Question

This is for my electric machines class.

It’s a four loop mesh analysis problem

Please show work, and I think open circuit/ short circuit is required to do it

Namc a. Solve resistor the mesh equations for the network in Figure 1, Determine the power dissipated by the 6 Ω Oce 20Ω 80Ω 10Ω a6Ω 15Ω Road 200 522 5Ω Figure 1 b. Show that the net complex power dissipated (by the passive elements) is equal to the net complex power generated by the sources.(Conservation of Energy] c. Determine the ohmic value of Ra that will dissipate the maximum power that this circuit can possibly deliver. What is the value in watts of the maximum power? d. Using open-circuit conditions and short-circuit conditions at terminals a-b, determine Thevenin's equivalent circuit, including Rh and Va

Answer 1

Given network is Apply KVL to Loop-1, we get 10 O_1 + 20 I_1 + 20(I_1-I_3)+ 6(I_1-I_2) = 0 56 I_1-6I_2-20I_3 = 0 equation 1 Apply KVL to loop-2, we get -25 angle 0 degree + 6(I_2-I_1)+ 10(I_2-I_4)+ 5I_2 = 0 -6I_1 + 21 I_2-10 I_4 = 25 angle 0 degree equation 2 Apply KVL to loop-3, we get 80I_3 + 25 angle 20 degree + 15(I_3-I_4)+ 20(I_3-I_1) = 0 -20 I_1 + 115 I_3-15 I_4 = -25 angle 20 degree equation 3 Apply KVL to Loop-4, we get 15(I_4-I_3)+ 20I_4 + 5I_4 + 10(I_4-I_2) = 0 -10 I_2-15 I_3 + 50 I_4 = 0 equation 4 solving 1,2,3,4 equations we get in matrix form, [56 -6 -20 0 -6 21 0 -10 -20 0 115 -15 0 -10 -15 50] [I_1 I_2 I_3 I_4] = [0 angle 25 0 degree angle -25 20 degree 0] [I_1 I_2 I_3 I_4] = [56 -6 -20 0 -6 21 0 -10 -20 0 115 -15 0 -10 -15 50]^-1 [0 angle 25 0 degree angle -25 20 degree 0] [I_1 I-2 I-3 I_4] = [0.019 0.006 0.003 0.002 0.006 0.055 0.002 0.001 0.003 0.002 0.009 0.003 0.002 0.011 0.003 0.02][0 angle 25 0 degree angle -25 20 degree]