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enter image description here

I can't get the correct mesh analysis equations for the circuit above. The end result I got for \$i_x = 0.111\$... but that does not work in simulation. Iused the equations below, the approximate value that works in simulation for \$i_x\$ is around 8.3A, but I can't get to that value.

$$ i_x = i_a - i_b $$

$$1,5i_x = (i_b -i_c) $$

$$i_a = 2$$ $$i_d = -5$$

$$-(i_b - i_a) \times 10 - i_b \times 20 - i_c \times 25 - (i_c-i_d) \times 5 = 0 \Rightarrow -30i_b-30i_c = 5$$

And got this for substitution:

$$ 3 - 2.5i_b = i_c \Rightarrow i_b = 19/9$$

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  • \$\begingroup\$ First off based on your selected current directions id = -5A. That should be it. Your KVL looks right. \$\endgroup\$ Mar 19 at 17:39
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    \$\begingroup\$ And $$1.5Ix=Ic-Ib$$ \$\endgroup\$
    – user215805
    Mar 19 at 17:45
  • \$\begingroup\$ @user215805 Yeah. I think the OP wrote that one down, already. \$\endgroup\$
    – jonk
    Mar 19 at 17:46
  • \$\begingroup\$ @user215805 If I use $$ix = ic - ib$$ I get ib = 6.33 and ix = -4.33, which does not work on simulations either \$\endgroup\$
    – Tempo
    Mar 19 at 17:57
  • \$\begingroup\$ @StainlessSteelRat The -5 was a typo, thanks I edited it, I don't know if the equation is $$ -30ib -30ic = 5 $$ \$\endgroup\$
    – Tempo
    Mar 19 at 18:00
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It helps to redraw the schematic, a little, and to include some other unknowns in the schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

One of the things that earlier students often forget about is that current sources have applied voltages across them and that these may be required in order to work out the KVL loops.

With the above, we can write out the loops:

$$\begin{align*} 0\:\text{V}+V_{I_1}-R_1\cdot\left(I_A-I_B\right)&=0\:\text{V}\\ 0\:\text{V}-R_1\left(I_B-I_A\right)-R_2\cdot I_B-V_{I_2}&=0\:\text{V}\\ 0\:\text{V}+V_{I_2}-R_3\cdot I_C-R_4\cdot\left(I_C-I_D\right)&=0\:\text{V}\\ 0\:\text{V}-R_4\cdot\left(I_D-I_C\right)-V_{I_3}&=0\:\text{V} \end{align*}$$

This suggests two unknown loop currents and three unknown voltages, but only four equations.

You know a few other details:

$$\begin{align*} I_A&=2\:\text{A}\\ I_D&=-5\:\text{A}\\ I_X&=I_A-I_B \\ I_2&=I_C-I_B=1.5\cdot I_X=1.5\cdot\left(I_A-I_B\right)\\\therefore \\I_C&=\frac32\cdot I_A-\frac12\cdot I_B\\&=3\:\text{A}-\frac12\cdot I_B \end{align*}$$

Only the last equation is the important addition. Now you have five unknowns and five equations.

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  • \$\begingroup\$ @aconcernedcitizen R1 and R4 can both be Theveninized. But the equations do work out as shown. I verified. \$\endgroup\$
    – jonk
    Mar 19 at 19:09
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    \$\begingroup\$ @aconcernedcitizen Oh shoot! Thanks. I will fix that. It was just my own failure to adjust the value from its default. (Appears someone didn't like my answer. Hehe. That's their problem, not mine.) \$\endgroup\$
    – jonk
    Mar 19 at 19:11
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    \$\begingroup\$ People forget the purpose of this Q&A site: to provide solutions to problems. Downvoting is a personal vendetta, not something that tells the OP "this answer is questionable". For this reason I would have made a separate section of comments that must be filled whenever someone downvotes, because the OP ends up as the target, despite the personal motive. I would also eliminate the points and most of the badges & co, leave only those that tell what & how well the answers from a user are. Otherwise, they have become a reason for attention, for pride (which is selfish). Endrant (sorry about that). \$\endgroup\$ Mar 20 at 17:25
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    \$\begingroup\$ @aconcernedcitizen I do think that down-voting should require a comment. But I've noticed that this site "stackexchange," and not electronics in particular, is marketing itself as a source of employees to be harvested by those looking to fill technical positions and that they specifically mention to the head-hunters the idea of using someone's points as a means of deciding who to contact. So they are marketing the numbers, it appears and monetizing it. But yes, I'd get rid of the points. I don't believe they will, though. Money is money and hard to turn away. \$\endgroup\$
    – jonk
    Mar 20 at 17:29
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First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit:

enter image description here

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

$$ \begin{cases} \text{I}_\text{a}=\text{I}_1+\text{I}_2\\ \\ \text{I}_3=\text{n}\cdot\text{I}_1+\text{I}_2\\ \\ \text{I}_4=\text{I}_\text{b}+\text{I}_3\\ \\ \text{I}_4=\text{I}_\text{b}+\text{I}_6\\ \\ \text{n}\cdot\text{I}_1=\text{I}_5+\text{I}_6\\ \\ \text{I}_1=\text{I}_\text{a}+\text{I}_5 \end{cases}\tag1 $$

