Recent content by MrKriss

all i have found that voltage rquired to provide same values as 230V is 180V not sure how to mathematically show with correct calculations

Homework Statement For a balanced star connection i have given angles for V1=90 V2=0 V3=-90 i have to find the voltage required to produce the same phase currents as in standard three phase voltage source e.g V1=120 V2=0 V3 =-120 Homework EquationsThe Attempt at a Solution Im not even sure...

Yes thank you for correcting me . done as u previously mentiond and started from definition of J overall got to the answer which is 20piad =62... and got my answer exatly as needed. have it attached if any has interest . I am just having hard time as i missed some classes regarding these. i do...

Sorry about my mistake been sitting too long at this task i assume, but i think i have solved it so when i derrive (PI*d^4)/32 i get (PI*d^3)/8 then i integrate 6(high), 2(low)for (PI*d^3)/8 and i aquaire antiderrivative which i do not need which is PI*d^4/32 + C and the correct answer of...

Thank you both for reply and help i found how to derive that... which was suprisingly obvious and simple he he pi*(d^4)/32 = pi^((2*r)^4)/32 = (16*pi*r^4)/32 = (pi*r^2^2)/2 then i can simply integrate values. Thanks

Hya thanks for reply i know its is all in one line not rly sure how to add symbols here etc . The disk is laying flat with outside diameter 6cm and inner diamter 2cm. the answe to your question is yes(it is hollow) . the 32 part means i have to prove that 2nd moment of area "J" abouz zz axis is...

Homework Statement D1=6cm D2=2cm ( i have to prove that 2nd moment of area (J) of a circulal plate abouts its polax axis(zz) is equal to piD^4/32 ) Homework EquationsThe Attempt at a Solution J=pi6^4/32 - pi2^4/32 = 125.66 J=r1^2x2pirdr J=Integral 2pir^3d2 = 2pi integral(high 3 , low 1)...