2
\$\begingroup\$

I did a nodal analysis on this circuit where I put the reference node at the middle part.

enter image description here

The question was to find the current passing through 10V source using nodal analysis. I tried answering it like this in the picture but the answer key says it should be 3.55 A

enter image description here

I tried other equations to get the 3.55A and I got it by I1 - I3 = 3.5537A but I think its wrong. What is the error that I did?

\$\endgroup\$
6
  • \$\begingroup\$ But how can this be true? The current that passes through 10V source is equal to (10V - 7.107V)/5Ω = 578.6mA. And how can it be otherwise? \$\endgroup\$
    – G36
    Jul 20, 2023 at 16:34
  • \$\begingroup\$ @G36 that’s what I’ve been trying to find because the current that I always got is 578.6mA. The answer key in the book was 3.55A but I still don’t get it \$\endgroup\$
    – mama b
    Jul 20, 2023 at 16:39
  • \$\begingroup\$ The book's answer is wrong, nothing new here. The 2 Ohms resistor is equal to 7.107V/2Ω = 3.5535A. \$\endgroup\$
    – G36
    Jul 20, 2023 at 16:42
  • \$\begingroup\$ Rounding a number (My calculator displayed 3.553719008, by the way) to three significant digits doesn't necessarily make it "wrong." Although, if the book answer was rounded to three digits, it would have been polite for the author to tell you so. \$\endgroup\$ Jul 20, 2023 at 17:02
  • \$\begingroup\$ @SolomonSlow yes, but the answer 3.55 was supposedly for the 2 ohm resistor. Maybe the book was wrong here. \$\endgroup\$
    – mama b
    Jul 20, 2023 at 17:13

1 Answer 1

0
\$\begingroup\$

Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_2&=\text{I}_1+\text{I}_4\\ \\ 0&=\text{I}_0+\text{I}_1+\text{I}_5\\ \\ 0&=\text{I}_0+\text{I}_2+\text{I}_3\\ \\ \text{I}_5&=\text{I}_3+\text{I}_4 \end{alignat*} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_\text{a}-\text{V}_1}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_3-0}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle\text{R}_4}\\ \\ \text{I}_5&=\frac{\displaystyle0-\text{V}_3}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag2 $$

We also know that \$\displaystyle\text{V}_2-\text{V}_3=\text{V}_\text{b}\$.

Using \$(2)\$ we can rewrite \$(1)\$ as follows:

$$ \begin{cases} \begin{alignat*}{1} \frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}&=\frac{\displaystyle\text{V}_\text{a}-\text{V}_1}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle\text{R}_4}\\ \\ 0&=\text{I}_0+\frac{\displaystyle\text{V}_\text{a}-\text{V}_1}{\displaystyle\text{R}_1}+\frac{\displaystyle0-\text{V}_3}{\displaystyle\text{R}_5}\\ \\ 0&=\text{I}_0+\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{V}_3-0}{\displaystyle\text{R}_3}\\ \\ \frac{\displaystyle0-\text{V}_3}{\displaystyle\text{R}_5}&=\frac{\displaystyle\text{V}_3-0}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle\text{R}_4} \end{alignat*} \end{cases}\tag3 $$

Solving this for your circuit we get:

In[1]:=Clear["Global`*"];
Va = 10;
Vb = 20;
R1 = 5;
R2 = 2;
R3 = 4;
R4 = 3;
R5 = 2;
FullSimplify[
 Solve[{I2 == I1 + I4, 0 == I0 + I1 + I5, 0 == I0 + I2 + I3, 
   I5 == I3 + I4, I1 == (Va - V1)/R1, I2 == (V1 - 0)/R2, 
   I3 == (V3 - 0)/R3, I4 == (V2 - V1)/R4, I5 == (0 - V3)/R5, 
   V2 - V3 == Vb}, {I0, I1, I2, I3, I4, I5, V1, V2, V3}]]

Out[1]={{I0 -> -(310/121), I1 -> 70/121, I2 -> 430/121, I3 -> -(120/121), 
  I4 -> 360/121, I5 -> 240/121, V1 -> 860/121, V2 -> 1940/121, 
  V3 -> -(480/121)}}

In[2]:=N[%]

Out[2]={{I0 -> -2.56198, I1 -> 0.578512, I2 -> 3.55372, I3 -> -0.991736, 
  I4 -> 2.97521, I5 -> 1.98347, V1 -> 7.10744, V2 -> 16.0331, 
  V3 -> -3.96694}}
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.