A 240 V, dc shunt motor draws 15 A while supplying the rated load at a speed of 80 rad/s. The armature resistance is 0.5 Ω and the field winding resistance is 80 Ω. The external resistance to be added in the armature circuit to limit the armature current to 125% of its rated value is
Right Answer is:
31.1 Ω
SOLUTION
Given Data:-
Supply Voltage V = 240 V
Line Current IL = 15 A
Shunt Resistance Rsh = 80Ω
Armature Resistance Ra = 0.5 Ω
Armature Current Ia =?
Back EMF Eb =?
(i) Armature current
The Line current of DC shunt motor is the sum of Armature current and Shunt field current
IL = Ia + Ish
And Shunt Field Resistance is given as
Ish = V/Rsh = 240/80 = 3 A
∴ Ia = IL − Ish
Ia = 15 − 3 = 12A
Back EMF
From the voltage equation, the Back EMF of DC Motor is given by
Eb = V − IaRa
= 240 − 12 × 0.5
= 240 − 6
Eb = 234 V
External resistance to be added in the armature circuit to limit the armature current upto 125%.
Eb = V − Ia(Ra + Rext)
234 = 240 − Ia(0.5 + Rext)
$\begin{array}{l}12 \times 1.25 = \dfrac{{474}}{{0.5 + {{\mathop{\rm R}\nolimits} _{ext}}}}\\\\15\left( {0.5 + {{\mathop{\rm R}\nolimits} _{ext}}} \right) = 474\\\\7.5 + 15{{\mathop{\rm R}\nolimits} _{ext}} = 474\\\\{{\mathop{\rm R}\nolimits} _{ext}} = 31.1\Omega \end{array}$