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A DC series motor develops a torque of 20 Nm at 3 A of load current. If the current is increased to 6 A, the torque developed will be
1. 10 Nm
2. 20 Nm
3. 80 Nm
4. 40 Nm

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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,

\({T_2} = {\left( {\frac{6}{3}} \right)^2} \times 20 = 80\;Nm\)

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