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Thevenin Resistance

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hupsenfg

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upload_2017-9-25_2-29-50.png
Hi i got this question off a youtube video.

In this video, the person found the equivalent resistance to be 10 Ohms.

He started off with solving 60 Ohms // 12 Ohms

However, I am not used to his method, and I rather redraw the circuit so that it is easier for me to calculate the equivalent value.

Can somebody teach me how can I trace the current flow in the entire circuit?

Shouldnt there be a short circuit at the 30 V branch?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
It's 18 ohms :D

But only because the ratios of the two series parts are the same, which means you can totally ignore the 10 ohm resistor. So you've got 72 ohms in parallel with 24 ohms.

I can't be bothered to work it out otherwise! :D
 

AnalogKid

Well-Known Member
Most Helpful Member
So you've got 72 ohms in parallel with 24 ohms.
The is true for the load on the 30 V source. But if you are asking about the Thevenin equivalent output voltage and impedance at terminals a and b, that is something else.

ak
 

Les Jones

Well-Known Member
Most Helpful Member
Hi Nigel,
I do not agree with your answer of 18 ohms for the source resistance between a and b. I agree with the value of 10 ohms quoted in post #1.
Here is the reasoning. The 60 ohm in parallel with the 12 ohm gives a value of 10 ohms. So that 10 ohms is in series with the 10 ohm resistor giving 20 ohms. this 20 ohms in in parallel with the 20 ohm resistor giving the final value of 10 ohms. My result for the open circuit voltage between a and b is 17.5 volts. so the current through the 4 ohm load would be 17.5/(10 + 4) = 1.25 amps.

Les.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Hi Nigel,
I do not agree with your answer of 18 ohms for the source resistance between a and b. I agree with the value of 10 ohms quoted in post #1.
Here is the reasoning. The 60 ohm in parallel with the 12 ohm gives a value of 10 ohms. So that 10 ohms is in series with the 10 ohm resistor giving 20 ohms. this 20 ohms in in parallel with the 20 ohm resistor giving the final value of 10 ohms. My result for the open circuit voltage between a and b is 17.5 volts. so the current through the 4 ohm load would be 17.5/(10 + 4) = 1.25 amps.

Les.
How do you get the 60 ohms in parallel with the 12 ohm? - it's in series with it.

You appear to be calculating the resistance between points a and b?, and presuming the battery is a short circuit (rather than a battery). I was calculating the resistance ACROSS the battery, which makes a lot more sense than trying to calculate the resistance of a network when it's got a spurious 30V battery powering it.
 

JimB

Super Moderator
Most Helpful Member
How do you get the 60 ohms in parallel with the 12 ohm?
By application on Thevenins Theorem, from a random Google Grab:
Thevenin's Theorem. Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r. The value of e is the open circuit voltage at the terminals, and the value of r is e divided by the current with the terminals short circuited.
And I also get 10 Ohms.

JimB
 

hupsenfg

New Member
I was hoping to calculate the thevenin's resistance. Shouldnt there be a shorted wire or something where we ignore a resistor?

I do not which one though..
By application on Thevenins Theorem, from a random Google Grab:


And I also get 10 Ohms.

JimB
 

dknguyen

Well-Known Member
Most Helpful Member
I was hoping to calculate the thevenin's resistance. Shouldnt there be a shorted wire or something where we ignore a resistor?

I do not which one though..
You're asking the wrong question. The a and b terminals and arrow in your circuit indicate that the you are looking into the system at terminals a and b such that the 4 ohm resistor is a load. That means you remove the 4 ohm resistor. You REMOVE it, you don't ignore it. Removing means just that. You remove it and terminals and and b become open.

We turn the voltage source off when calculating. We "ignore" it and set it to 0V which is the same as a short. Turning off a current source is setting it to 0A and that would be an open circuit instead.

I get 10 ohm if I do all this.

I would not get too hung up on not needing to redraw the circuit. We are all redrawing it, whether it is in our head or on paper. The method is no different. It just takes some experience to recognize what something is without needing to draw it down.

If two resistors share the same node at each end, they are in parallel and we just successively block resistors together like this until we get something where all the remaining resistors are in a simple series or parallel combination that we can do math on. Just like you do when you re-draw, we just don't write down all the steps because we are familiar enough that we don't need to.

In this case, I stared at it for a bit and went:
0. Set voltage source to 0V or a short and stare at it for a bit searching for any two resistors can be blocked together immediately in parallel or series.
1. I only find one: 60 || 12 so I block them together.
2. Try to block (60||12) with 20 in series or parallel...nope that won't work.
3. Notice that 60||12 and 10 can be immediately blocked together in series so I do that.
4. Now all that is left is 20 in parallel with the block (60||12 + 10)
5. I end up with 20 || [10 + (60 || 12)]

I could have redrawn it or written it down, but I kept track of the single group of resistors on a calculator and kept adding it to things I come across while remembering what I already accounted for in the diagram. So don't worry about needing to redraw it.
 
