Recent content by justadaftspark

Homework Statement Homework EquationsThe Attempt at a Solution Stator Line voltage 220 164 112 88 42 Stator Line Current 6.8 5.4 3.9 3.8 3.7 Stator power 470 360 278 244 232 By plotting a suitable graph from these results, determine the total of windage and friction...

Thank you both... I have applied a supernode to the circuit which gave me the equation: (120+j0-VA/2+j0)+(0-VA/0-J5)+(0-VB/0+J4)+(0+J120-VB/4+J0)=0 VA-VB=V3 Therefore VA/Z4 = -9.15+j17.28A :) I now have to double check my loop analysis. Thanks again for the nudge much appreciated.

Homework Statement Determine using nodal analysis the current in branch Z4 Homework Equations [/B]The Attempt at a Solution [/B] Not sure how to handle the source v3 my solution so far has came up with answers that are not the same as the answers that I got for mesh analysis. Node A...

Hi Folks, Could anyone help please? I'm trying to find the same loop currents as His_Tonyness and this is what I have so far LOOP 1 V1=Z1(i1)+Z4(i1-i2) = 120+J0 = i1(2-J5)-i2(0-J5) LOOP 2 -V3=Z5(i2-i3)+Z4(i2-i1) = -14.14+J14.14 = -i1(0-J5)+i2(0-J1)-i3(0+j4) LOOP 3...

Awesome, Thanks a lot

Thanks for getting back, so arcos( 0.72 ) = 43.95deg Tanθ o/a Tan 43.95 * XL/10 transpose for XL XL = Tan 43.95*10 = 9.64 ohm using trig Z^2 = R^2 + XL^2 = Drum roll..... 13.89 ohm Does this seem closer to the mark? it works when i put it into a known equation P was from a earlier...

Homework Statement Find Z R=10Ω Vs=120v (PF)=0.72 Lagging Homework Equations I=P/Vs(PF) Z=Vs/I The Attempt at a Solution I have I=P/Vs(PF) and transpose for I to get 4.34A Then to find Z Z=Vs/I which gives 27.65ohm When...

Eddievic, I got the same answers as bensm0, If you double the answer that you got in a) 0.5164 use that in your equations and transpose for mass. i think :)

Ahh... big help with t = mg Not sure where you are in the world but you may have heard the penny drop. Thanks again Sunil Simha.

The mass is allowed to fall at a rate of 0.44ms^-2

Not an inconvenience at all mate, thank you for all your patience and help. I am trying to understand why the equation is the way it is. Yes my diagram is similar but without the rope on the left hand side being pulled to re cap T= up mg= down ma=down thus T-mg-ma = 0 correct?

I maybe should have mentioned that the rope is on a flywheel would it be that acceleration (0.44) is working against gravity as it is slower?

So we are saying that ma=mg-T T= m(g-a) T= 0.5(9.81-0.44) = 4.685N in an upward direction Sunil Simha, where did i go wrong in post #3?

Sorry i mean to say that the person on the ground ma would equal mg

For a person standing on the ground ma would equal m, correct? no acceleration Transpose ma=mg-T for T T=mg-ma ? can someone explain why it is either g-a or g+a. This may be simple but I am not getting it