# Answer check to average power (simple)

Discussion in 'Homework Help' started by ihaveaquestion, Sep 20, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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I've checked my work at least 4 times now and I see absolutely nothing wrong, but apparently my answer is wrong.

http://img23.imageshack.us/img23/2991/imgdi.jpg

The answer in the back of the book is 11.3 W through R1.. I'm obviously not getting that.... what's up?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Taking a different approach I calculated the impedance seen by the source.

I get Z=4.6-j1.2 ohms

This gives an RMS current of 2.38A from the source.

So the current in the 1 ohm resistor is 2.38A rms.

Hence power in 1 ohm, P = 1x(2.38)^2=5.66 Watts

So we have even more confusion with 3 different answers!

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Interestingly 5.66 W is close to half of 11.3W .... this might be worth considering - maybe the book is wrong ....???

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I have run a simulation of the circuit to check values - 5.66W looks about right. Have you double-checked that you have correctly transcribed everything from the book problem statement to your working notes?

5. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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I have double and triple checked, haha...

6. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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I've caught the book being wrong at least 3 times now....

I still can't find out what's wrong with my approach though...

7. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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I get the same result that tnk does, calculating the impedance seen by the source.

One error I see in your calculations is in the 4th equation down under the heading "For I1".

You have:

j6 I2 = -12.3 + 10.3 + I1(1 + j6)

You divide left and right side by j6 and get:

I2 = -1.72 - j2.05 + I1(-1 + j0.167)

but you have sign errors; the result should be:

I2 = +1.72 + j2.05 + I1(+1 - j0.167)

This propagates to your final result causing a substantial error. There may be other errors, but I stopped looking when I found that one.

8. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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The line I think you're looking at is 6i I2 = -12.3 + 10.3i + I1(1+6i)

I just plugged into my calculator (-12.3 + 10.3i) / 6i for the part without I1

and it gives me -1.72 - 2.05i

Then plugging in again (1+6i) / 6i for the I1 part

it gives me -1 + 1/6i

So I2 = -1.72 - 2.05i + I1(-1 + 1/6i) which is my reasoning for having that written

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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Let's try doing it by hand.

We know that 1/(6i) = -i/6

then, for example, taking just the real part of (-12.3 + 10.3i) we would have:

-12.3/6i = -12.3 * -i/6 = (12.3/6)i = (2.05)i = 2.05i, not -2.05i

And for the imaginary part:

10.3i/6i = 10.3i * -i/6 = -(10.3i/6)i = -(1.72)i*i = 1.72, not -1.72

So your calculator is saying that (1+6i) / 6i = -1 + 1/6i

Let's do it part by part.

(1+6i) / 6i = (6i + 1)/6i = 6i/6i + 1/6i

Where does the -1 come from? 6i/6i = 1, not -1.

Next consider the second part. 1/6i = -i/6 = -0.167i, not +0.167i, as you have in the image attached to post #1.

Unbelievable as it may seem, apparently your calculator is getting the signs wrong. What kind of calculator are you using?

10. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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What you're saying makes sense Electrician... very frustrating.... it's a Casio 115ES

11. ### ELECTRONERD Senior Member

May 26, 2009
1,146
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I have the exact same calculator. Funny it isn't working like it should...

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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What do you get if you calculate 6i/6i?

13. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
For some reason I can't use "i" in my calculator. It isn't working...I need to know what settings the calculator should be on...

Last edited: Sep 21, 2009
14. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
I just pressed the Mod/Setup button and pressed 2 for complex numbers.

15. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
How stupid... you have to put parenthesis around (6i) to get the right answer... i.e. (1+6i) / (6i)

16. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Pretty sure I'm still not getting right answer with right calculations...

17. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Ok, I have the CMPLX on the screen but I press A and then the <Shift> ENG button to get i, but it won't come out.

Last edited: Sep 21, 2009
18. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Nevermind, I figured it out. When you do (1 + 6i) / (6i) I get 1 / (-1/6i). Is that what you get? I do have the right "i" don't I?

19. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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You should get (1 + 6i)/(6i) = 1/(6i) + 6i/6i = -.0167i + 1 = 1 - 0.167i

I redid the calculations under the heading "For I1"

When you get down to the place where you had the sign error, you should have:

I2 = +1.72 + j2.05 + I1(1 - j0.167)

Then substituting I2 = (j6 I1)/(2 + j4) you should have:

j6 I1 = (2 + j4)(1.72 + 2.05 + I1(1 - j0.167)

and finally:

I1 = (-4.76 + j10.98)/(-2.67 + j2.33) = (3.05 - j1.44)

The magnitude of I1 is 3.375 amps.

The average power is 1/2*(3.375)*(3.375) = 5.695 watts

Your numbers may be slightly different due to rounding error from using only 3 digits in the arithmetic, but this is essentially the same result tnk and I got using the impedance seen by the voltage source.

Last edited: Sep 24, 2009
20. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Anyone who is going to do a lot of complex arithmetic should have a calculator that isn't going to have weird behavior that trips you up like this Casio. Imagine if you were trying to solve a problem on an exam and didn't know about this peculiarity!

I would recommend something like a TI89 or HP50. You can get those on eBay for a quite reasonable price. Instead of the current production HP50, get the just barely obsolete HP49G+ for less money, but the same performance.