Question

A 230 V dc shunt motor is used for adjustable speed drive over the range of 0 to 2000 rpm. Speed from zero to 1000 rpm are obtained by adjusting the armature terminal voltage from 0 to 230 V with field current kept constant. Speeds from 1000 to 2000 rpm are obtained by decreasing the field current with armature terminal voltage held constant at 230 V. Neglect armature reaction effects and losses, where $V_T=230Vand{I}_a=rated$ armature current at 1000 rpm. If torque required by load remains constant over the entire speed range. The curve for Power (P) vs Speed (N) is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

T = Constant for entire speed range

(For 0 to 1000 rpm)

$V_T$ series from 0 to 230 V.

and ${\varphi }_F$ is constant

as losses is ignored

$E_b=V_t$

$T\propto {\varphi }_F{I}_a$

as ${\varphi }_F$ is constant T is constant

$I_a$ will be constant

So, $P=E_b{I}_a$

$P=V_t{I}_a$

$V_t$ varies

$P\propto V_t\propto N$

P also increases from 0 to $V_t{I}_a$

Now,

For 1000 to 2000

$V_t$ is constant

T is constant.

${\varphi }_F$ is decreasing

$T\propto {\varphi }_F{I}_a$

${\varphi }_F{I}_a=Constant\Rightarrow I_a\propto \frac{1}{{\varphi }_F}$ ....(i)

So, $I_a$ will increase.

as $V_T$ is constant

$E_b\propto {\varphi }_{F}N$

$N\propto \frac{1}{{\varphi }_F}$ ....(ii)

From (i) and (ii),

$N\propto I_a$

So, as N increases, $I_a$ increases

$P=E_b{I}_a$

$P=V_T{I}_a$

$P\propto I_a\propto N$

P also increases linearly,

So correct option is B.

Alternate Solution:

$because$ $P=T\omega \propto TN$

$\because$ T constant over entire speed range $P\propto N$

Graph will be a straight line due to linear proportionality