Thevenise circuit problem

Hi everyone

I have to thevenise this circuit. I have done the problem but would like your input since I have not done this in a very long time.



Finding Rth
(1/20 + 1/8 ) ^-1 = 7.714ohm.
( 1/7.714 + 1/5 )^-1 = 3.03ohm
3.03 + 2 = 5ohm
soo Rth = 5ohm



Find Vth or voltage over 5ohm.

Rtotal = 20 + (1/8 + 1/(5+2))^-1 = 23.73ohm
Itotal = 20 / 23.73 = 0.85579 Amp.
V-20ohm = IR = 0.85579 * 20 = 17.115V
V-8ohm = 20-17.115 = 3.73V
So V-5ohm = 3.144( 5 / 2+5 ) = 2.2457V



Am I correct in assuming that the voltage over 5-ohm is what they are looking for in the question ?
Could this be a legitimate Rth and Vth answer?

Thanks

I would start at the other end.
5 ohms (R) + 2 ohms = 7 ohms

The 7 ohms is now paralleled with 5 Ohms

(7*5)/(7+5) = 35/12=2.9166666

now we have a 2 ohm in series, so

2+2.9166666=4.91666666 ohms

now we have 8 ohms in parallel with that

(8*4.91666666)/(8+4.916666)=39.3333333/12.916666666= 3.04516129 ohms

finally, you have a 20 ohm in series
=23.045... ohms

RobertFalbo said:

I would start at the other end.
5 ohms (R) + 2 ohms = 7 ohms
=23.045... ohms

Click to expand...

Awesome thanks for the reply. Although is my method wrong though?
I am just a bit cautious.

Thanks

Your method and results are correct (unless the rounding error)

The exact values are: Rth = 448/89 ≈ 5.033707865 , Vth = 200/89 ≈ 2.247191011

RobertFalbo said:

I would start at the other end.
5 ohms (R) + 2 ohms = 7 ohms

The 7 ohms is now paralleled with 5 Ohms

(7*5)/(7+5) = 35/12=2.9166666

now we have a 2 ohm in series, so

2+2.9166666=4.91666666 ohms

now we have 8 ohms in parallel with that

(8*4.91666666)/(8+4.916666)=39.3333333/12.916666666= 3.04516129 ohms

finally, you have a 20 ohm in series
=23.045... ohms

Click to expand...

I didn't get a chance to finish

20 V / 23.045.. ohms = 867.861 ma
through the 20 ohm gives an IR drop of 17.357..Volts

leaving 2.642..Volts agross the 8 ohm

2.642../8 ohms = 330.347 ma into the 8 ohm.

867.861ma - 330.347ma = 537.513 ma through the 2 ohm.
that gives us an IR drop of 2ohms * 537.513 ma = 1.075.. volts...That is one answer.


2.624..V - 1.075..V = 1.5677.. volts across the 5 ohm.

1.5677 / 5 = 313.5498.. ma

537.513 ma - 313.5498 ma = 223.964 ma

That is the other answer

- - - Updated - - -

Another approach would be to start with the 20 V and the 20 and 8 ohm resister

20 ohm parallel 8 ohm

(8*20)/(8+20)= 5.714... ohms

and a voltage divider value
20V *8 ohm / (8 ohms + 20 ohms) = 5.714 volts

So we have a 5.71 voltage source with a internal impedance of 5.714.. ohms

we tie a 2 ohm in series
so now we have a 5.71 volt source with a 7.714... ohm impedance.

next we take the 5 ohm and we have another voltage divider

5.71..V *5 ohms/(5 ohms + 7.714..ohms)= 2.247...V
and a impedance of (5*7.714)/(5+7.714) = 3.0337... ohms

Now we have a 7 ohm load on that (2 ohms + R) R=5 ohms

So for I, we have 2.247V / 10.0337 ohms (load + impedance) = 223.964... ma

That is one of the answers

now we take 223.964 ma * 7 ohms to get 1.5677... volts across 5 ohms
So the current in the 5 ohms is 1.5677../ 5 = 313.549...ma

Add that to the current in the 2 and 5 ohms
313.549 ma + 223.964 ma = 537.514.. ma

So the IR drop of the 2 ohm is 537.514 ma * 2 ohms = 1.075... volts
the answer to the second question V

Hi,

From post#2:

I would start at the other end.
5 ohms (R) + 2 ohms = 7 ohms

The 7 ohms is now paralleled with 5 Ohms

(7*5)/(7+5) = 35/12=2.9166666

Click to expand...

It should be "8 Ohms" instead of "5 Ohms"

Klaus

Another approach would be to start with the 20 V and the 20 and 8 ohm resister

20 ohm parallel 8 ohm

(8*20)/(8+20)= 5.714... ohms

and a voltage divider value
20V *8 ohm / (8 ohms + 20 ohms) = 5.714 volts

So we have a 5.71 voltage source with a internal impedance of 5.714.. ohms

we tie a 2 ohm in series
so now we have a 5.71 volt source with a 7.714... ohm impedance.

next we take the 5 ohm and we have another voltage divider

5.71..V *5 ohms/(5 ohms + 7.714..ohms)= 2.247...V
and a impedance of (5*7.714)/(5+7.714) = 3.0337... ohms

Now we have a 7 ohm load on that (2 ohms + R) R=5 ohms

So for I, we have 2.247V / 10.0337 ohms (load + impedance) = 223.964... ma

That is one of the answers

now we take 223.964 ma * 7 ohms to get 1.5677... volts across 5 ohms
So the current in the 5 ohms is 1.5677../ 5 = 313.549...ma

Add that to the current in the 2 and 5 ohms
313.549 ma + 223.964 ma = 537.514.. ma

So the IR drop of the 2 ohm is 537.514 ma * 2 ohms = 1.075... volts
the answer to the second question V

- - - Updated - - -

KlausST said:

Hi,

From post#2:


It should be "8 Ohms" instead of "5 Ohms"

Klaus

Click to expand...

No, I started on the far right
R=5 ohms and there is a 2 ohm resistor in series with that.

Hi,

No, I started on the far right
R=5 ohms and there is a 2 ohm resistor in series with that.

Click to expand...

You are correct.
I had the third pictire of post#1 in mind.

Sorry.

Klaus

KlausST said:

Hi,


You are correct.
I had the third pictire of post#1 in mind.

Sorry.

Klaus

Click to expand...

Hey, I'm just happy that I got the same answer, doing it from two directions. Usually , you make at least one mistake.