Another from: Fundamentals of Electric Circuits by Alexander and Sadiku 5th edition

Just to verify, for Vx, I'm getting 6.67V, the answers in the book report 2V.

For power absorbed by the 12ohm resistor, I got 21.33W, matching the book.

 

Oh. I thought this was some new public health issue. Nevermind.

 

6.67 looks right to me.
Just try to keep the VD out of your hobby work.

 

I think there is an editor out there who is removing the phrase

Edited "Fundamentals of Electric Circuits by Alexander and Sadiku 5th edition"

from his resume.

 

I think there is an editor out there who is removing the phrase

Edited "Fundamentals of Electric Circuits by Alexander and Sadiku 5th edition"

from his resume.

+1

I get the same answer you did (for the voltage), DYLH.

One very useful thing you can do is to bound your answers, preferably before you start.

So if we figure that the resistance on two resistors in parallel is equal to the smaller of the two resistors (which is the absolute most it could be), then we can do something like this in our heads:

8Ω||12Ω = 8Ω
8Ω + 1.2Ω = 9.2Ω
3Ω||6Ω = 3Ω
3Ω + 4Ω = 7Ω
7Ω || 9.2Ω = 7Ω
7Ω + 1Ω = 8Ω

So now we want the current in the 8Ω that is in parallel with 2Ω, so it will have 1/4 the current, or 1/5 of the 20A which would be 4A. Since that was with overstating all of the resistances, we know that the actual current will be more than that. So that right there rules out their 2A answer from consideration.

We can bound it the other way by noting that the resistance of two parallel resistors can't be any smaller than half the value of the smaller resistor. So using that we have:

8Ω||12Ω = 4Ω
4Ω + 1.2Ω = 5.2Ω
3Ω||6Ω = 1.5Ω
1.5Ω + 4Ω = 5.5Ω
5.5Ω || 5.2Ω = 2.6Ω
2.6Ω + 1Ω = 3.6Ω

With that in parallel with the 2Ω resistor, the current is going to be less than the 10A it would be if it were 2Ω (instead of 3.6Ω). If it were 3Ω then we would have 8A. So we know the answer must be between 4A and 8A.

So now if we work the problem and get 3.5A or 11.9A, we KNOW the answer is wrong.

If we wanted, we could tighten up that lower bound by noting that the maximum value of two resistors in parallel is also no more than half the larger resistor. That results in

8Ω||12Ω = 6Ω
6Ω + 1.2Ω = 7.2Ω
3Ω||6Ω = 3Ω
3Ω + 4Ω = 7Ω
7Ω || 9.2Ω = 4.6Ω
4.6Ω + 1Ω = 5.6Ω

Since we are doing an upper bound on the resistance, call it 6Ω meaning that 1/4 of the current is flowing in this resistor, or 5A. So now our bounds are 5A to 8A.

 

Here is my method.
The math goes:
I1 = 2X
I2 = X
20/3 = 6.67
I2 = 6.67
V = 6.67 x 1 ohm