# Thevenins equivalent resistance with only dependent sources

#### hoyyoth

Joined Mar 21, 2020
282
Dear Team,

What is the Rth of the below circuit Given R=R1=R2=10 ohms and g=10 siemens

#### crutschow

Joined Mar 14, 2008
33,325
Good question.
Since this appears to be school homework, you need to give us your best effort at solving the problem, and we will help you after that.
We are not here to do your homework for you.

#### crutschow

Joined Mar 14, 2008
33,325
Does that mean it's not really homework so we should do his work for him?

#### ericgibbs

Joined Jan 29, 2010
18,212
Hi,
All it means he is not registered as a Student, but as a Design Engineer, so technically it is not school or a college assignment.
If you don't want to help him, you are not obliged to do so, if there is any doubt in your mind about the purpose of his request for help.

Moderation.
@bertus

#### WBahn

Joined Mar 31, 2012
29,481
Dear Team,

What is the Rth of the below circuit Given R=R1=R2=10 ohms and g=10 siemens

View attachment 298444
Student or not, it's for more useful for you to work the problem yourself, with some guidance from us as needed, than just to be told the answer or even see one of use work the problem.

There are a few reasonable ways to tackle this problem, so give it some thought and perhaps describe the ones that come to mind. We can then discuss the pros and cons of each. Or pick the first one that comes to mind and work the problem as far as you can. Show your work so that we can see where you are going right and where you are going off track.

The first thing you might do is see if there are any easy simplifications that can be applied (hint -- there are).

Here's another hint -- Rth is between 1/10 Ω and 1/4 Ω.

#### hoyyoth

Joined Mar 21, 2020
282
Student or not, it's for more useful for you to work the problem yourself, with some guidance from us as needed, than just to be told the answer or even see one of use work the problem.

There are a few reasonable ways to tackle this problem, so give it some thought and perhaps describe the ones that come to mind. We can then discuss the pros and cons of each. Or pick the first one that comes to mind and work the problem as far as you can. Show your work so that we can see where you are going right and where you are going off track.

The first thing you might do is see if there are any easy simplifications that can be applied (hint -- there are).

Here's another hint -- Rth is between 1/10 Ω and 1/4 Ω.
Below is my solution.I am getting a value of 0.2 Ohm

#### WBahn

Joined Mar 31, 2012
29,481
You got it! The actual result is 20/101 ohms, but this is within about 1%, which would likely be well within the tolerance of the components.

I like that you kept everything symbolic until near the end.

I would make a couple of recommendations. First, go ahead and finish of the result symbolically -- you were so close. You end up with

$$R_{TH} \; = \; \frac{R_1 \; + \; R_2}{1 \; + \; gR_2}$$

Second, properly track your units at every step. Most mistakes you make will mess up the units, so by tracking them you can usually catch mistakes right when they are made. Otherwise, you can easily miss a mistake and it might not get caught until something fails, possibly with catastrophic consequences (I watched a guy die because he couldn't be bothered to track his units).

As you can see in the above equation, the units work out. The numerator is a resistance plus a resistance and the denominator is a unitless number added to the product of a conductance and a resistance, which is also unitless. So life is probably good. The fact that the units work out doesn't guarantee a correct result, but if they don't work out, you know the answer is wrong.

When you made the transition to using the numbers instead of symbols, you divided both sides by Vtest. But your work shows that the right hand side is supposedly still equal to Itest, not Itest/Vtest.

That first right hand side should have been written as

$$\frac{}{} \; = \; 10 \; S \; + \; \frac{1}{10 \; \Omega} \; - \; \frac{\left( 1 \; + \; 10 \; S \times 10 \; \Omega \right) \; 10 \; \Omega}{10 \; \Omega \; \times \; 20 \; \Omega}$$

Now it's a simple matter to confirm that every term works out to units of Siemens.

Continue working with the units just like they were variable -- they follow the exact same rules of algebra -- all the way to the final result, which will naturally end up with the correct units, including any scaling prefixes.

#### hoyyoth

Joined Mar 21, 2020
282
You got it! The actual result is 20/101 ohms, but this is within about 1%, which would likely be well within the tolerance of the components.

I like that you kept everything symbolic until near the end.

I would make a couple of recommendations. First, go ahead and finish of the result symbolically -- you were so close. You end up with

$$R_{TH} \; = \; \frac{R_1 \; + \; R_2}{1 \; + \; gR_2}$$

Second, properly track your units at every step. Most mistakes you make will mess up the units, so by tracking them you can usually catch mistakes right when they are made. Otherwise, you can easily miss a mistake and it might not get caught until something fails, possibly with catastrophic consequences (I watched a guy die because he couldn't be bothered to track his units).

As you can see in the above equation, the units work out. The numerator is a resistance plus a resistance and the denominator is a unitless number added to the product of a conductance and a resistance, which is also unitless. So life is probably good. The fact that the units work out doesn't guarantee a correct result, but if they don't work out, you know the answer is wrong.

When you made the transition to using the numbers instead of symbols, you divided both sides by Vtest. But your work shows that the right hand side is supposedly still equal to Itest, not Itest/Vtest.

That first right hand side should have been written as

$$\frac{}{} \; = \; 10 \; S \; + \; \frac{1}{10 \; \Omega} \; - \; \frac{\left( 1 \; + \; 10 \; S \times 10 \; \Omega \right) \; 10 \; \Omega}{10 \; \Omega \; \times \; 20 \; \Omega}$$

Now it's a simple matter to confirm that every term works out to units of Siemens.

Continue working with the units just like they were variable -- they follow the exact same rules of algebra -- all the way to the final result, which will naturally end up with the correct units, including any scaling prefixes.
Hi ,

Thank you very much.

Regards
HARI