Question

Can anyone help me with this problem set... i got to do many assignment on my other subjects, can anyone help me to answer this...ill give medal.

Answer 1

@osprey

Answer 2

7,8,9 :D

Answer 3

for No7, combining the impedances is just some complex number algebra

For the parallel set up, \(\dfrac{1}{Z_p} = \dfrac{1}{3 + 4i}+ \dfrac{1}{8 - 6i}\)

Realise the denominators using conjugates

\(\dfrac{1}{Z_p} = \dfrac{1}{3 + 4i}\dfrac{3 - 4i}{3 - 4i}+ \dfrac{1}{8 - 6i}\dfrac{8 + 6i}{8 + 6i}\)

\(= \dfrac{3 - 4i}{3^2 + 4^2}+ \dfrac{8 + 6i}{8^2 + 6^2}\)

\(= 0.2 - 0.1 i = \dfrac{1}{Z_p} \)

same trick again

\(Z_p = \dfrac{1}{0.2 - 0.1 i} \dfrac{0.2 + 0.1 i}{0.2 + 0.1 i} \)

\(= \dfrac{0.2 + 0.1 i}{0.2^2 + 0.1^2 } =( 4 + 2i )\Omega\)

then combine the series:

\(Z_T = Z_ P + Z_1 = 4 + 2i + 10 = (14 + 2i )\Omega\)

So in polar coordinates \(Z_T = (4.47 \Omega, 26.57^o )\)

I'd need you or @osprey to tell me again what you are calling Total Power. Is it power across the pure resistor?!?! Instantaneous or RMS??

but that's what you need for the phasor diagram

8 seems to have disappeared?!?!

And 10 to start with is a whole loada that same old algibree, i'd shove it into Wolfram

then some calculations.....with some metrics/ calculations, some of which that I am not too fluent with right now :(

Answer 4

Correction for polar coords \(\Large (Z_T=(14.14~Ω,8.13^o)\)

Answer 5

@IrishBoy123 I was thinking about 000---000 to you and @radar over this one.

Answer 6

OK :))

Radar's tagged so that's good. i could dig about and guess something (!!) but i think the phasor is the key and it'll all fall into place somehow .....

Answer 7

@IrishBoy123 The key is having the mental stamina ! Also, the poster has "offered" a tight deadline, which, of course, they all do. Doesn't help in formulating answers, though.

Answer 8

@Caerus Further scribbles added to enclosed

Answer 9

@osprey @IrishBoy123 , guys... whats the correct answer in number 7? your answer is not vary to each other lol..which one is correct?

Answer 10

@osprey

Answer 11

@Caerus @IrishBoy123 I'd defer to the shamrock

Answer 12

ok

Answer 13

so irush got the correct answer?

Answer 14

I've added a few more bits in yellow.

Answer 15

can you check this for me @osprey

Answer 16

@osprey hmm.. i think you just done a topographic error ,you plug 8−i instead of 8−6i

Answer 17

@Caerus typographic errors are one of my "strong points". There's so much algebra flying around.
This is for an Elec Eng course ?

Answer 18

yup.. ^.^ .... hmmm.. so how can we get for total power?

Answer 19

if ZT=(14.14 Ω,8.13) how can we get total power, lol i really dont know

Answer 20

we have a phase of 60 Hz..which we didnt use at all..

Answer 21

first Hz are Hertz which is a frequency. Phase is usually in degrees or radians.

Could be that the 60Hz is saying that it's your mains frequency.

Is the "deadline" for these answers Monday ? - How "serious" a deadline is that, and what's a good score ?

You'll presumably get "support" from the course in terms of maths and complex numbers and things ?

Answer 22

yup, its for tommorow..... yet i didnt understand some of this stuff because im working for statics, dif. equations, basic mechanical stuffs... hmmm... we will have exam for Wednesday...its totally catastrophe :'(

Answer 23

this problem set serve for our class standing for this term..

Answer 24

serve as*

Answer 25

how many hours do you have left ?
Catastrophes happen when things actually fall apart.
Do you have any idea how your classmates are doing ?

Answer 26

im lucky to have you and irish...

Answer 27

that sounds calmer.

how many hours do you have left ?

Answer 28

its 10:20 pm here, 2ill pass 7:30 in the morning, im starting to transfer the answers... i need to be sure that all problems are correct...its the only class standing in this term..

Answer 29

Done some corrections, hopefully shown by red borders ... The numbers bother me a little since I'd sort of hope that with all the algebra that seems to be involved, the numbers would turn out to be "easy", as they often seem to be designed to do in Applied Maths problems.
But, your time is short ...