Answer to Question #93756 in Electrical Engineering for Shawn Dickson

Question

DC series motor has 3.5 ohm resistance in starter, no load current of 0.3 amps and a full load current of 2 amps rotating at 1800 RPM find speed at full load?

Answer 1

The speed of the motor is directly proportional to its back EMF:

$$n=\frac{E}{c\Phi}=\frac{V_m-Ir}{c\Phi}.$$

Here $V_m$ - supply voltage on the terminals of the motor.

For both cases (no-load and full-load) we have:

$$n_1=\frac{V_m-I_1r}{c\Phi_1},\\ \space\\ n_2=\frac{V_m-I_2r}{c\Phi_2}.$$

Divide one by another:

$$\frac{n_2}{n_1}=\frac{\Phi_1(V_m-I_2r)}{\Phi_2(V_m-I_1r)},$$

$$n_2=n_1\frac{\Phi_1(V_m-2\cdot3.5)}{\Phi_2(V_m-0.3\cdot3.5)}=1800\frac{\Phi_1(V_m-7)}{\Phi_2(V_m-1.05)}.$$

In this problem we assumed that "3.5 ohm resistance in starter" is the total resistance of stator and armature.