# Finding the Thevenin equivalent

Joined May 21, 2015
5
Just want to know if my answer is correct or not. the guidline is important to understand my mistakes.

the question and my answer are attached.

R1 = 2
R3 = 4
R4 =5

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#### Jony130

Joined Feb 17, 2009
5,243
Your Vth and Rth looks good, but your power calculations are wrong.
If Rth = Rsurce = 1.64Ω

Joined May 21, 2015
5
Your Vth and Rth looks good, but your power calculations are wrong.
If Rth = Rsurce = 1.64Ω
Yes .. it was a silly mistake

So, shall I calculate the current (I = Vth / Rth) and find the power ?

#### WBahn

Joined Mar 31, 2012
26,398
R1=2 is not a resistance. That's like saying that a hole is 2 deep.

You need to track your units. Most mistakes you make will mess up your units and let you catch them almost immediately. But only if they are there to be messed up.

A case in point is your power calculation.

You have

$$Power \; = \; I^2R \; = \; (1.64)^2 \times 10.36 \; = ; 27.86W$$

You just grabbed the numbers 1.64 and 10.36 and threw them into a formula and then tacked the units that you wanted the answer to have onto the end. Had you bothered to track your units, even superficially, you would have had:

$$Power \; = \; I^2R \; = \; (1.64 \Omega)^2 \times 10.36 V\; = ; 27.86W$$

And it should jump out at you that 1.64Ω is not a current and 10.36V is not a resistance. So you KNOW the answer is wrong and there is no point going any further until you resolve the problem.

But even if you didn't recognize that and insisted on turning the crank you would have ended up with units of

$$\Omega^2 \cdot V \; = \; \( \frac{V}{A}$$^2 \cdot V \; = \; \frac{V^3}{A^2} \; = \; W \cdot \frac{V^2}{A^3}
\)

Which clearly can never be equal to W.

You also need to learn how to check your own work. One of the nice things about most fields in engineering is that you can verify the correctness of the answer from the answer itself. You do this by assuming your answers are correct and then seeing if those results are consistent with the problem itself.

#### WBahn

Joined Mar 31, 2012
26,398
Yes .. it was a silly mistake

So, shall I calculate the current (I = Vth / Rth) and find the power ?
Why do you think that that current has any meaning?

You are practicing analysis by happening, meaning that you are just throwing values that you've come up with into formulas hoping that something will happen to work.

Joined May 21, 2015
5
because the formula of power is P = 0.5 I2 R
so, we have to find the current. I am learning and trying to understand not throwing numbers and formulas

#### WBahn

Joined Mar 31, 2012
26,398
because the formula of power is P = 0.5 I2 R
so, we have to find the current. I am learning and trying to understand not throwing numbers and formulas
Which current? Which resistance?

Why do you think that Vth divided by Rth will give you the current you are looking for? You must have a reason, otherwise you are just throwing numbers at formulas hoping for a happening to occur.