Engineering Find the Thevenin Equivalent circuit

[Fatima Hasan]

Find Thevenins Equivalent circuit external to RL, then find I.

Here's my work :
5Ω & 5Ω series → R(1) = 5+5 = 10 Ω
10Ω & R(1) parallel→ R(2) = (10*10)/(10+10) = 5 Ω
2Ω & R(2) series → R(3) = R Th = 5+2 = 7Ω

V = I R
R = (R Th * RL)/(R Th + RL)
R = 70/17 Ω
V = 5 * (70/17) = 350 / 17 V = 20.6 V

Could somebody check my answer please ?

 

[AlexCaledin]

. . .
2Ω & R(2) series → R(3) = R Th = 5+2 = 7Ω
. . .

Hi ! Will you please explain what is 2Ω here?

 

[Fatima Hasan]

Hi ! Will you please explain what is 2Ω here?

I forgot to put it ...

 

[AlexCaledin]

These are the currents at RL=0 as calculated by a simulation program;
the "Thevenin current" is 1.786A ( to be then multiplied by 70/17 Ω).

( 357.143m = 0.357143A )

 

[Fatima Hasan]

These are the currents at RL=0 as calculated by a simulation program; the Thevenin current is 1.786A, to be multiplied by 70/17 Ω.

View attachment 221969

The current in a series circuit is everywhere the same , why I₂ doesn't equal to I₁ ?

 

[AlexCaledin]

Why is it series?
5 amperes are rather applied to parallel resistors:
one is the R1 = 5ohm and the other is {R2 + R3-and-R4} = 5 + 20/12 = 40/6 = 6.66666etc ohm;
so in the R2 there is less current.

____________

Thanks to the Electrician, now I know about the "Norton", as opposed to "Thevenin". I have been taught about "The Equivalent Source Theorem", combining both of them.

 

[The Electrician]

Fatima Hasan, you are calculating a Norton equivalent, not a Thevenin equivalent: https://en.wikipedia.org/wiki/Norton's_theorem

Is that what you really want?

 

[The Electrician]

See the section on this page: https://en.wikipedia.org/wiki/Thévenin's_theorem

with the heading "Calculating the Thevenin equivalent".

What do you get if you follow that method?

 

[Fatima Hasan]

See the section on this page: https://en.wikipedia.org/wiki/Thévenin's_theorem

with the heading "Calculating the Thevenin equivalent".

What do you get if you follow that method?

R_No = R_Th = 7 Ω
I_No * R_No = V_Th
V_Th=5*7 = 35 V
V_No = 1.786*70/17 =7.354 V
I_Th = V_No / R_No =7.354/7 = 1.051 A

 

[cnh1995]

VTh=5*7 = 35 V

That is incorrect.
You need to remove R_L and calculate the open circuit voltage V_AB.

RNo = RTh =

Right.

 

[Fatima Hasan]

You need to remove RL and calculate the open circuit voltage VAB.

VAB = V3 (parallel)
VAB= I3 * R3
= 0.357143 * 10 = 3.57 V

 

[cnh1995]

VAB = V3 (parallel)
VAB= I3 * R3
= 0.357143 * 10 = 3.57 V

No. You get I3*R3=3.57V when you 'short circuit' the terminals A and B and V_AB=0 in that çase.

Do you know the steps for finding the Thevenin equivalent of a network? It looks like you are mixing Thevenin and Norton without knowing the fundamental differences between them.

 

[AlexCaledin]

. . . you are mixing Thevenin and Norton . . .

- the thing is, they may be mixed in the textbooks. In our Russian textbooks, there is typically one "Equivalent Source Theorem" mixing Thevenin and Norton straight away, never mentioning any of them, like here:

http://energ2010.ru/Toe/Toe_lekcii_VUZ/Toe_lekcii_belarus/2_11_Teorema_ekvivalentnom_generatore.html

- and in some textbooks that same mixed theorem is called also "Thevenin theorem", as its second name. And of course lots of people - including teachers - from many countries have studied here in Russia so this confusion may well be typical.

