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*I converted the wye network into a delta and found the equivalent resistances in parallel and series and got it to be 10 ohms. However i am stuck at trying to find Vth. I tried applying mesh analysis and found the two mesh currents(source conversion to the center branch was done to obtain a single branch of 50V in series with 10 ohms)

Please do guide me on how to solve this problem.*

[The circuit is given below]

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    \$\begingroup\$ First, show us how you found Rth = 10Ohms. And then, show us your effort on finding Vth and where you got stuck at. \$\endgroup\$ Oct 6, 2020 at 13:22
  • \$\begingroup\$ I converted the wye network into a delta and found the equivaient resistances in parallel and series and got it to be 10 ohms. However i am stuck at trying to find Vth. I tried applying mesh analysis and found the two mesh currents(source conversion to the center branch was done). \$\endgroup\$
    – Zuzu
    Oct 6, 2020 at 13:27
  • \$\begingroup\$ Alright, but show these in the question body. You can use the Edit button and edit your question by adding the details. People here may not read your explanation under the Comments section. \$\endgroup\$ Oct 6, 2020 at 13:28
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    \$\begingroup\$ You can use superposition. \$\endgroup\$ Oct 6, 2020 at 13:33
  • \$\begingroup\$ I got the answer to be 10 V. Please do confirm if someone's tried it. Thanks \$\endgroup\$
    – Zuzu
    Oct 6, 2020 at 13:41

2 Answers 2

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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:

$$ \begin{cases} \text{I}_1=\text{I}_2+\text{I}_7\\ \\ \text{I}_2=\text{I}_4+\text{I}_6+\text{I}_8\\ \\ \text{I}_5=\text{I}_4+\text{I}_6+\text{I}_7\\ \\ \text{I}_1=\text{I}_5+\text{I}_9\\ \\ \text{I}_3=\text{I}_\text{k}+\text{I}_9\\ \\ \text{I}_3=\text{I}_\text{k}+\text{I}_8 \end{cases}\tag1 $$

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

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_2-\text{V}_4}{\text{R}_6}\\ \\ \text{I}_7=\frac{\text{V}_1-\text{V}_3}{\text{R}_7} \end{cases}\tag2 $$

And we also know that \$\text{V}_3-\text{V}_4=\text{V}_\text{p}\$.

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

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}+\frac{\text{V}_1-\text{V}_3}{\text{R}_7}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}+\frac{\text{V}_2-\text{V}_4}{\text{R}_6}+\text{I}_8\\ \\ \frac{\text{V}_3}{\text{R}_5}=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}+\frac{\text{V}_2-\text{V}_4}{\text{R}_6}+\frac{\text{V}_1-\text{V}_3}{\text{R}_7}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_3}{\text{R}_5}+\text{I}_9\\ \\ \frac{\text{V}_2}{\text{R}_3}=\text{I}_\text{k}+\text{I}_9\\ \\ \frac{\text{V}_2}{\text{R}_3}=\text{I}_\text{k}+\text{I}_8 \end{cases}\tag3 $$

Now, we can solve for using your values:

  • $$\text{V}_\text{th}=\lim_{\text{R}_7\to\infty}\left(\text{V}_1-\text{V}_3\right)=10\space\text{V}\tag4$$
  • $$\text{I}_\text{th}=\lim_{\text{R}_7\to0}\text{I}_7=1\space\text{A}\tag5$$
  • $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=10\space\Omega\tag6$$

Where I used the following Mathematica-codes:

In[1]:=Vi = 30;
Vp = 20;
Ik = 5;
R1 = 10;
R2 = 10;
R3 = 10;
R4 = 20;
R5 = 10;
R6 = 20;
FullSimplify[
 Solve[{I1 == I2 + I7, I2 == I4 + I6 + I8, I5 == I4 + I6 + I7, 
   I1 == I5 + I9, I3 == Ik + I9, I3 == Ik + I8, I1 == (Vi - V1)/R1, 
   I2 == (V1 - V2)/R2, I3 == V2/R3, I4 == (V2 - V3)/R4, I5 == V3/R5, 
   I6 == (V2 - V4)/R6, I7 == (V1 - V3)/R7, V3 - V4 == Vp}, {I1, I2, 
   I3, I4, I5, I6, I7, I8, I9, V1, V2, V3, V4}]]

Out[1]={{I1 -> 5/(10 + R7), I2 -> -(5/(10 + R7)), I3 -> 3, 
  I4 -> (5 + R7)/(20 + 2 R7), I5 -> 2 + 5/(10 + R7), 
  I6 -> (25 + 3 R7)/(20 + 2 R7), I7 -> 10/(10 + R7), I8 -> -2, 
  I9 -> -2, V1 -> 30 - 50/(10 + R7), V2 -> 30, 
  V3 -> 20 + 50/(10 + R7), V4 -> 50/(10 + R7)}}

In[2]:=FullSimplify[
 Limit[(30 - 50/(10 + R7)) - (20 + 50/(10 + R7)), R7 -> Infinity]]

Out[2]=10

In[3]:=FullSimplify[Limit[10/(10 + R7), R7 -> 0]]

Out[3]=1

In[4]:=FullSimplify[%2/%3]

Out[4]=10
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Simplify and add a label.

enter image description here

The sum of the currents into c:

(30-Vc)/20 + (50-Vc)/10 + (-10-Vc)/20 = 0

Vc = 30, Va = 30, Vb = 20, Va-Vb = Vth = 10

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