for those need the calculation 5:45 : J^2 = -1 , 1/J = -J , -1/J = J -50J / 5 - 10j >> multiply by conjugate of (5+10J) / (5 + 10J) >> you should get (-250J -500J^2)/(25+50J-50J-100J^2) >>>> now apply J^2 = -1, you should get 4 - 2J as rectangular form... then convert them to polar form r = sqrt(x^2 + y^2) >>>>>>> final answer is 4.47 ..(-26.6*).
can you tell me how you are doing the calculations purely on the calculator? (what mode is it degrees, radians, any other calc settings that must be changed. also can you write out what you are typing on the calculator if you don't mind, please and thanks!)
for those who are asking how to do without calculator, answer is you cant. when it comes to how to do it with calculator, first you gotta calculate -j50/5-j10 by hand you can do it, its 4-j2. For making it phasor you open a root (4^2 + 2^2) and for the angle you use tan-1 (reverse tangent) function on your calculator and type tan-1 (-2/4) = -26.56 degree. but the prof rolled it into -26.6 for more nice value.
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for those need the calculation 5:45 : J^2 = -1 , 1/J = -J , -1/J = J
-50J / 5 - 10j >> multiply by conjugate of (5+10J) / (5 + 10J) >> you should get (-250J -500J^2)/(25+50J-50J-100J^2) >>>> now apply J^2 = -1, you should get 4 - 2J as rectangular form... then convert them to polar form r = sqrt(x^2 + y^2) >>>>>>> final answer is 4.47 ..(-26.6*).
bruh thank you, i was about to go and review this
You make this very clear! Helped me understand this months ago in my intro to circuits class, and helping me again now that I am tutoring the class.
can you tell me how you are doing the calculations purely on the calculator? (what mode is it degrees, radians, any other calc settings that must be changed. also can you write out what you are typing on the calculator if you don't mind, please and thanks!)
juanvera94 complex mode dear
to get the angle, you multiply the value by 180/pi to get from radians to degrees
@@bajza6046 you saved me !!
This depends on the calcualor. Some will automatically interpret. Some will let you use rectangular and polar together. Others won't.
excellent -excellent -excellent - my lord -please solve by transfer impedence method also --please make lecture of 10 minutes i.e short time
How did you come up with the angle -26.6
you are the best sir , thanks
Fabulous!
it is helpful but you should do all by hand not by calculator
4.47 (-26.6) and 2+2i are parallel, so the total resistance should be 4.47 (-26.6) *(2+2i)/(4.47 (-26.6)+(2+2i))
he totally forgot that, thanks god those are the basic consepts so we can still catch up the missing calls
You don't need total resistance when using current division.
Keep going dr , u r awesome
thanks you so much!!!
Sir, u are my hero :)
How about a 4th answer? needs to download open source Octave (~= MATLAB)
>> Is = 4*cos(20/180*pi)+4*sin(20/180*pi)*i Is = 3.7588 + 1.3681i
>> Zp1 = -5*i/(100*0.001)/( 5-i/(100*0.001) ) Zp1 = 4 - 2i
>> Zs=2+0.02*100*i Zs = 2 + 2i
>> I = Is*Zp1/(Zp1+Zs) I = 2.96187 - 0.34087i Why memorize this equation? Why not just use Ohms law?
>> Imag = abs(I) Imag = 2.9814 >> Iang = angle(I)*180/pi Iang = -6.5651
I’m getting stuck on the phasor math. Do you convert to from rectangular to polar if you want to do the division by hand?
If I recall, one form is easier for adding, the other for multiplication.
Decent calculator lol
God bless you.
Correct me if Im wrong but when I did it I got: 2.47
I got the same answers that he got .
In engineering we are not alowed to use these calculators
Why negative -10j. Why the negative?
how is the capacitor is negative?
by the formula -j/(wc)
what a calc!
22:04 what happened to the 'j' after you multiplied 10j to 1.153?????????????????????????????????????
Its converted into polar
It is converted rectangular to polar
is there a part 2
where is Part 2?
Can you do the calculations without a calculator...come on
When I try to calculate this on the calculator it gives me an error
Thank you so much sir
for those who are asking how to do without calculator, answer is you cant.
when it comes to how to do it with calculator, first you gotta calculate -j50/5-j10 by hand you can do it, its 4-j2. For making it phasor you open a root (4^2 + 2^2) and for the angle you use tan-1 (reverse tangent) function on your calculator and type tan-1 (-2/4) = -26.56 degree.
but the prof rolled it into -26.6 for more nice value.
how did tan-1(-2/4) = -26.56???? isn't it -.463??
zahi how why didnt combine the all the load to become 1 load
it’s the current divider formula
Isn't the current divider formula i(source) * (i(total) / i(target))?
NO, it is (source current)*(offset resistor)*(total resistance)
You don't know this topic
Please first get clear of it