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, $V_{t}$ = 400V
Speed, N = 1000 r. p. m. = $(\frac{1000}{60}) r . p . s .$
= 16.667 r. p.s.

Angular speed, $ω_{m}$ = $2 π N = 2 π \times 16.667$
= 104.72 rad/sec

Armature Resistance, $R_{A} = 0.25 Ω$

Armature current $I_{A}$ = 100A

Back emf = E_b

Electrical Power Converted to mechanical power = P_m

Torque, T = ?

$E_{b} = V_{t} - I_{A} R_{A} = 400 - 100 \times 0.25 = 375 V$

$P_{m} = E_{b} I_{A} = 375 \times 100 = 37500 W$

$T = \frac{P_{m}}{ω_{m}} = \frac{37500}{104.72} \approx 358.1 N m (A n s)$

3 weeks ago

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