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

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2-\text{V}_3}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_3}{\text{R}_4} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \text{I}_\text{a}=\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \frac{\text{V}_2-\text{V}_3}{\text{R}_3}=\text{n}\cdot\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \frac{\text{V}_3}{\text{R}_4}=\text{I}_\text{b}+\frac{\text{V}_2-\text{V}_3}{\text{R}_3}\\ \\ \frac{\text{V}_3}{\text{R}_4}=\text{I}_\text{b}+\text{I}_6\\ \\ \text{n}\cdot\frac{\text{V}_1}{\text{R}_1}=\text{I}_5+\text{I}_6\\ \\ \frac{\text{V}_1}{\text{R}_1}=\text{I}_\text{a}+\text{I}_5 \end{cases}\tag3 $$

Solving for \$\text{I}_1\$, gives:

$$\text{I}_1=\frac{\text{I}_\text{b}\text{R}_4+\text{I}_\text{a}\left(\text{R}_2+\text{R}_3+\text{R}_4\right)}{\text{R}_1+\text{R}_2+\left(1-\text{n}\right)\left(\text{R}_3+\text{R}_4\right)}\tag4$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{Ia == I1 + I2, I3 == n*I1 + I2, I4 == Ib + I3, I4 == Ib + I6, 
   n*I1 == I5 + I6, I1 == Ia + I5, I1 == V1/R1, I2 == (V1 - V2)/R2, 
   I3 == (V2 - V3)/R3, I4 == V3/R4}, {I1, I2, I3, I4, I5, I6, V1, V2, 
   V3}]]

Out[1]={{I1 -> (Ib R4 + Ia (R2 + R3 + R4))/(R1 + R2 - (-1 + n) (R3 + R4)), 
  I2 -> (-Ib R4 + Ia (R1 - n (R3 + R4)))/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  I3 -> (Ia (R1 + n R2) + Ib (-1 + n) R4)/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  I4 -> (Ia (R1 + n R2) + Ib (R1 + R2 + R3 - n R3))/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  I5 -> (-Ia R1 + Ib R4 + Ia n (R3 + R4))/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  I6 -> (Ia (R1 + n R2) + Ib (-1 + n) R4)/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  V1 -> (R1 (Ib R4 + Ia (R2 + R3 + R4)))/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  V2 -> (Ib (R1 + R2) R4 + Ia (R1 + n R2) (R3 + R4))/(
   R1 + R2 - (-1 + n) (R3 + R4)), 
  V3 -> ((Ia (R1 + n R2) + Ib (R1 + R2 + R3 - n R3)) R4)/(
   R1 + R2 - (-1 + n) (R3 + R4))}}

Using your values, we get:

$$\text{I}_1=\frac{25}{3}\approx8.33\space\text{A}\tag5$$

The code gives the rest of the answers:

In[2]:=Clear["Global`*"];
Ia = 2;
Ib = 5;
R1 = 10;
R2 = 20;
R3 = 25;
R4 = 5;
n = 3/2;
FullSimplify[
 Solve[{Ia == I1 + I2, I3 == n*I1 + I2, I4 == Ib + I3, I4 == Ib + I6, 
   n*I1 == I5 + I6, I1 == Ia + I5, I1 == V1/R1, I2 == (V1 - V2)/R2, 
   I3 == (V2 - V3)/R3, I4 == V3/R4}, {I1, I2, I3, I4, I5, I6, V1, V2, 
   V3}]]

Out[2]={{I1 -> 25/3, I2 -> -(19/3), I3 -> 37/6, I4 -> 67/6, I5 -> 19/3, 
  I6 -> 37/6, V1 -> 250/3, V2 -> 210, V3 -> 335/6}}

In[3]:=N[%2]

Out[3]={{I1 -> 8.33333, I2 -> -6.33333, I3 -> 6.16667, I4 -> 11.1667, 
  I5 -> 6.33333, I6 -> 6.16667, V1 -> 83.3333, V2 -> 210., 
  V3 -> 55.8333}}
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    \$\begingroup\$ Not sure why this got a downvote, it answers in a different way, but it does answer. At the very least it presents the forming of the system of equations and the way to solve it. \$\endgroup\$ Apr 4 at 21:27
  • \$\begingroup\$ @aconcernedcitizen thank you for noticing, I really appreciate that. \$\endgroup\$
    – Jan
    Apr 5 at 7:41

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