Can you explain the details of how we arrive at this specifi

Can you explain the details of how we arrive at this specifically for the Co-Energy? Finding the electromagnetic torque from there is simple enough.

From the Flux Linkage equations, what portions are contributing to the resultant Co-Energy (Wc), which are being discounted, why?

The flux linkage equations for the machine shown to the right can be expressed as: A, = (Ls + Lms)Isa-Lms sin().-Lms cos@rx, rb ra ra rSa msra COS Sa Sa ra where Lms-50 mH and 7s Determine the value of the electromagnetic torque Sa 6, -36.9° ra ra rb rb Sa sa

Solution

given that

i_sa=8A,i_ra^\'=-4A,i_rb\'=3A

L_ms=50mH,L_ls=L_lr^\'=1mH,angle=36.9^o

^{angular speed is given by}

^W_e=0.5L_msi_sa²+0.5L_msi_ra²+0.5L_msi_rb^2-L_ms(i_sai_ra-i_rbi_sa)

by substituting all the values we can get

^W_e=0.826

torque=(2/p)J(dW/dt)

here p=2

by substituting all the values we can get

T=0.826J

here torque depends on inertia.