How is the armature current of the shunt motor found?

Ia = IL - Ish = 26 - 230/130 = 24.23 A

Which expression gives the armature torque?

Ta = P phi Z Ia/(2 pi A) = 75.59 N·m

How is the back emf used to verify the torque?

Eb = V - Ia Ra = 227.09 V, and Ta = Eb Ia/omega gives the same value

2081 Kartik (new course) · 4 marks

DC shunt motor armature torque

Short answer

Armature current is24.23 A. Torque is computed using Ta = (P φ Z)/(2π A) Ia, giving75.59 N·m.

A 4 pole 230 V dc shunt motor, lap wound has 980 conductors. The flux per pole is 20 mWb. Determine the torque developed by the armature in Nm when the current drawn by the motor is 26 A. The armature resistance is 0.12 Ω and the field resistance is 130 Ω.

Given: $P=4$ , lap winding so $A=P=4$ , $Z=980$ , $\phi=20$ mWb, $V=230$ V, line current $I_L=26$ A, $R_a=0.12\ \Omega$ , $R_{sh}=130\ \Omega$ .

Shunt field current:

I_{sh}=\frac{V}{R_{sh}}=\frac{230}{130}=1.769\ \text{A}

Armature current:

I_a=I_L-I_{sh}=26-1.769=24.23\ \text{A}

Armature torque (using $T_a=\frac{P\phi Z}{2\pi A}I_a$ ):

T_a=\frac{4\times0.02\times980}{2\pi\times4}\times24.23=\frac{78.4}{25.13}\times24.23=3.119\times24.23=75.59\ \text{N·m}

Check through back emf and speed:

E_b=V-I_aR_a=230-24.23\times0.12=227.09\ \text{V}

N=\frac{60AE_b}{P\phi Z}=\frac{60\times4\times227.09}{4\times0.02\times980}=695.2\ \text{rpm}

T_a=\frac{E_bI_a}{2\pi N/60}=\frac{227.09\times24.23}{72.80}=75.59\ \text{N·m}

Answer: Armature torque $T_a=75.59$ N·m (at about 695 rpm).

✓ Answer checked by Claude Opus 5.5 (high effort)

Back emf in dc motor · Basic Electrical and Electronics Engineering, chapter 2
25 kVA ideal transformer currents, flux · Basic Electrical and Electronics Engineering, chapter 2
Max torque condition and torque-slip curve · Basic Electrical and Electronics Engineering, chapter 2
RL series circuit current, pf, power · Basic Electrical and Electronics Engineering, chapter 1
Nodal analysis current in 6 Ω · Basic Electrical and Electronics Engineering, chapter 1
Nodal analysis current in branch a-b · Basic Electrical and Electronics Engineering, chapter 1
Reverse current (plugging) braking of 20 kW DC shunt motor · Utilization of Electrical Energy, chapter 3
Shunt motor 480 V, 740 conductors · Electrical Machine, chapter 4
Shunt motor speed with 1 Ω armature resistance added · Electrical Machines I, chapter 4 · asked 5×
Shunt motor speed with 1.2 Ω armature resistance added · Electrical Machines I, chapter 4 · asked 5×

All Basic Electrical and Electronics Engineering chapter 2 questions →

DC shunt motor armature torque

Related questions