Question

a shunt motor is running off a 220 - V supply taking an armature current of 15A , the resistance of the armature circuit being 0.8ohms . calculate the value of the generated e.m.f . jf the flux were suddenly reduced by 10 percent , to what value would the armature current increase momentarily ?

Answer 1

Given quantities:

$U = 220V$ $I = 15A$ $R = 0.8\varOmega$

The electromotive force (e.m.f) is defined as the energy per unit charge that converted from other forms of energy to electrical energy to move a charge across the whole circuit.

$E = \large\frac{W}{Q} = \frac{\frac{U^2t}{R}}{It}=\frac{U^2}{IR} = \frac{220*220}{15*0.8} =$ 4033N

$\varPhi_2 = 0.9\varPhi_1$ $X_L = wL$

$\large\frac{L_1I_1}{2} =0.9* \frac{L_2I_2}{2}$

$\frac{R_1I_1}{2w} =0.9* \frac{R_2I_2}{2w}$

$U = 0.9*0.8*I_2 \to I_2 = 306A$