Find the Power consumed by 90 Ohm Resister by using Thevenin’s Theorem

I got the power as 2.5 watts. But I got the current as -0.166 A That's why I thought it may be wrong. Can you please do and check if my answer is correct or not?

 

As this is homework, the purpose of which is to acquaint yourself with analysis of such circuits, how about providing us with your worksheets so that we can see the method you used in getting your answer to the power across the 90 ohm resistor.

hgmjr

 

As hjmgr says, you need to show some work before we can help you.

To motivate you, I will tell you that the result you have so far is incorrect.

 

Alright! I don't have a scanner so I took the pics from my phone. You have to zoom.

 

Your first 3 equations are correct. Vth is -10 volts.

Equation 4 has a problem. You seem to be using the bottom edge of your circuit for the reference, so when you replace the 150 volt source with a short, V1 becomes zero; there is no node V1 anymore. There is no longer any need for KCL at V1, because V1 is now ground; V1 can't possibly be 165 volts. So, get rid of that equation 4.

Also, you aren't treating the 3I dependent current source correctly, even if there were a voltage at V1. (The left end of the 3I dependent current source is grounded now, anyway.)

Get rid of V1 in your next two equations and try again.

 

I haven't looked over your work carefully, but...when you use KCL at node v1, it looks like you ignore one of the branches.

 

Consider your second equation ...

-3I+(VA-V2)/60+VA/30=0 ****** equation (2)

You should keep in mind that VA/30=I or VA=30I

make the substitution for VA into you equation (2) above and you will
then have V2 simply as a function of I. You'll get V2=constant*I where the constant could be negative.

Substitute V2 as a function of I into your first equation ....

(V2-V1)/45 +V2/15 + (V2-VA)/60=0 ****** equation (1)

Again remember VA=30I and you know V1=150. So you'll then have an equation with the only unknown being I - for which you can then solve.

Knowing the value of 'I' will give you VA.

You can then find the Thevenin resistance to give the complete equivalent circuit.

 

Incidentally, you shouldn't be surprised if the Thevenin resistance is negative in this case. I'm still getting my head around that one. But it all seems to work out OK - mathematically speaking.

 

Consider your second equation ...

-3I+(VA-V2)/60+VA/30=0 ****** equation (2)

This is his equation 3 (before simplification). He has a circled 3 to indicate the simplified version.

You should keep in mind that VA/30=I or VA=30I

make the substitution for VA into you equation (2) above and you will
then have V2 simply as a function of I. You'll get V2=constant*I where the constant could be negative.

Substitute V2 as a function of I into your first equation ....

(V2-V1)/45 +V2/15 + (V2-VA)/60=0 ****** equation (1)

This is his equation 2. His equation 1 is V1 = 150

Again remember VA=30I and you know V1=150. So you'll then have an equation with the only unknown being I - for which you can then solve.

Knowing the value of 'I' will give you VA.

He actually did ok finding VA.

You can then find the Thevenin resistance to give the complete equivalent circuit.

This is the part he is having trouble with.

 

Thanks Electrician - I missed that he had found VA. Good work on sonutulsiani's part!

Sometimes it's a challenge to wade through pages of handwritten work. You are more patient than I am.

Yep - finding Rth is deceptive.

 

Your first 3 equations are correct. Vth is -10 volts.

Equation 4 has a problem. You seem to be using the bottom edge of your circuit for the reference, so when you replace the 150 volt source with a short, V1 becomes zero; there is no node V1 anymore. There is no longer any need for KCL at V1, because V1 is now ground; V1 can't possibly be 165 volts. So, get rid of that equation 4.

Also, you aren't treating the 3I dependent current source correctly, even if there were a voltage at V1. (The left end of the 3I dependent current source is grounded now, anyway.)

Get rid of V1 in your next two equations and try again.

So now I got Vab after after solving for V2 and Va.. Va = -19 V - Vab
So then for current I got this equation:


10 - 19I + 90I = 0

I = -0.1408 A

Power = 1.785 watts ?

 

That's right, but don't forget that you're supposed to use Thevenin's theorem to find the answer. So, you need to show the Thevenin equivalent.

 

Yes I did that already
Thanks