Why is N proportional to Eb in this shunt motor problem?

The field current is fixed since V and Rf are fixed

2071 Magh · 6 marks

Shunt motor speed at armature current 30 A

Short answer

Field current is 2 A. At 600 rpm, Ia=58 A, Eb1=218.4 V. When armature current is 30 A, Eb2=224 V. Using N∝Eb, speed = 615.4 rpm.

A dc shunt motor runs at 600 RPM taking 60 A from a 230 V supply. Armature resistance is $0.2\ \Omega$ and field resistance is $115\ \Omega$ . Find the speed when the current through the armature is 30 A.

The field current is constant ( $V$ and $R_f$ are fixed), so the flux is constant and $N\propto E_b$ .

Field current: $I_f = \dfrac{230}{115} = 2$ A.

Case 1 (600 rpm, line current 60 A):

I_{a1} = 60 - 2 = 58\ \text{A}, \qquad E_{b1} = 230 - 58(0.2) = 218.4\ \text{V}

Case 2 (armature current 30 A):

E_{b2} = 230 - 30(0.2) = 224.0\ \text{V}

N_2 = N_1\frac{E_{b2}}{E_{b1}} = 600\times\frac{224.0}{218.4} = 615.4\ \text{rpm}

Answer: speed at armature current 30 A $= 615.4\ \text{rpm}$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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