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A 400 V DC motor takes an armature current of 100 A when its speed is 1000 rpm. If the armature resistance is 0.25 ohm, calculate the torque produced in Nm. 

Created: 3 weeks ago | Updated: 3 weeks ago
Updated: 3 weeks ago

Given, 

Terminal Voltage, Vt = 400V 
Speed, N = 1000 r. p. m. = (1000 
                                             = 16.667 r. p.s. 

Angular speed, ωm = 2πN = 2π× 16.667 
                                       = 104.72 rad/sec 

Armature Resistance, RA=0.25 Ω

Armature current IA = 100A 

Back emf = E

Electrical Power Converted to mechanical power = Pm 

Torque, T = ?

 

Eb= Vt-IA RA = 400-100 × 0.25 = 375 V 

Pm=EbIA= 375 × 100 = 37500 W 

T = Pm ωm=37500 104.72 358.1 Nm Ans

3 weeks ago

ইলেকট্রিক্যাল এন্ড ইলেকট্রনিক্স ইঞ্জিনিয়ারিং (EEE)

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