# A 220 V, 15 kW, 1000 RPM shunt Motor with an armature resistance of 0.25Ω, has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 RPM while drawing a line current of 52.8 A and a field current of 1.8 A is

A 220 V, 15 kW, 1000 RPM shunt Motor with an armature resistance of 0.25Ω, has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 RPM while drawing a line current of 52.8 A and a field current of 1.8 A is

### Right Answer is: 36. 36 % decrease

#### SOLUTION

Given Data

Supply Voltage V = 220 V
Speed N1 = 1000 RPM
Armature Resistance Ra = 0.25 Ω
Line Current IL = 68A & 52.8A
Field current IF = 2.2 A & 1.8 A
Speed N2 = 1600 RPM The Line current of DC shunt motor is the sum of Armature current and Shunt field current

IL = Ia + Ish

∴ Ia = IL − Ish

Ia1 = 68 − 2.2 = 65.8 A

From the voltage equation, the Back EMF will be

Eb1 = V − Ia1Ra

= 220 × 65.8 × 0.25

Eb1 = 203.55

Similarly for the line current of 52.8 A and a field current of 1.8 A the armature current will be

Ia2 = 52.8 − 1.8 = 51 A

Hence Eb2 = V − Ia2Ra

= 220 × 51 × 0.25

Eb2 = 207.25

Let the on load speed be N. As we know that back EMF of DC motor is directly proportional to the flux and speed.

Eb ∝ Nφ

$\begin{array}{l}\dfrac{{{E_{b1}}}}{{{E_{b2}}}} = \dfrac{{{N_1}}}{{{N_2}}} \times \dfrac{{{\Phi _1}}}{{{\Phi _2}}}\\\\\dfrac{{203.55}}{{207.25}} = \dfrac{{1000}}{{1600}} \times \dfrac{{{\Phi _1}}}{{{\Phi _2}}}\\\\\dfrac{{{\Phi _2}}}{{{\Phi _1}}} = 0.6364\\\\\therefore \% {\text{ Decrease = }}\dfrac{{{\Phi _1} – {\Phi _2}}}{{{\Phi _1}}} \times 100\\\\ = 1 – \dfrac{{{\Phi _2}}}{{{\Phi _1}}} = 1 – 0.6364\\\\ = 0.3636 \times 100 = 36.36\end{array}$

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