A 280 V, separately excited DC motor with armature resistance of 1 Ω and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10 Ω is connected in series with the armature, is ______. (round off to nearest integer)

The correct answer is 483.

Step-by-Step Explanation:

For a separately excited DC motor, we have the following relationships under constant field excitation (constant magnetic flux, $\varphi$):

1. The electromagnetic torque ($T$) developed by the motor is directly proportional to the armature current ($I_a$):
$T\propto I_a$

2. The load torque is given as proportional to the speed ($N$):
$T\propto N$

From these two proportionalities, we can write:
$I_a\propto N$

Therefore, comparing the two operating states:
$\frac{I_{a2}}{I_{a1}}=\frac{N_2}{N_1}$

Given the initial values:
$I_{a1}=30\text{A}$
and
$N_1=1000\text{rpm}$

We can express the new armature current ($I_{a2}$) in terms of the new speed ($N_2$):
$I_{a2}=\frac{30}{1000}{N}_2=0.03N_2$

The back EMF ($E_b$) of a DC motor is proportional to the speed when flux is constant:
$E_b\propto N$

Which gives:
$\frac{E_{b2}}{E_{b1}}=\frac{N_2}{N_1}$

Let's calculate the initial back EMF ($E_{b1}$) where terminal voltage $V=280\text{V}$ and armature resistance $R_a=1\text{Ω}$:
$E_{b1}=V-I_{a1}{R}_a=280-(30\times 1)=250\text{V}$

When an additional resistance of $R_{\text{ext}}=10\text{Ω}$ is connected in series with the armature, the new back EMF ($E_{b2}$) is:
$E_{b2}=V-I_{a2}(R_a+R_{\text{ext}})=280-I_{a2}(1+10)=280-11I_{a2}$

Substituting $I_{a2}=0.03N_2$ into this expression:
$E_{b2}=280-11\times 0.03N_2=280-0.33N_2$

Now, substituting the values into the speed ratio equation:
$\frac{280-0.33N_2}{250}=\frac{N_2}{1000}$

Solving for $N_2$:
$280-0.33N_2=\frac{250}{1000}{N}_2$
$280-0.33N_2=0.25N_2$
$280=0.58N_2$
$N_2=\frac{280}{0.58}\approx 482.76\text{rpm}$

Rounding to the nearest integer, the motor speed is 483 rpm.