Question

A 20 kW, 300 V dc shunt motor is driving a constant torque load with line current of 100 A at a speed of 1600 rpm. Motor has armature and field winding resistance as 0.02 $\Omega$ and 100 $\Omega$ respectively. If 50 $\Omega$ external resistance is added in field circuit, then motor is running at (Assume linear magnetzation curve)

• A. 1812.9 rpm
• B. 2100 rpm
• C. 2392.18 rpm
• D. 1594.8 rpm

Answer 1

The correct option is C 2392.18 rpm

$\text{Case - I:}$


$\text{Armature current,}{I}_{a1}=100-3=97A$

$\text{back emf,}{E}_{a1}=300-97\times 0.02$

$E_{a1}=298.06V$

$\text{Case - II:}$

$\text{When}50\Omega \text{external resistance added in field circuit,}$


$\text{Load torque is constant}$

$T\propto \varphi I_a$

${\varphi }_2{I}_{a2}={\varphi }_1{I}_{a1}$

$\varphi \propto I_f$

$I_{a2}=I_{a1}\frac{{\varphi }_1}{{\varphi }_2}=97\times \frac{3}{2}=145.5A$

$E_{a2}=300-145.5\times 0.02$

$E_{a2}=297.09V$

$\text{We know that,}{E}_a\propto \varphi {\omega }_m$

$\frac{E_{a2}}{E_{a1}}=\frac{{\varphi }_2}{{\varphi }_1}\times \frac{N_2}{N_1}$

$\frac{297.09}{298.06}=\frac{2}{3}\times \frac{N_2}{1600}$

$N_2=2392.18rpm$