pls help me solve this question it’s confusing me
3 Answers
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\$ how do you get 2 ohm for RTH i got 6 ohm.For VTH i got 3V \$\endgroup\$– KaizerCommented Dec 19, 2020 at 9:02
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\$\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\$ Commented Dec 19, 2020 at 9:30
I think that simple circuits like this should be simply solved by inspection. Let's note the node A in figure.
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
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:
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\$ Commented Jun 9, 2021 at 11:14