5A \cdot \dfrac{20}{20 + 25} = 2.222A\$

Now, zero that source and enable the 5Iy source.

Then, by inspection:

\$V_x = 0V \$

\$I_y = -\dfrac{5I_y}{20 + 25} = -0.1111 \cdot I_y\$

Finally, zero that source and enable the 0.1Vx source.

Then, by inspection:

\$V_x = 0.1V_x \cdot 20\Omega || 10 \Omega = 0.6667 V_x \$

\$I_y = 0A \$

The sums of the associated four terms give the answers:

\$V_x = -33.33V + 33.33V + 0V + 0.6667V_x \rightarrow V_x = 0V \$

\$I_y = 1.111A + 2.222A +0A - 0.1111 \cdot I_y \rightarrow I_y = 3A\$

  • \$\begingroup\$ Interesting approach but can you explain why my way is wrong? \$\endgroup\$ – user29568 Sep 24 '13 at 14:39
  • \$\begingroup\$ @Justin, I just caught a silly sign error in my first equation so we now agree. \$\endgroup\$ – Alfred Centauri Sep 24 '13 at 14:55

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