I am looking at some circuit solving using the superposition principle and I came across this one.

enter image description here

We need to find \$v_0\$. My thought would be to use superposition and voltage division in the following way:

\$v_0=V_{ss}\cdot\frac{2R_a}{2R_a+R}+2V_a\cdot \frac{2R_a}{2R_a+R}\$

We in both cases find the voltage across \$2R_a\$ with the voltage divider formula. However, it turns out the solution is actually:

enter image description here

So my denominator and numerator should switch places in the second part of my solution, but I don't understand why this is. Aren't we then finding the voltage across \$R\$? And does that have an effect on \$v_o\$?

I hope someone can clarify this for me.

  • \$\begingroup\$ The 2nd voltage divider (2Va etc.) is wrong. Redraw the circuit. \$\endgroup\$
    – Chu
    Dec 21, 2019 at 14:55

1 Answer 1



First I remove \$2V_a\$ source. So we left with this:

enter image description here

And we can find \$V_{O1}\$

$$V_{O1} = V_{SS} \frac{2R_a}{R+ 2R_a} = 0.081V = 81mV$$

Nowe we turn-off the \$V_{SS}\$ source and we are ending with this circuit:

enter image description here

And now we can solve for \$V_{O2}\$

As you can see this time \$V_{O2}\ = I\cdot R\$ or \$V_{O2} = 2V_a - I\cdot 2R_a\$


\$I = \frac{2V_a}{R+2R_a}\$ therefore:

$$V_{O2} =I\cdot R =\frac{2V_a}{R+2R_a}\cdot R = 2V_a \frac{R}{R+2R_a} = 1.167V $$

and finally

$$V_O =V_{O1}+V_{O2} = V_{SS} \frac{2R_a}{R+ 2R_a} + 2V_a \frac{R}{R+2R_a} =1.248V $$

Your circuit looks like this:

enter image description here

I mark the voltage drop by the arrows and the arrow tip is pointing the "positive" side.

And from KVL we have:

$$2V_a = V_{2R_a} + V_R$$ and also notice that \$V_o2\$ is a voltage drop betwenn \$V_A\$ node and GND.

$$V_A = 2Va - V_{2R_a} = 2Va - I*{2R_a} $$ or becouse \$V_A = V_R = I*R\$

Do you see it?

  • \$\begingroup\$ Hmm, I don't quite see why \$V_{o2}=I*R\$. Why isn't it \$V_{o2}=I*2R_a\$? \$\endgroup\$
    – Carl
    Dec 21, 2019 at 15:11
  • \$\begingroup\$ Because \$I*2R_a\$ is a voltage drop across \$2R_a\$ resistor only. \$\endgroup\$
    – G36
    Dec 21, 2019 at 15:29
  • \$\begingroup\$ Yeah, but isn't \$V_{o2}\$ determined by the voltage drop across \$2R_a\$? \$\endgroup\$
    – Carl
    Dec 21, 2019 at 15:33
  • \$\begingroup\$ The voltage you're looking for is across R, not across 2Ra. \$\endgroup\$
    – Chu
    Dec 21, 2019 at 15:50
  • 1
    \$\begingroup\$ I edit my answer. Is it any better now? Do you see why \$Vo2 = I*R = 2Va - I*2Ra\$ \$\endgroup\$
    – G36
    Dec 21, 2019 at 15:57

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