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thevenin equivalent circuit

pls help me solve this question it’s confusing me

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    \$\begingroup\$ which part is confusing you? \$\endgroup\$
    – jsotola
    Dec 19 '20 at 7:42
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For finding the Thevenin equivalent resistance, you need to short circuit the independent voltage sources and open circuit the independent current sources. For finding the Thevenin voltage, find the open-circuit voltage Voc. A nodal analysis will get you there. You'd get Rth as 2 Ohm and VTh as 3 V In which part are you stuck in?

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  • \$\begingroup\$ thankyou so much \$\endgroup\$
    – Kaizer
    Dec 19 '20 at 8:14
  • \$\begingroup\$ how do you get 2 ohm for RTH i got 6 ohm.For VTH i got 3V \$\endgroup\$
    – Kaizer
    Dec 19 '20 at 9:02
  • \$\begingroup\$ 3 Ohm and 6 Ohm resistors are in parallel. The resulting combination is in series with 2 Ohm resistor. This resultant is again parallel with the 4 Ohm resistor. \$\endgroup\$
    – Kaswechiha
    Dec 19 '20 at 9:30
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I think that simple circuits like this should be simply solved by inspection. Let's note the node A in figure. enter image description here

Now, looking the right side of the circuit, from A down to (-) terminal, you can note a parallel of 6//(4+2) = 3 ohms. At this point, considering the left side too, we will have a very simple circuit with 9 V and two 3 ohms series resistance; therefore, the voltage on point A is 9/2 V. Knowing the voltage at node A, it's easy to calculate the voltage level Vo, considering the voltage divider : Vo = (9/2)* 4/(2+4) = 3 V

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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:

schematic

simulate this circuit – Schematic created using CircuitLab

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

$$ \begin{cases} \text{I}_1=\text{I}_2+\text{I}_3\\ \\ \text{I}_3=\text{I}_4+\text{I}_5\\ \\ \text{I}_6=\text{I}_4+\text{I}_5\\ \\ \text{I}_1=\text{I}_2+\text{I}_6 \end{cases}\tag1 $$

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

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

Now, we can set up a Mathematica-code to solve for all the voltages and currents:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{I1 == I2 + I3, I3 == I4 + I5, I6 == I4 + I5, I1 == I2 + I6, 
   I1 == (Vi - V1)/R1, I2 == V1/R2, I3 == (V1 - V2)/R3, I4 == V2/R4, 
   I5 == V2/R5}, {I1, I2, I3, I4, I5, I6, V1, V2}]]

Out[1]={{I1 -> (((R2 + R3) R4 + (R2 + R3 + R4) R5) Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  I2 -> ((R4 R5 + R3 (R4 + R5)) Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  I3 -> (R2 (R4 + R5) Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  I4 -> (R2 R5 Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  I5 -> (R2 R4 Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  I6 -> (R2 (R4 + R5) Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  V1 -> (R2 (R4 R5 + R3 (R4 + R5)) Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5), 
  V2 -> (R2 R4 R5 Vi)/(
   R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
    R1 (R2 + R3 + R4) R5)}}

Now, we can find:

  • \$\text{V}_\text{th}\$ we get by finding \$\text{V}_2\$ and letting \$\text{R}_5\to\infty\$: $$\text{V}_\text{th}=\frac{\text{R}_2\text{R}_4\text{V}_\text{i}}{\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)}\tag3$$
  • \$\text{I}_\text{th}\$ we get by finding \$\text{I}_5\$ and letting \$\text{R}_5\to0\$: $$\text{I}_\text{th}=\frac{\text{R}_2\text{V}_\text{i}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)}\tag4$$
  • \$\text{R}_\text{th}\$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_4\left(\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)\right)}{\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)}\tag5$$

Where I used the following Mathematica-codes:

In[2]:=FullSimplify[
 Limit[(R2 R4 R5 Vi)/(
  R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
   R1 (R2 + R3 + R4) R5), R5 -> Infinity]]

Out[2]=(R2 R4 Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4))

In[3]:=FullSimplify[
 Limit[(R2 R4 Vi)/(
  R2 R3 R4 + R1 (R2 + R3) R4 + R2 (R3 + R4) R5 + 
   R1 (R2 + R3 + R4) R5), R5 -> 0]]

Out[3]=(R2 Vi)/(R2 R3 + R1 (R2 + R3))

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

Out[4]=((R2 R3 + R1 (R2 + R3)) R4)/(R2 (R3 + R4) + R1 (R2 + R3 + R4))

So, using your values we get:

  • $$\text{V}_\text{th}=3\space\text{V}\tag6$$
  • $$\text{I}_\text{th}=\frac{3}{2}=1.5\space\text{A}\tag7$$
  • $$\text{R}_\text{th}=2\space\Omega\tag8$$
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  • \$\begingroup\$ Although this is a possible solution, it is worth pointing out that it is absolute overkill and no sane engineer would use it in practice (it could be a learning experience, though). The whole point of the kind of exercises like the one in the OP is to make the future engineer develop the skills to do back-of-the-envelope calculations. In particular, with the nice values in the OP, the problem can be solved almost entirely "in one's head" by repeatedly applying the formulas for parallel/series resistors and for voltage dividers, which any engineer MUST get to know by heart. \$\endgroup\$ Jun 9 at 11:14

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