 

[The Electrician]

The link you have provided shows in Fig 27 two circuits. The one on the left, designated а) is known on this forum as the Thevenin equivalent. The one on the right, designated б) is known on this forum as the Norton equivalent. They are shown as two different circuits. Are they not given different names in that Russian text?

At any rate, I pointed out the two different circuits and their names used here in posts #7 and #8.

 

[AlexCaledin]

- they are called equivalent voltage and equivalent current sources - making the point that you can always use this or that, whatever is better for your problem.

 

[The Electrician]

So far Fatima Hasan hasn't calculated either one correctly. Let's hope he (or she) is still with us and tells us which one is needed.

 

[Fatima Hasan]

So far Fatima Hasan hasn't calculated either one correctly. Let's hope he (or she) is still with us and tells us which one is needed.

I want to calculate the Thevenin's Equivalent .

 

[The Electrician]

I want to calculate the Thevenin's Equivalent .

OK. You have already correctly calculated Rth. Next you need to respond to post #12. You must calculate the voltage across terminals A & B when RL is not present. That will be Vth.

 

[Fatima Hasan]

OK. You have already correctly calculated Rth. Next you need to respond to post #12. You must calculate the voltage across terminals A & B when RL is not present. That will be Vth.

R = R1 + R2 = 5 + 5 = 10 Ω
I_s = 5 A
V_s = I_s * R = 5*10 = 50 V
V3= V_s /(R+R₃) * R3
V3 = 50/(10+10) * 10 = 25 V
V3 = V_Th = 25 V
I_sc = R_Th * V_oc
=7 *25 = 175 A

 

[The Electrician]

The resistance R should be the resistance seen by the current source, but it's not just R1+R2. There is some current also in R3, so R2+R3 is in parallel with R1; that equivalent resistance is the load on the source.

Also, the next to last thing in your calculations is Isc = RTh * Voc. That can't be right because the units are wrong.

 

[Fatima Hasan]

Isc = RTh * Voc

I sc = Voc / RTh

 

[The Electrician]

I sc = Voc / RTh

That's good. Now calculate V5 with the R1, R2, R3 combination load.

 

[Fatima Hasan]

That's good. Now calculate V5 with the R1, R2, R3 combination load.

 

[The Electrician]

View attachment 222096

You have correctly calculated the voltage V5 across the current source, which is also the voltage across R1. Your last calculation is wrong. You should be calculating the voltage V3 across the resistor R3. You have labeled the left side of that equation V5, but it should be V3, which is also Vth.

If you know the voltage at the left end of R2, which is V5, then what would be the voltage at the right end of R2 considering that R2 and R3 form a voltage divider?

 

[Fatima Hasan]

You have correctly calculated the voltage V5 across the current source, which is also the voltage across R1. Your last calculation is wrong. You should be calculating the voltage V3 across the resistor R3. You have labeled the left side of that equation V5, but it should be V3, which is also Vth.

If you know the voltage at the left end of R2, which is V5, then what would be the voltage at the right end of R2 considering that R2 and R3 form a voltage divider?

V2 = R2 /(R1+R2)*Vs
= 5/(10+5) * 18.75
= 6.25 V

 

[The Electrician]

V2 = R2 /(R1+R2)*Vs
= 5/(10+5) * 18.75
= 6.25 V

I'm not sure why you would want the voltage across R2 (which we would call V2). Vth is the same as V3 which is the voltage across R3.

If you replace the current source and R1 with an 18.75 volt battery, could you calculate the voltage across R3? If all you had was a circuit with an 18.75 volt battery on the left side, with R2 and R3 forming a voltage divider, what would be the voltage across R3?

 

[Fatima Hasan]

I'm not sure why you would want the voltage across R2 (which we would call V2). Vth is the same as V3 which is the voltage across R3.

