Electric Motors and Drives

The motor is shunt, it means armature and field resistances are in parallel.

$Ia*Ra = If*Rf$

$Ia/If = Rf/Ra = 100/0.1 =1000$

Total current is

$If/Ia =10^{-3}$

$If =120*10^{-3} =0.120 A$

Total starting current  is $I = Ia+If =120+0.120 =120.12 A$

Total internal resistance of the motor is

$R_{eq} = Rf*Ra/(Rf+Ra) =0.1*100/100.1 =0.0999 ohm$

we add an external resistance R to have 120.12 A maximum current

$R_{tot} = R_{eq}+R$

$R_{tot} = U/I = 200/120.12 =1.665 ohm$

R = 1.665-0.0999 =1.5651 ohm

b) The field current resistance and current remains the same.

The power of the armature is $P = U^2/Ra (=E/t)$

the energy of the armature is $E = I*omega^2/2 = K*F^2$  where K is a constant and F the freq.

it means $U^2/Ra = K*F^2$

$(Ra+R)/Ra =(F1/F2)^2 = (1000/800)^2 =1.5625$

$Ra+R = 1.5625*Ra$

$R = 0.5625*Ra =0.5625*0.1 =0.056 ohms$

c) delivered power $P = U^2/R (=E/t)$

R is the same before and after reducing voltage.

Again energy $E = I*omega^2/2 = K*F^2$

it means $(U1/U2)^2 = (F1/F2)^2$

$F2/F1 =U2/U1 = 0.75U1/U1 =0.75$

$F2 = 0.75*F1 =0.75*1000 =750 rpm$