Question

A dc series motor is driving a fan load where the load torue is proportional to square of speed. The rsistance of armature and field are 0.03 $\Omega$ and 0.05 $\Omega$ respectively. If the motor takes 15 A and runs at 1500 rpm with supply voltage of 200 V. The value of resistance to be added in series with the armature to reduce the speed to 750 rpm, is ______

• A. 20.04 $\Omega$
• B. 20.12$\Omega$
• C. 19.88$\Omega$
• D. 19.96 $\Omega$

Answer 1

The correct option is D 19.96 $\Omega$

Load characteristic is

$T_L\alpha N^2$


For dc series motor, torque-currrent relation is given by

$T_d\alpha I_a^2$

$\therefore \frac{I_{a2}}{I_{a1}}=\frac{N_2}{N_1}$

$\frac{I_{a2}}{15}=\frac{750}{1500}$

$I_{a2}=7.5A$

Case-I:


$E_{a1}=200-15\times (0.03+0.05)$

$E_{a1}=198.8V$

Case-II:

When additional resistance added in series with the armature circuit,

$I_{a2}=7.5A,$

$N_2=750rpm$

Now, $E_a\alpha \varphi {\omega }_m$

$\frac{E_{a2}}{E_{a1}}=\frac{N_2}{N_1}\frac{I_{a2}}{I_{a1}}$

(In dc series motor $\varphi \infty I_a$)

$\frac{E_{a2}}{198.8}=\frac{750\times 7.5}{1500\times 15}$

$E_{a2}=49.7V$

$E_{a2}=200-7.5(0.08+R_{ext})=49.7$

$0.08+R_{ext}=\frac{200-49.7}{7.5}$

$R_{ext}=20.04-0.08$

$=19.96\Omega$