If you replace the current source and R1 with an 18.75 volt battery, could you calculate the voltage across R3? If all you had was a circuit with an 18.75 volt battery on the left side, with R2 and R3 forming a voltage divider, what would be the voltage across R3?

V3 = 10/15*18.75 = 12.5 V

 

[The Electrician]

Congratulations! This is correct. You now have Rth and Vth. Do you understand all the steps now?

 

[Fatima Hasan]

Congratulations! This is correct. You now have Rth and Vth. Do you understand all the steps now?

The voltage across R2 is 6.25 V
V2= I2 * R2 --> = 5 *2.173 = 10.865
Why I didn't get the same answer ?

 

[The Electrician]

The voltage across R2 is 6.25 V
V2= I2 * R2 --> = 5 *2.173 = 10.865
Why I didn't get the same answer ?

I don't know where that value of 2.173 came from, so I can't explain. You'll have to tell me how you calculated I2.

 

[Fatima Hasan]

I don't know where that value of 2.173 came from, so I can't explain. You'll have to tell me how you calculated I2.

R4 and R3 in parallel --> R* = 2*10/(2+10) = 5/3Ω
R* and R2 in series --> R total = 5/3 +5 = 20/3Ω
I1 = R total / (R total + R 1)* i_s
I1 = (20/3) / ((20/3)+5) * 5 = 2.857 A
I_s = I1 + I2
I2 = 5 - 2.857 = 2.143 A

 

[The Electrician]

We got the value of 6.25 volts across R2 with RL removed. If RL is not there then R3 and R4 are not in parallel; R3 and R4 are only in parallel if RL is replaced with a short. You are trying to compare the voltage across R2 with RL removed to the voltage across R2 with RL replaced with a short. You can't expect the voltage in the two cases to be the same.

 

[Fatima Hasan]

We got the value of 6.25 volts across R2 with RL removed. If RL is not there then R3 and R4 are not in parallel; R3 and R4 are only in parallel if RL is replaced with a short. You are trying to compare the voltage across R2 with RL removed to the voltage across R2 with RL replaced with a short. You can't expect the voltage in the two cases to be the same.

OK , Thank you for your help

 

[gneill]

Since the OP has reached a correct result I can offer an alternative method to get to the solution.

You can use successive source transformations (essentially Norton to Thevenin or Thevenin to Norton transformations), working your way from left to right through the circuit while "absorbing" new resistances along the way. For example, you can convert the original current source and its parallel resistor to a voltage source with a series resistor:

As you can see, the "new" 5 Ω resistor can be combined with R2 into a single 10 Ω resistor and the result is a 25 V voltage source in series with 10 Ω. That pair can in turn be converted to its Norton equivalent of a current source in parallel with a resistor and you'll then find that its resistor will be in parallel with R3, so you can combine them. Continue in such a fashion until you've absorbed all the components into a single Thevenin or Norton model driving the load.

 

[Fatima Hasan]

Since the OP has reached a correct result I can offer an alternative method to get to the solution.

You can use successive source transformations (essentially Norton to Thevenin or Thevenin to Norton transformations), working your way from left to right through the circuit while "absorbing" new resistances along the way. For example, you can convert the original current source and its parallel resistor to a voltage source with a series resistor:

View attachment 222127

As you can see, the "new" 5 Ω resistor can be combined with R2 into a single 10 Ω resistor and the result is a 25 V voltage source in series with 10 Ω. That pair can in turn be converted to its Norton equivalent of a current source in parallel with a resistor and you'll then find that its resistor will be in parallel with R3, so you can combine them. Continue in such a fashion until you've absorbed all the components into a single Thevenin or Norton model driving the load.

Thanks for the explanation.
Can I know what software did you use to draw the circuits?

 

[gneill]

Thanks for the explanation.
Can I know what software did you use to draw the circuits?

I used an ancient copy of Microsoft Visio. Still works great for quickly putting together a drawing to accompany a physics problem.