Last edited:

Ratchit

Well-Known Member
View attachment 108188
Hi i got this question off a youtube video.

In this video, the person found the equivalent resistance to be 10 Ohms.

He started off with solving 60 Ohms // 12 Ohms

However, I am not used to his method, and I rather redraw the circuit so that it is easier for me to calculate the equivalent value.

Can somebody teach me how can I trace the current flow in the entire circuit?

Shouldnt there be a short circuit at the 30 V branch?
Be careful with those youtubies. They are not always correct, or they sometimes show the hard way of doing something.

I prefer finding out the Thevenin and Norton equivalents by first finding the open circuit voltage and dividing by the short circuit current. This gives the correct answer even if the circuit contains dependent sources, which this circuit does not. OK, let's get started.

Method 1) Directly calculate the resistance. ( (60||12)+10)||20) = 10 ohms. But, what is the Thevenin voltage or Norton current?

Method 2) Use either the loop method or node method to calculate the open circuit voltage and short circuit current. The open circuit or Thevenin voltage is 35/2 and the short circuit or Norton current is 7/4. Dividing, we get 10 ohms. Ask if you have trouble doing node or loop analysis.

Method 3) Applied source method. Apply a 1 amp source to the output and see what the voltage is. Make sure you turn off all independent sources like the 30 source and substitute their resistance in the circuit. For voltage sources, the resistance is zero and for current sources the resistance is infinite. This circuit calculates to 10 volts . You can imaging what that tells you about the Thevenin resistance. Ask if you have trouble doing the applied source method.

Method 4) GIT with it and look at post #11 of http://www.electro-tech-online.com/...and-shunt-shunt-feedback.146597/#post-1242558 . You can find current in the loop containing the 4 ohm resistor by loop analysis. This is a transfer function. It is i3=(35/2)/(10+r4) , which tells you that the output resistance is 10 ohms. Again, ask if you want a detailed description.

So there are lots of ways to calculate something. It is limited only by your imagination.

Ratch
 

Les Jones

Well-Known Member
Most Helpful Member
The way I started was to just deal with the 30 volt source, and the 60 ohm an 12 ohm resistors. The thevenin equivalent circuit (At the junction of the 60 ohm and 12 ohm resistors) is 10 ohm source resistance (The two resistors in parallel) and a voltage of 5 volts ( 30 x 12/(60 + 12) ) So this can be drawn as below.
250917.jpg
The left hand 10 ohm resistance is the Thevenin equivalent resistance from the first step.
We now have a potential divider between the + 5 volts and the + 30 volts formed by the three resistors. Point "a" will be at 5 + 25 x (10 + 10)/(10 + 10 + 20) = 5 + 25 x (20/40) = 17.5 volts. The Thevenin source resistance is 20 ohms in parallel with the two 10 ohms in series so it is 10 ohms. So the current through the 4 ohm load will be 17.5 /(10 +4) = 17.5/14 = 1.25 amps.
So we applied Thevenins theorem twice.

Les.
 

hupsenfg

New Member
is it wise to trace the direction of current from point a ?

Hence, from point a, current splits into 10 Ohms and 20 Ohms
Current in 10 Ohms then split into 60 Ohms and 10 Ohms.
Current in 60 Ohms then join together with the 20 Ohms, which then joins to 12 Ohms
 

Ratchit

Well-Known Member
is it wise to trace the direction of current from point a ?
No, that method won't work if dependent sources are present.

Hence, from point a, current splits into 10 Ohms and 20 Ohms
Current in 10 Ohms then split into 60 Ohms and 10 Ohms.
Current in 60 Ohms then join together with the 20 Ohms, which then joins to 12 Ohms
Unless the circuit is absurdly simple, it is wise to use the open circuit voltage divided by the shorted circuit current. It works all the time under all conditions provided the circuit is linear. It gives you the Thevenin voltage, Norton current, and the Thevenin resistance in one method. Tracing currents can lead to confusion, especially if delta or Y circuits are encountered.

Ratch
 

MikeMl

Well-Known Member
Most Helpful Member
Look at these three figures:
open.png
Open circuit Thevenin voltage V(a) is 17.5V
short.png
The 0V supply is a convenient means to measure the short-circuit current.
thev.png
Short Circuit current I(V2) is 1.75A, therefore, the effective series Thevenin resistance is R = Eoc/Isc = 17.5/1.75 = 10Ω
Note that in the third figure, V(a)=V(t), and I(R7)=I(R5), ergo, with respect to the 4Ω load, the upper circuit is equivalent to the lower.
 
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