# help on loop analysis

Discussion in 'Homework Help' started by Hawkeye87, Oct 20, 2008.

1. ### Hawkeye87 Thread Starter Active Member

Oct 7, 2008
52
0
Ok, my teacher is really being difficult and assigning a review then a quiz. Both of which i'm having difficulty with. The quiz says to find the power in R2 and R3 using loop analysis. I've already found the loop currents but I don't know a way to find the power through those resistors.

File size:
6.7 KB
Views:
21
File size:
3.8 KB
Views:
25
2. ### alwayslearning Member

Feb 27, 2008
22
0
The negative sign on your current values mean the current is flowing the opposite direction you assummed.
Note that both currents i1 and i3 travel in the same direction through R2, according to your answers.

Power R2:
PR2= (i1^2)*R2 + (i3^2)*R2

Power R3:
PR3= (i3^2)*R3

Last edited: Oct 20, 2008
3. ### Hawkeye87 Thread Starter Active Member

Oct 7, 2008
52
0
could you check and make sure i got the right currents? i'm pretty sure i got the right loop currents but don't quote me on that.

4. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Greetings hawkeye87,

A quick check of your equations revealed no errors. I then checked the resulting currents that you calculated and again I found no errors. Looks like you have succeeded with your analysis up to that point.

It would seem that all you would need to do at this point is take the net currents flowing in R2 and R3 and use I*I*R to get the power in those resistors. I*I will get rid of any negative signs that you have obtained so you will end up with power bearing a positive sign which is proper.

hgmjr

5. ### Hawkeye87 Thread Starter Active Member

Oct 7, 2008
52
0
by saying I*I, what exactly do you mean? multiplying by the current twice seeing that two negatives make a positive? or could you just multiply by either a positive or negative 1 to get the same result?

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
What I mean is the formula for power when the resistance and the current flowing in the resistance is given.

7. ### Hawkeye87 Thread Starter Active Member

Oct 7, 2008
52
0
ahh gotcha. that's right. let me do the quick math and repost

8. ### steveb Senior Member

Jul 3, 2008
2,431
469
When using loop currents to calculate power with the I^2*R formula, make sure to add the loop currents before you square the value. Loop currents are not real currents in the general case, so you can't square them and then add them.

For example: If I1 and I2 are loop currents flowing in the same direction thru a resistor R, the power dissipated by R is (I1+I2)^2*R not I1^2*R+I2^*R

9. ### Hawkeye87 Thread Starter Active Member

Oct 7, 2008
52
0
Why is it you have to add the loop currents?

10. ### steveb Senior Member

Jul 3, 2008
2,431
469
Basically, the loop currents are artificial, and used as a mathematical trick to solve problems. The real current in any particular resistor is the sum of all loop currents that pass through it. Note that when I say sum, I really mean either addition or subtraction based on the direction of the loop current.

When calculating power, it is necessary to use the actual real current that flows through the resistor in order to use I^2*R.

11. ### Hawkeye87 Thread Starter Active Member

Oct 7, 2008
52
0
so you wouldn't happen to know much about Thevenin's, max power transfer, Superpostion, Nodal or mesh? i've got a test friday and i need a good review on these ones.

12. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
The material on network analysis contained in the AAC ebook offers a good review of the various techniques you have mentioned as well as others. There is also a worksheet section available that provides example problems along with the answers.

hgmjr

13. ### alwayslearning Member

Feb 27, 2008
22
0
steveb wrote:

When using loop currents to calculate power with the I^2*R formula, make sure to add the loop currents before you square the value. Loop currents are not real currents in the general case, so you can't square them and then add them.

For example: If I1 and I2 are loop currents flowing in the same direction thru a resistor R, the power dissipated by R is (I1+I2)^2*R not I1^2*R+I2^*R
Yesterday 07:54 PM
--------------------------------------------------------------------------

Steveb:

Thank-you for directing us on the math error:

(i1^2)*R2 + (i3^2)*R2 not equal to (i1 + i3)^2*R2

expand and cancel R2 both sides:

(i1^2) + (i3^2) not equal to (i1 + i3)(i1 + i3)

finally:

(i1^2) + (i3^2) not equal to (i1^2 + 2*(i1^2)*(i3^2) + i3^2)

Though when you say "...make sure to add the loop currents before you square the value. Loop currents are not real currents..."

They are real tangible currents steveb, it is just that the two formulas are for two different circuits-It has nothing to do with "not real currents". The 1st formula would be used for two equal resistors in parallel and the 2nd formula for the circuit in this posting.

Referring to your second post; that loop currents are artificial. Again the loop currents are actual measurable values. You can easily prove this by measuring the voltage across R1, R2 and R3. The current that travels through R3 and R1 physically travel through R2. They consume real power into heat-nothing artificial about it!

14. ### steveb Senior Member

Jul 3, 2008
2,431
469
Yes, I see your point. It can be misleading to call them artificial, or not-real. Basically, I was just trying to call attention to the fact that you have to be careful when you use them, but I was unclear. If only one loop current flows thru a resistor, then that current is certainly real. However, if a resistor current is said to be the sum of two or more loop currents, the meaning is less clear. But, still those loop currents may be measurable somewhere else. However, that is not always the case, and there are exceptions to this.

To me the loop current approach is a mathematical device to get solutions. A good example is an extensive two dimensional resistor ladder network. You can have many, or even most, loop currents that do not show up as a current in any single resistor, or even a wire. I seem to remember some examples from school where no loop current showed up in any device. That is, the actual current in any resistor was the sum of two or more loop currents. This is all I mean by artificial. There are some cases where a loop current is not measurable.

Last edited: Oct 21, 2008