Neglecting saturation, if current taken by a series motor is increased from 8 A to 10 A, the percentage increase in its torque is:

Correct option is C

$$\bullet \ \text{For DC machine, torque is directly proportional to armature current and flux.} \\\bullet \ T \propto \phi I_a \ \text{In DC shunt motors, flux is constant (not dependent on armature current).} \\\bullet \ \text{Hence torque is proportional to square of armature current. } T \propto I_a^2 \ \text{In DC series motors, flux is directly proportional to armature current.} \\\bullet \ \text{Hence torque is directly proportional to square of the armature current. } T_a \propto I_a^2 \ (\because \text{ flux saturation is neglected).} \\\text{In DC series motor at saturation condition, flux is constant.} \\\bullet \ \text{Hence torque is proportional to armature current. } T \propto I_a \\[6pt]\text{Where,} \\T = \text{torque}, \\I_a = \text{armature current}, \\\phi = \text{field flux} \\[10pt]\text{Given} \\\Rightarrow \ T_1 \ \text{at } I_1 = 8~\text{A}, \\\Rightarrow \ T_2 \ \text{at } I_2 = 10~\text{A} \\[6pt]\Rightarrow \ \frac{T_1}{T_2} = \left(\frac{I_1}{I_2}\right)^2 \\[6pt]\Rightarrow \ \frac{T_1}{T_2} = \left(\frac{8}{10}\right)^2 = \frac{64}{100} \\[8pt]\text{Percentage increase in torque} = \frac{100 - 64}{64} \times 100 \\[6pt]\text{Therefore there is increase of } 56.25\%$$​​