Schaums BEE prob 3.38

[jdd18vm]

I think, this should be an easy one but I am not getting the numbers 48 Watts for (c.) and 480 Watts for (d.) no matter what I do. Got a. and b.

I must be missing something !

John

 

[Ilan]

Hope this helps. Gurus, please correct me if I am wrong.

a. Ia = Ian = Van / Z = 120/(36+j48) = 2 ang -53.1 A

b. Va'n = Van + R * Ia = 120 + 4*2 ang -53.1 = 124.96 ang -2.93 V

c. P total loss = 3 * |Ia|^2 * R = 3*2^2*4 = 48 W

d. P total input = 1.732 * |Va'b'| * |Ia'| * cos(theta) = 1.732 * (124.96*1.732) * 2 * cos(53.1-2.93) = 480.21 W

Ilan.

PS: what happens when I copy and paste the symbols from MS word, will it work?

 

[benbo]

That looks right to me! I had no idea how to do part D. I can never tell when to multiply by sqrt(3). That's why I took another module.

 

[Dark Knight]

I cannot see the problem but the SQRT(3) is usually associated to 3 phase power systems. Will dig in my stuff to see if I have that Schaum book. I might have it.

Looks like Ilan got it.

 

[jdd18vm]

Thanks Ilan thats it

I was trying to incorporate the 36 Ohms

Benbo, Im beginning to think *I* am taking the wrong module. I canuse you help on that Logig crap when i get there...lol

Luis sorry its from the Basic Electrical Emgineering Schaums and it is a 3 phase problem.

Can anyone tell me why its no longer viewable? Is is not legal to post scanned of problems? Im going to ask the general forum.

John