Electronic – Calculating Thevenin’s open circuit voltage

circuit analysiskirchhoffs-lawsthevenin

I am unsure how to calculate \$V_{Th}\$ with the circuit below.


I have attempted to reduce the circuit below into the following circuit, but I am unsure if this is correct. Can anyone confirm that I am taking the appropriate steps?

Best Answer

Well, I am 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:

$$ \begin{cases} \text{I}_\text{a}=\text{I}_1+\text{I}_5\\ \\ \text{I}_5=\text{I}_2+\text{I}_3\\ \\ \text{I}_4=\text{I}_\text{b}+\text{I}_3\\ \\ \text{I}_1=\text{I}_\text{a}+\text{I}_6\\ \\ 0=\text{I}_2+\text{I}_6+\text{I}_7\\ \\ \text{I}_4=\text{I}_\text{b}+\text{I}_7 \end{cases}\tag1 $$

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

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

Because you already have the answer I will present a method that uses Mathematica to solve this problem. I used the following code to solve the two systems from above:

In[1]:=FullSimplify[
 Solve[{Ia == I1 + I5, I5 == I2 + I3, I4 == Ib + I3, I1 == Ia + I6, 
   0 == I2 + I6 + I7, I4 == Ib + I7, I1 == V1/R1, I2 == V1/R2, 
   I3 == (V1 - V2)/R3, I4 == V2/R4, Ib == (V3 - V2)/R5}, {I1, I2, I3, 
   I4, I5, I6, I7, V1, V2, V3}]]

Out[1]={{I1 -> (R2 (Ib R4 + Ia (R3 + R4)))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I2 -> (R1 (Ib R4 + Ia (R3 + R4)))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I3 -> (Ia R1 R2 - Ib (R1 + R2) R4)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I4 -> ((Ia + Ib) R1 R2 + Ib (R1 + R2) R3)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I5 -> (-Ib R2 R4 + Ia R1 (R2 + R3 + R4))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I6 -> (Ib R2 R4 - Ia R1 (R2 + R3 + R4))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I7 -> (Ia R1 R2 - Ib (R1 + R2) R4)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V1 -> (R1 R2 (Ib R4 + Ia (R3 + R4)))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V2 -> (((Ia + Ib) R1 R2 + Ib (R1 + R2) R3) R4)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V3 -> Ia R4 + Ib R4 - ((R1 + R2) R4 (Ib R4 + Ia (R3 + R4)))/(
    R2 (R3 + R4) + R1 (R2 + R3 + R4)) + Ib R5}}

In order to find the Thevenin equivalence, we need to take a loot at the open-circuit voltage \$\text{V}_\text{th}=\text{V}_1-\text{V}_2\$ (when \$\text{R}_3\to\infty\$) and at the short circuit current \$\text{I}_3\$ (when \$\text{R}_3\to0\$):

In[2]:=FullSimplify[
 Solve[{Ia == I1 + I5, I5 == I2 + I3, I4 == Ib + I3, I1 == Ia + I6, 
   0 == I2 + I6 + I7, I4 == Ib + I7, I1 == V1/R1, I2 == V1/R2, 
   I3 == (V1 - V2)/Infinity, I4 == V2/R4, Ib == (V3 - V2)/R5}, {I1, 
   I2, I3, I4, I5, I6, I7, V1, V2, V3}]]

Out[2]={{I1 -> (Ia R2)/(R1 + R2), I2 -> (Ia R1)/(R1 + R2), I3 -> 0, I4 -> Ib,
   I5 -> (Ia R1)/(R1 + R2), I6 -> -((Ia R1)/(R1 + R2)), I7 -> 0, 
  V1 -> (Ia R1 R2)/(R1 + R2), V2 -> Ib R4, V3 -> Ib (R4 + R5)}}

In[3]:=FullSimplify[(Ia R1 R2)/(R1 + R2) - Ib R4]

Out[3]=(Ia R1 R2)/(R1 + R2) - Ib R4

In[4]:=Limit[(Ia R1 R2 - Ib (R1 + R2) R4)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)),
  R3 -> 0]

Out[4]=(Ia R1 R2 - Ib (R1 + R2) R4)/(R2 R4 + R1 (R2 + R4))

So, using your values we get:

  • $$\text{V}_\text{th}=-\frac{11}{3}\approx-3.66667\space\text{V}\tag3$$
  • $$\text{I}_\text{th}=-\frac{11}{5}=-2.2\space\text{A}\tag4$$
  • $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{5}{3}\approx1.66667\space\Omega\tag5$$
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