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I am struggling to solve the circuit given in the attachment. I have found the current values in the circuit and I am not sure about it(I do not have the answer sheet). I would like to ask whether it is true or wrong. If wrong, where have I done wrong?Circuit


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\$9V-90\Omega I_{1}+90\Omega I_{2}=0\$

\$12V-210\Omega I_{1}-90\Omega I_{3}=0\$

\$9V+120\Omega I_{1}-30\Omega I_{2}=0\$

I tried to convert these equation to matrix as follows:

\$A=\$\begin{bmatrix}90 & -90 & 0\\210 & 0 & 90\\ -120 & 30& 0\end{bmatrix}

\$b=\$\begin{bmatrix}90\\12\\9\end{bmatrix}

And solved the equation in MATLAB by the command x=A\b an got the results:

\$I_{1}=-0,1333\$

\$I_{2}=-0,2333\$

\$I_{3}=0,4444\$

I do not get 0 when I put the values back to the equation nor when I run A*x-b on MATLAB. I am still struggling on where I do wrong. Is the problem on the assignments of the currents or the equation or some numerical errors while solving?

I know that my handwriting is not really clear but I have added my paper which I solved on just in case as attachment with the name "Attachment 2". Attachment 2

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    \$\begingroup\$ Now that you have found the currents you can calculate the voltages across all of the resistors. Then you can use KVL to check your own work! It's a great skill to have. \$\endgroup\$ May 5, 2019 at 23:34
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    \$\begingroup\$ Thanks Elliot. I have checked all 3 Loops. Loop 1 and loop 2 seems to be correct but 3 does not gives 0 when summed. I supposed its because of the miscalculation of I1 but even though I have calcuşated several times I got the same result. I am trying to find the problem. \$\endgroup\$
    – E.Canberk
    May 5, 2019 at 23:48

1 Answer 1

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Thanks for providing all your work effort. It's not as commonly done as many of us would wish.

You are able to use the schematic editor to produce a more readable version and you should learn how to do that as soon as you can, if you plan to ask more questions here. And you can use MathJax to write out your equations, as well. (As I have to, in fact.)

To keep my time down, I'm not creating a schematic for you. It would be nice, but I'm short on time. I'll use \$I_1\$ for the current in loop 1, etc. (Please do note that this means I will not be following all of your designations on your paper as you have, for example, an arrow with \$I_1\$ labeled on it that has nothing to do with the current I will label \$I_1\$. Just be aware.) And I'll also start at the tail end of your loop arrows and work myself around in the indicated directions, as well. Just so you can follow along with my reasoning:

$$\begin{align*} 0\:\text{V} + 9\:\text{V} - \left(I_1+I_2\right)\cdot 90\:\Omega - I_1\cdot 60\:\Omega-I_1\cdot 30\:\Omega&=0\:\text{V}\\\\ 0\:\text{V}+12\:\text{V}-\left(I_2-I_3\right)\cdot 120\:\Omega-\left(I_1+I_2\right)\cdot 90\:\Omega&=0\:\text{V}\\\\ 0\:\text{V} + 9\:\text{V} - \left(I_3-I_2\right)\cdot 120\:\Omega-I_3\cdot 30\:\Omega&=0\:\text{V} \end{align*}$$

You should be able to compare the results of solving those three equations with what you wrote down and solved. (I can't read your blurry figures towards the end of your photo, sorry.) From the above currents, you can add and subtract them as required and work out various voltage drops across the resistors and from there pretty much solve anything of interest in the schematic.


Just added this since you added more work to your question.

The solution to the above is (I'm using sympy):

var('i1 i2 i3')
eq1=Eq(0+9-(i1+i2)*90-i1*60-i1*30,0)
eq2=Eq(0+12-(i2-i3)*120-(i1+i2)*90,0)
eq3=Eq(0+9-(i3-i2)*120-i3*30,0)
solve([eq1,eq2,eq3],[i1,i2,i3])
{i1: -13/230, i2: 49/230, i3: 53/230}

The above checks out with a Spice simulation, perfectly.

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