Finding load current using kirchoffs

[Andrew187]

Hi there attached is a task where i had to calculate the value of the load using kirchhoffs law I would be grateful if someone can check this to see if its correct, its part of an assignment question.

 

[gneill]

Looks okay to me.

 

[KasraMohammad]

Yup, looks right. Nice work.

 

[The Electrician]

You did get the correct final answer, but I wouldn't call it nice work.

Under the heading "Loop B E F C" you have a number of algebra errors.

Your next to last equation has inconsistent units, a problem inherited from above.

Furthermore, in the next to last equation, you didn't divide the 3V by 4Ω without which you can't get a numerical value of .851

Don't turn this paper in.

 

[Andrew187]

So where have I gone wrong? It really is bugging me because I know my answer is right as I have done this circuit with nortons theorum and it gave me 0.444a, but with this was a stab in the dark with kirchhoffs but the paper requires me to use kirchhoff I'm all over the place:(

 

[The Electrician]

The first equation you have after the heading "Loop B E F C" is:

2v = 6ΩI2 + 1ΩI1(I2-I1)

The I1 in red shouldn't be there. Then next you have:

2V = 1ΩI1 + 7ΩI2 but it should be:

2V = -1ΩI1 + 7ΩI2

Next you have :

3V = 4ΩI1 - 1ΩI2 which you brought down from Loop A B C E; it is correct.

But next you rewrote equation (2) introducing a typo. The coefficient of I1 should be -1Ω (although you had +1Ω; you didn't even copy what you had correctly, but it was wrong anyway. But, at the very least you should be able to copy it without introducing a typo). You have:


2V = -4ΩI1 + 7ΩI2

but it should be:

2V = -1ΩI1 + 7ΩI2

Then you copied equation 1 again:


3V = 4ΩI1 - 1ΩI2

Next, you multiplied your incorrect version of equation 2 by 4:

8V = -16ΩI1 + 28ΩI2

but it should actually be:

8V = -4ΩI1 + 28ΩI2

Apparently, what you did at the end was to add these two equations:

3V = 4ΩI1 - 1ΩI2

8V = -16ΩI1 + 28ΩI2

and got as a result:

11V = 27ΩI2

But if you had done the algebra correctly, you would have gotten:

11V = -12ΩI1 + 27ΩI2

which would have given you a wrong answer.

The two equations you should have added are:


3V = 4ΩI1 - 1ΩI2 This is the correct equation (1) from Loop A B C D

8V = -4ΩI1 + 28ΩI2 This is the correct equation (2)

Now, if you add these two equations, you get:

11V = 27ΩI2 which is correct. You somehow got this even though your algebra was incorrect.
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Next, you aren't keeping track of your units properly.

Under the heading "Sub I2 into equation (1)", you have:

3V = 4ΩI1 - 1ΩI2 (0.407a)

which then becomes:

3V = 4ΩI1 - 0.407a

but the second term on the right hand side isn't amps; you got it by multiplying 1Ω times .407 amps which gives .407 volts, not .407 amps.

The units problem still exists in the next equation:

3V + 0.407a = 4ΩI1

It should be 3V + 0.407V = 4ΩI1

Then you would have:

I1 = 3V/4Ω + 0.407V/4Ω = 0.851a

I think you need to practice your algebra skills some more, and exercise greater care with your units.

 

[gneill]

So where have I gone wrong? It really is bugging me because I know my answer is right as I have done this circuit with nortons theorum and it gave me 0.444a, but with this was a stab in the dark with kirchhoffs but the paper requires me to use kirchhoff I'm all over the place:(

The Electrician is correct. I paid more attention to your initial loop equations and the final expressions for the result values than to the mechanics of the math lying in between. Shame on me

Clean up your algebra as The Electrician suggests.

 

[Andrew187]

First of all let me thankyou guys for your help, You are right electrician my algebra is attrocious:( But I've had another go hopefully it is right this time thankyou one more time.

 

[The Electrician]

Under the heading "Loop A B C D", you are not handling your units in a consistent way; you should have:

3v = 3ΩI1 + 1Ω(I1 - I2)
3v = 3ΩI1 + 1ΩI1 - 1ΩI2
3v = 4ΩI1 - 1ΩI2

and under the heading "Loop B E F C", the 5th and 6th equations down, you should have:

3v = 4ΩI1 -1ΩI2 You copied this wrong; the 4 is not negative.
8v = -4ΩI1 + 28ΩI2 You forgot to multiply the -1 by 4.

You've got to be more careful or you're going to have trouble come exam time!