Answer check to average power (simple)

t_n_k

Joined Mar 6, 2009
5,455
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!
 

t_n_k

Joined Mar 6, 2009
5,455
Interestingly 5.66 W is close to half of 11.3W .... this might be worth considering - maybe the book is wrong ....???
 

t_n_k

Joined Mar 6, 2009
5,455
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?
 

The Electrician

Joined Oct 9, 2007
2,986
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.
 

Thread Starter

ihaveaquestion

Joined May 1, 2009
314
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.
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
 

The Electrician

Joined Oct 9, 2007
2,986
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
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

Then plugging in again (1+6i) / 6i for the I1 part
it gives me -1 + 1/6i
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.

it gives me -1 + 1/6i
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?
 

ELECTRONERD

Joined May 26, 2009
1,147
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...
 
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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? :D
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.
 
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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.
 
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