I'm trying to figure out \$V_x = V_x' + v_x''\$

For \$V_x'\$, I had open the current source, and tried to solve using KVL.

Obviously, \$i_2 = 0.1v_x\$. The equation I concocted was \$-10 + 24I -.4V_x = 0\$. However, I am stuck at this equation.

For \$V_x''\$, I shorted the voltage source, and solved by simplifying the two parallel resistors to \$3.33\Omega\$. Thus, I could just do nodal analysis at \$V_x\$. is this correct?



1 Answer 1


Despite what most textbooks claim, superposition of dependent sources is valid if done correctly.

There are three sources in this circuit so there will be three terms in the superposition.

For the first term, the two current sources are zeroed (opened) so \$V_x\$ is given by voltage division:

\$V_x = 10V \cdot \dfrac{4}{4 + 20} = \dfrac{5}{3}V\$

For the second term, the voltage source is zeroed (shorted), so the two resistors are now in parallel, and the 2A source is activated. Thus:

\$V_x = 2A \cdot 4\Omega || 20 \Omega = \dfrac{20}{3}V\$

Since the third, dependent source is in parallel with the 2A source, the last term has the same form:

\$V_x = 0.1 V_x \cdot 4\Omega || 20 \Omega = \dfrac{1}{3}V_x\$

Now, it's crucial at this point to not try and solve the previous equation (you'll only get \$V_x = 0\$ if you do.)

Rather, proceed with the superposition sum and then solve.

\$V_x = \dfrac{5}{3}V + \dfrac{20}{3}V + \dfrac{1}{3}V_x\$

Grouping terms:

\$V_x (1 - \frac{1}{3}) = \dfrac{25}{3}V\$


\$V_x = 12.5V\$

  • \$\begingroup\$ I get your solution, but I think your answer may not be explaining your method. Your first equation has all of the nitty gritty in it, it might be worth expanding on that, solving the actual equation after is probably less important to the user. \$\endgroup\$
    – Kortuk
    Dec 2, 2012 at 4:47
  • \$\begingroup\$ @Kortuk, I've added additional detail. \$\endgroup\$ Dec 2, 2012 at 5:03

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