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GATE EE 2017 Official Paper: Shift 2

Option 2 : 30 A

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

Voltage (V) = 220 V

Torque (τ) = 70 N-m

Speed (N) = 900 rpm

Armature resistance (R_{a}) = 0.02 Ω

Power (P) = 10 kW

⇒ E_{b }I_{a} = 10 × 10^{3}

⇒ (V - I_{a }R_{a}) I_{a} = 10,000

⇒ [220 – I_{a}(0.02)] I_{a} = 10,000

⇒ I_{a} = 45.75 A

P = τω

\(\Rightarrow 10 \times {10^3} = {\rm{\tau }}\left( {\frac{{2{\rm{\pi N}}}}{{60}}} \right)\)

\(\Rightarrow {\rm{\tau }} = \frac{{10 \times {{10}^3} \times 60}}{{2{\rm{\pi }} \times 900}} = 106.1{\rm{\;N}} - {\rm{m}}\)

Torque α ϕ I_{a}

For separately excited motor

T α I_{a}

T_{1} = 106.1 N-m, I_{a1} = 45.75 A

T_{2} = 70 N-m

\({{\rm{I}}_{{{\rm{a}}_2}}} = \frac{{{{\rm{T}}_2} \times {{\rm{I}}_{{{\rm{a}}_1}}}}}{{{{\rm{T}}_1}}} = \frac{{70 \times 45.75}}{{106.1}} = 30.18{\rm{\;A}}\)