Correct Answer - Option 3 : 80 Nm
Concept:
In a DC motor, T ∝ ϕIa
In a DC series motor, for smaller values of armature current, ϕ ∝ Ia
\(T \propto I_a^2\)
\( \Rightarrow \frac{{{T_1}}}{{{T_2}}} \propto {\left( {\frac{{{I_{a1}}}}{{{I_{a2}}}}} \right)^2}\)
Calculation:
Given that, load current (IL1) = 3 A
In a dc series motor, armature current (Ia1) = IL1 = 3 A
Torque (T1) = 20 Nm
Increased current (Ia2) = 6 A
Torque developed at this increased current is,
\({T_2} = {\left( {\frac{6}{3}} \right)^2} \times 20 = 80\;Nm\)