Question

A dc series motor is driving a fan load where the load torque is proportional to square of speed. The resistance of armature and field windings are 0.03$\Omega$ and 0.05 $\Omega$ respectively. 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. 19.96$\Omega$
• B. 19.88$\Omega$
• C. 20.04$\Omega$
• D. 20.12$\Omega$

Answer 1

The correct option is A 19.96$\Omega$

$\text{Load characteristic is}$
$T_L\propto N^2$
$\text{For dc series motor, torque-current relation is given by}$
$\therefore$ $T_d\propto I_a^2$

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

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

$I_{a2}=7.5A$

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

$E_{a1}=198.8V$

$\text{Case -II}$
When additional resistance added in series with the armature circuit,
$I_{a2}=7.5A,$

$N_2=750rpm$

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

$\frac{E_{a2}}{E_{a1}}=\frac{N_2}{N_1}\frac{I_{a2}}{I_{a1}}$ $\text{(}\because \text{In dc series motor}\varphi \propto I_a)$

$E_{a2}=49.7$

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

$R_{ext}=20.04-0.08$

$=19.96\Omega$