Thevenin and Norton Problem

Question

I have been trying to solve this circuit for an hour. I am doing something wrong, could you please verify my step i am not sure what i did wrong. VOc/Isc should be 0.51 ohms and Vo should be -2.72 V.

Answer 1

First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is (way) more complicated than this one. Also, this method will check your work.

Well, we are trying the analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

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

$$ \begin{cases} \begin{alignat*}{1} \text{n}\cdot\text{V}_3&=\text{I}_1+\text{I}_2\\ \\ \text{I}_0&=\text{I}_2+\text{I}_3\\ \\ 0&=\text{I}_0+\text{I}_4+\text{I}_5\\ \\ \text{n}\cdot\text{V}_3&=\text{I}_1+\text{I}_6\\ \\ 0&=\text{I}_3+\text{I}_6+\text{I}_7\\ \\ \text{I}_7&=\text{I}_4+\text{I}_5 \end{alignat*} \end{cases}\tag1 $$

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

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_\text{a}-\text{V}_1}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_3}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_4}{\displaystyle\text{R}_4}\\ \\ \text{I}_5&=\frac{\displaystyle\text{V}_4}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag2 $$

We also know that \$\displaystyle\text{V}_2-\text{V}_3=\text{V}_\text{b}\$ and \$\displaystyle\text{V}_3-\text{V}_4=\text{V}_\text{c}\$.

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

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{n*V3 == I1 + I2, I0 == I2 + I3, 0 == I0 + I4 + I5, 
   n*V3 == I1 + I6, 0 == I3 + I6 + I7, I7 == I4 + I5, 
   I1 == (Va - V1)/R1, I2 == (V2 - V1)/R2, I3 == V3/R3, I4 == V4/R4, 
   I5 == V4/R5, V2 - V3 == Vb, V3 - V4 == Vc}, {I0, I1, I2, I3, I4, 
   I5, I6, I7, V1, V2, V3, V4}]]

Out[1]={{I0 -> ((R4 + R5) ((R1 + R2) Vc + 
      R3 (-Va + Vb + Vc + n R1 Vc)))/((R1 + R2) R3 R4 + (R1 + 
       R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  I1 -> ((R4 R5 + R3 (R4 + R5 + n R4 R5)) (Va - Vb) + (-1 + 
       n R2) R3 (R4 + R5) Vc)/((R1 + R2) R3 R4 + (R1 + 
       R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  I2 -> (-((R4 R5 + R3 (R4 + R5)) (Va - Vb)) + (1 + n R1) R3 (R4 + 
       R5) Vc)/((R1 + R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + 
       n R1 R3) R4 R5), 
  I3 -> (R4 R5 (Va - Vb) + (R1 + R2) (R4 + R5) Vc)/((R1 + 
       R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  I4 -> (R3 R5 (Va - Vb) - (R1 + R2 + R3 + n R1 R3) R5 Vc)/((R1 + 
       R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  I5 -> (R3 R4 (Va - Vb) - (R1 + R2 + R3 + n R1 R3) R4 Vc)/((R1 + 
       R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  I6 -> (-((R4 R5 + R3 (R4 + R5)) (Va - Vb)) + (1 + n R1) R3 (R4 + 
       R5) Vc)/((R1 + R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + 
       n R1 R3) R4 R5), 
  I7 -> ((R4 + 
      R5) (R3 (Va - Vb) - (R1 + R2 + R3 + n R1 R3) Vc))/((R1 + 
       R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  V1 -> (R2 (R4 R5 + R3 (R4 + R5)) Va + R1 R4 R5 Vb - 
      n R1 R2 R3 (R4 + R5) Vc + 
      R3 (R4 R5 (Va + n R1 Vb) + R1 R4 (Vb + Vc) + 
         R1 R5 (Vb + Vc)))/((R1 + R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + 
         R2 + R3 + n R1 R3) R4 R5), 
  V2 -> ((R1 + R2) R4 R5 Vb + 
    R3 (R4 R5 (Va + n R1 Vb) + (R1 + R2) R4 (Vb + Vc) + (R1 + 
          R2) R5 (Vb + Vc)))/((R1 + R2) R3 R4 + (R1 + 
       R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  V3 -> (R3 R4 R5 (Va - Vb) + (R1 + R2) R3 (R4 + R5) Vc)/((R1 + 
       R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5), 
  V4 -> (R4 R5 (R3 (Va - Vb) - (R1 + R2 + R3 + n R1 R3) Vc))/((R1 + 
       R2) R3 R4 + (R1 + R2) R3 R5 + (R1 + R2 + R3 + n R1 R3) R4 R5)}}

Now, we can find:

• \$\text{V}_\text{th}\$ we get by finding \$\text{V}_4\$ and letting \$\text{R}_5\to\infty\$: $$\text{V}_\text{th}=\frac{\displaystyle\text{R}_3\text{R}_4\left(\text{V}_\text{a}-\text{V}_\text{b}\right)-\text{R}_4\text{V}_\text{c}\left(\text{R}_1\left(1+\text{n}\text{R}_3\right)+\text{R}_2+\text{R}_3\right)}{\displaystyle\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1\left(1+\text{n}\text{R}_3\right)+\text{R}_2+\text{R}_3\right)}\tag3$$
• \$\text{I}_\text{th}\$ we get by finding \$\text{I}_5\$ and letting \$\text{R}_5\to0\$: $$\text{I}_\text{th}=-\left(\frac{\displaystyle\text{V}_\text{c}}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{V}_\text{c}\left(1+\text{nR}_1\right)+\text{V}_\text{b}-\text{V}_\text{a}}{\displaystyle\text{R}_1+\text{R}_2}\right)\tag4$$
• \$\text{R}_\text{th}\$ we get by finding: $$\text{R}_\text{th}=\frac{\displaystyle\text{V}_\text{th}}{\displaystyle\text{I}_\text{th}}=\frac{\displaystyle\text{R}_3\text{R}_4\left(\text{R}_1+\text{R}_2\right)}{\displaystyle\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1\left(1+\text{n}\text{R}_3\right)+\text{R}_2+\text{R}_3\right)}\tag5$$

Where I used the following Mathematica codes:

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

Out[2]=(R3 R4 (Va - Vb) - (R1 + R2 + R3 + n R1 R3) R4 Vc)/((R1 + 
    R2) R3 + (R1 + R2 + R3 + n R1 R3) R4)

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

Out[3]=-(Vc/R3) - (-Va + Vb + Vc + n R1 Vc)/(R1 + R2)

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

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

Using your values, we find:

$$\text{V}_\text{th}=-\frac{1043}{383}\approx-2.7\space\text{V}\space\wedge\space\text{I}_\text{th}=-\frac{149}{26}\approx-5.7\space\text{A}\space\wedge\space\text{R}_\text{th}=\frac{182}{383}\approx0.48\space\Omega\tag6$$

Answer 2

I transformed all the voltage generators into current generators with Norton's theorem, then I analyzed the circuit with the topology of the networks. I need to check the results better.