# Superposition principle and voltage division

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

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:

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.

• The 2nd voltage divider (2Va etc.) is wrong. Redraw the circuit.
– Chu
Dec 21 '19 at 14:55

Hmm.

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

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:

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\$$

Where:

$$\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

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

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?

• Hmm, I don't quite see why $V_{o2}=I*R$. Why isn't it $V_{o2}=I*2R_a$?
– Carl
Dec 21 '19 at 15:11
• Because $I*2R_a$ is a voltage drop across $2R_a$ resistor only.
– G36
Dec 21 '19 at 15:29
• Yeah, but isn't $V_{o2}$ determined by the voltage drop across $2R_a$?
– Carl
Dec 21 '19 at 15:33
• The voltage you're looking for is across R, not across 2Ra.
– Chu
Dec 21 '19 at 15:50
• I edit my answer. Is it any better now? Do you see why $Vo2 = I*R = 2Va - I*2Ra$
– G36
Dec 21 '19 at 15:57