Question

Consider a dc motor with armature voltage = 125 v and armature resistance = 0.4 ohm. It is running at 1800 rpm under no load condition. The rated armature current is 30

a. Calculate the rated speed of the motor. 1500 rpm 1800 rpm 1627 rpm 900 rpm

Answer 1

The emf equation of DC motor is given by

∅$E=\frac{NP∅Z}{60A}$

Here, N = speed of rotation in rpm. P = number of poles. A = number of parallel paths. Z = total no. conductors in armature.

Thus speed of the rotation $N=\frac{60A}{PZ}\times\frac{E}{∅}\\\\\Rightarrow{N}=\frac{E}{k∅}$

where k=\frac{PZ}{60A} is a constant.

Consider in No load Condition the rpm of motor is $N_{1}$ and EMF is $E_{1}$

There fore $N_{1}=\frac{E_{1}}{k∅}$ equation ............1

Consider in load Condition the rpm of motor is $N_{2}$ and EMF is $E_{2}$

There fore $N_{2}=\frac{E_{2}}{k∅}$ equation ............2

Dividing Equation 1 by equation 2 we get

$\frac{N_{1}}{N_{2}}=\frac{E_{1}}{E_{2}}$

We also know that back emf E also depends on armature current and armature resistance,  

therefore $E=V-i_{a}\times{R}_{a}$

where $i_{a}$ is armature current and $R_{a}$ is armature resistance

.

Therefore $E_{1}=V-i_{a0}\times{R}_{a}=125-0\times0.4=125$

and $E_{2}=V-i_{al}\times{R}_{a}=125-30\times0.4=125-12=113$

Putting the value of $E_{1} and E_{2}$ in equation 3 we get

$\frac{N_{1}}{N_{2}}=\frac{125}{11</p><p>}\\\\N_{2}=N_{1}\times\frac{113}{125}\\\\1800\times{113}{125}=1627$