Question

A 220V, DC shunt motor is operating at a speed of 1440 rpm. The armature resistance is 1  and armature current is 10A. If the excitation of the machine is reduced by 10% the extra resistance to be put in the armature circuit to maintain the same speed and torque will be
1. 1.79 Ω
2. 2.1 Ω
3. 3.1 Ω
4. 18.9 Ω

Answer 1

Correct Answer - Option 1 : 1.79 Ω

\({T_2} = {T_1}\)           \({\phi _2} = 0.9{\phi _1}\)            \({\phi _1} \times {I_1} = {\phi _2} \times {I_2}\)

\(\begin{array}{l} {I_2} = 11.11A\\ {E_1} = 220 - 10 \times 1 = 210\ V\\ {E_2} = 220 - 11.1 \times \left( {1 + R} \right)\\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_2}}}{{{E_1}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\\ I = \frac{{220 - 11.1\left( {1 + R} \right)}}{{210}} \times \frac{{{\phi _1}}}{{0.9{\phi _1}}} \end{array}\)

R = 1.79 Ω