Voltage Difference in Superposition Theorem

Question

I have to find voltage across R₅ using superposition theorem but the problem is that I have a voltage difference of 4.16V,

Solution:-
Shorting battery V_S1

R₃₄ =8/3Ω =2.67Ω (Solved R₃ & R₄ in parallel)
R₂₃₄ =41/3Ω =13.67Ω (Solved R₃₄ & R₂ in series)
R₁₂₃₄ =205/56Ω =3.66Ω (Solved R₂₃₄ & R₁ in parallel)

Using Voltage Divider Formula on R₅
(1)V₅= 6.56V (Voltage when V_S1 is shorted)

Shorting Battery V_S2

R₃₄ =8/3Ω =2.67Ω (Solved R₃ & R₄ in parallel)
R₂₃₄ =41/3Ω =13.67Ω (Solved R₃₄ & R₂ in series)
R₂₃₄₅ =287/62Ω =4.62Ω (Solved R₂₃₄ & R₅ in parallel)

Using Voltage Divider Formula on R₂₃₄₅
Therefore, (2)V₅=V₂₃₄₅= 2.40V(Voltage when V_S2 is shorted)

Since the flow of voltage is opposite on R₅ the resultant voltage(V₅) can be obtained by taking the difference of the voltages obtained by shorting the batteries.
Resultant V₅ =(1)V₅ - (2)V₅
Resultant V₅ = 6.56 - 2.40
Resultant V₅ = 4.16V
x------------------------------------------------------------x--------------------------------------------------------------------x

According to my solution after taking the difference at the end I'm obtaining a value of 4.16V of V₅ but according to the given answer it says that the answer is 2.40V. I want to know that if there is something wrong with my solution or the answer provided is wrong?.

Answer 1

It seems as though you may need to prove something, given that your answer is correct and apparently a given "correct" answer is actually wrong.

To achieve a counter-proof, it's best to use a series of simple and unarguable transitions of the circuit that lead inevitably to the answer you have. This means using very basic steps, one at a time; steps that anyone can follow and will not be able to argue with:

^{simulate this circuit – Schematic created using CircuitLab}

Assuming you can get whomever the individual is that you need to convince to read this, have them follow along with the above steps:

1. Replace \$R_3\$ and \$R_4\$ with their parallel equivalent, \$R_X\$.
2. Replace \$R_X\$ formed in step 1, combined with \$R_2\$, to form a new series equivalent, \$R_Y\$
3. Redraw the resulting schematic in step 2, into a more readable form. It's the same circuit. Just drawn differently. (Getting rid of distracting wiring often helps to clarify a situation.)
4. Form the Thevenin source equivalent for the pair of resistors, \$R_1\$ and \$R_Y\$ (formed in steps 1 and 2) to form a new source voltage and source resistance as shown in the final diagram above, \$V_\text{TH}\$ and \$R_\text{TH}\$.

Once you have arrived at step 4, the rest is fairly easy.

$$\begin{align*} I_\text{TOTAL} &=\frac{10\:\text{V}-V_\text{TH}}{R_5+R_\text{TH}}\\\\ V_{R_5}&=I_\text{TOTAL}\cdot R_5\\\\ &=\left(10\:\text{V}-V_\text{TH}\right)\cdot\frac{R_5}{R_5+R_\text{TH}}\\\\ &= 4.16237\approx 4.16\:\text{V} \end{align*}$$

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When trying to convince someone, it's better to walk them through a very simple and easy to follow process and to avoid using general, powerful and abstract tools to get there (which may not be as crystal clear in their minds.) Just keep it very basic and I think you will get them to admit their error. It's inescapable.

Answer 2

I do not see any mistake in your solution.

So to check it I will use the nodal analysis and solve for the voltage at the node where R1, R5, and R2 are connected.

$$\frac{V_X}{R_{2}+R_3||R_4} + \frac{V_X - 5V}{R_1} + \frac{V_X - 10V}{R_5} = 0$$

And the solution is:

\$V_X = 5.8375V \$

http://www.wolframalpha.com/input/?i=X%2F(11+%2B+8%2F3)+%2B+(X+-+5)%2F5+%2B+(X+-+10)%2F7+%3D+0

Therefore the voltage across \$R_5\$ is:

\$V_5 = 10V - 5.8375V = 4.16V \$

Which means that your answer is correct indeed.