# A dc series motor has an armature resistance of 0.06 Ω and series field resistance of 0.08 Ω. The motor is connected to a 400 V supply. The line current is 20 A when the speed of the machine is 1100 rpm. When the line current is 50 A and the excitation is increased to 30%, the speed of the machine in rpm is

A dc series motor has an armature resistance of 0.06 Ω and series field resistance of 0.08 Ω. The motor is connected to a 400 V supply. The line current is 20 A when the speed of the machine is 1100 rpm. When the line current is 50 A and the excitation is increased to 30%, the speed of the machine in rpm is

### Right Answer is: 837 RPM

#### SOLUTION

Here

V = 400 V , I1 = 20 A, N1 = 1100

Ra = 0.06 ohm, Rse = 0.08 ohm

I2 = 50 A, N2 =?

Φ2 = 130 % of Φ1 = 1.3 Φ1

While drawing line current of 20 A

Eb1 = V – I1(Ra + Rse)

= 400 – 20(0.06 + 0.08)

= 397. 2 V

While drawing line current of 50 A

Eb2 = V – I2(Ra + Rse)

= 400 – 50(0.06 + 0.08)

= 393 V

We know that

$\begin{array}{l}\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{{E_{b2}}}}{{{E_{b1}}}} \times \dfrac{{{\Phi _1}}}{{{\Phi _2}}}\\\\\dfrac{{{N_2}}}{{1100}} = \dfrac{{393}}{{397.2}} \times \dfrac{{{\Phi _1}}}{{1.3{\Phi _1}}}\\\\{N_2} = 837RPM\end{array}$

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