Not really a homework question, but I need your assistance in understanding this practice problem from the book "Applied Circuit Analysis 1e". Every time I solve the problem step by step, I seem to be getting a different answer from the book. The book says the answer for Vo is 2.4V but I keep getting 2.07V as the solution for this circuit.

Here are the steps I used to arrive to my answer.

Like the book demanded I only tried to use mesh analysis to solve this. Anyways, the 2 equations I got from both loops are the following:

Loop 1 = 9i1 - 4i2 = 24
Loop 2 = -4i1 + 12i2 + 6Ix = 0

Now from my understanding, Ix = i1, so, if I plug i1 to the loop 2 equation I get:

-4i1 + 12i2 + 6i1 = 0
2i1 + 12i2 = 0 (Now I subtract both sides by 12i2 which gives me)
2i1 = -12i2 (Divide both sides by 2)
i1 = -6i2 (Now I use this equation and plug it to the loop 1 equation)

9 (-6i2) - 4i2 = 24
-54i2 - 4i2 = 24
-58i2 = 24 (Divide both sides by -58)
i2 = -0.414A

Now that I have i2, to find Vo on this circuit, don't I just have to multiply i2 by the 5 ohm resistor? When I do so, I get 2.07V. (Positive because of voltage polarity)

I been wondering what I'm doing wrong, or what the book is doing right to arrive to that answer. Anyways, any feedback on this problem would be appreciated, thanks

 

I am getting same answers. Have you checked the textbook errata? They usually publish the list of errors after the book has been printed and has been in circulation. Go to publisher's website and see if they have the errata available for download.

 

There should seldom, if ever, be any question as to which is right. You should get in the habit of checking your work and then, with rare exceptions, you will KNOW that you are right (or not), and when you disagree with the book's answers then you can check THEIR answers and KNOW that that are right (or not).

Let's check the book's answer.

They say Vo is 2.4 V. Okay, then that means that is 0.48 A flowing in both the 5 Ω resistor and the 3 Ω resistor right below it. Thus the voltage drop across those two resistors totals to 3.84 V and the voltage across the 4Ω resistor is 3.84V + 6 Ω·Ix. So the current in the 4 Ω resistor is (3.84 V + 6 Ω·Ix)/4 Ω. This current, plus the current in the 5 Ω resistor, equal Ix.

Ix = (3.84 V + 6 Ω·Ix)/4 Ω + 0.48 A = (0.96 A + 0.48 A) + 1.5·Ix
-0.5·Ix = 1.44 A

Ix = -2.88 A

Since this is the current flowing in the 2 Ω and the 3Ω resistor right below it and since Ix is negative, this would result in the voltage across the 4 Ω resistor being greater than the 24 V supplied by the leftmost source. But with Ix negative, the voltage across the 4 Ω resistor must be less than 3.84 V. So we know that the text's answer is wrong.

Let's try this again using your answer of 2.07 V.

You get a current in the 5 Ω resistor of 414 mA giving a voltage drop across the 4 Ω resistor of (3.312 V + 6 Ω·Ix). This gives us

Ix = (3.312 V + 6 Ω·Ix)/4 Ω + 414 mA = 1.242 A + 1.5·Ix

Ix = - 2.484 A

And you have the same problem as the textbook because, with a negative Ix, the voltage across the 4 Ω resistor has to be greater than 24 V when looking at the left loop and less than 3.312 V when looking at the right loop. So we know that YOUR answer is wrong, too.

But now we can look at your work (and thank you, thank you , thank you for providing it!) and see exactly where the problem (or at least A problem) is. Your i2=-0.414 A and, based on having to reverse engineer your definitions since you don't provide them, which is BAD!, your i2 is the current flowing clockwise in the right hand loop. This means that Vo = i2·5 Ω = -2.07 V.

So let's check THIS answer.

You get a current of -414 mA giving a voltage drop across the 4 Ω resistor of (-3.312 V + 6 Ω·Ix). This gives us

Ix = (-3.312 V + 6 Ω·Ix)/4 Ω + -414 mA = -1.242 A + 1.5·Ix

-0.5·Ix = -1.242 A => Ix = 2.484 A

This means that the voltage across the 4 Ω resistor looking at the left loop is 24 V - 2.484 A·5 Ω = 11.58 V

The voltage across the 4 Ω resistor looking at the right loop is 6 Ω·2.484 A - 414 mA·8 Ω = 11.59 V

Thus this answer is almost certainly correct.

My guess is that had you simply followed good practice and defined your terms before using them, you would not have made the mistake you had. It is easy enough to do and there is really no excuse for not doing it:



Note that you really do need to define BOTH mesh currents, because the Ix given is NOT a mesh current, it is simply a particular branch current. But it is perfectly fine and reasonable to define your mesh currents such that they are compatible with the given branch currents.

You also need to start tracking your units throughout your work. Most mistakes you make (though not the one here) will mess up the units, but only if the units are there to get messed up.

Always, always, ALWAYS track your units.

Always, always, ALWAYS ask if the answer makes sense.

Always, always, ALWAYS check your work (which is almost always possible, even in the "real world").

EDIT: Thanks to The Electrician for catching a typo.

 

Notice that the branch current Ix is shown as the current through the 2Ω resistor; the voltage across that resistor could have been labeled Vx. If the dependent voltage source in the right mesh had then been labeled "Vx", or "2 Ix", the book answer would have been correct. Apparently, whoever did the drawing wrote "6" instead of "2".

 

And you have the same problem as the textbook because, with a negative Ix, the voltage across the 4 Ω resistor has to be greater than 24 V when looking at the left loop and less than 3.312 V when looking at the right loop. So we know that YOUR answer is wrong, too.

But now we can look at your work (and thank you, thank you , thank you for providing it!) and see exactly where the problem (or at least A problem) is. Your i2=-0.414 A and, based on having to reverse engineer your definitions since you don't provide them, which is BAD!, your i2 is the current flowing clockwise in the right hand loop. This means that Vo = i2·5 Ω = -0.414 V. (Should be Vo = i2·5 Ω = -2.07 V.)

So let's check THIS answer.

You get a current of -414 mA giving a voltage drop across the 4 Ω resistor of (-3.312 V + 6 Ω·Ix). This gives us

Ix = (-3.312 V + 6 Ω·Ix)/4 Ω + -414 mA = -1.242 A + 1.5·Ix

-0.5·Ix = -1.242 A => Ix = 2.484 A

This means that the voltage across the 4 Ω resistor looking at the left loop is 24 V - 2.484 A·5 Ω = 11.58 V

The voltage across the 4 Ω resistor looking at the right loop is 6 Ω·2.484 A - 414 mA·8 Ω = 11.59 V

Thus this answer is almost certainly correct.

See correction above.

 

See correction above.

Thanks for looking at the details and catching that. I even remember the copy/paste that resulted in that error.

 

Thanks for looking at the details and catching that. I even remember the copy/paste that resulted in that error.

Did studiot also catch the error? Before me?

 

Sorry.