So for C1 i have the parametrisation as r(t) = <t,0,0>
I got this by doing:
x= t(1) + (1-t)(0) = t
and zeros for the others.
Is this the correct way to evaluate it?
Okay, so i form three integrals and combine them?
Also that's where i am confused, when i am doing C1 shouldn't the 3dz go away too as its treated as a constant?
Hi all,
I'm finding it difficult to start this line integral problem.
I have watched a lot of videos regarding line integrals but none have 3 line segments in 3D.
If someone can please point me in the right direction, it would help a lot.
I've put down the following in my workings:
C1...
So it's just the nature of the question that made both values close to each other?
All in all thank you so much for your thorough help in this problem, I can't tell you how much I appreciate you and am thankful for helping me understand this topic clearer.
ok here it is:
Ea for 260V: Vdc - IaRa
260 - 30(0.25) = 252.5V
Ea for 300V as calculated before is 292.5V
T (260V) = 252.5 (30) / (1276.36) (2pi/60) = 65.65 Nm
T (300V) = 292.5 (30) / (1276.36) (2pi/ 60) = 56.67 Nm
now : T (300V) / T (260) = I (f) 1 / I(f) 2
65.65 / 56.67 = 5/ I(f)2...
Yes so after doing the method i posted earlier, I am still obtaining an answer of 20 ohms.
So something I'm doing must not be right, but i can't understand what?
Okay so applying that method;
After i get the torques in the two cases is it correct for me to do this:
T(300V) / T (260V) = I (f) 1 / I (f) 2
where I (f) 1 is 5A and I (f) 2 is unknown.
Also is the Wm (mechanical speed in radians) the same for both volatges?
I think my confusion is lying in what i exactly equate at which step.
For the first step of the solution do i use the equation T = Ea * Ia / Wm
For both 300V and 260V to get the torques for both?
Then i would need to relate that to field current I(f).
So where does mechanical power come...