The equivalent armature resistance of a series motor is 0.1 ohm. When connected across a 110-V mains, the armature takes 20 A and its speed is 1000 rpm. Determine its speed when the armature takes 50 A from the same mains, with the field increased by 10%.
The correct answer and explanation is :
Given Data:
- Equivalent armature resistance (( R_a )) = 0.1 Ω
- Supply voltage (( V )) = 110 V
- Initial armature current (( I_{a1} )) = 20 A
- Initial speed (( N_1 )) = 1000 rpm
- New armature current (( I_{a2} )) = 50 A
- Field increased by 10% (( \phi_2 = 1.1 \phi_1 ))
Step 1: Calculate the Back EMF
The back EMF (( E )) of a series motor is given by:
[
E = V – I_a R_a
]
For the first condition (( I_{a1} = 20 ) A):
[
E_1 = 110 – (20 \times 0.1) = 110 – 2 = 108V
]
For the second condition (( I_{a2} = 50 ) A):
[
E_2 = 110 – (50 \times 0.1) = 110 – 5 = 105V
]
Step 2: Relationship between Speed, Back EMF, and Flux
The speed of a DC series motor is approximately given by:
[
N \propto \frac{E}{\phi}
]
So,
[
\frac{N_2}{N_1} = \frac{E_2 / \phi_2}{E_1 / \phi_1}
]
[
N_2 = N_1 \times \frac{E_2}{E_1} \times \frac{\phi_1}{\phi_2}
]
Substituting the values:
[
N_2 = 1000 \times \frac{105}{108} \times \frac{1}{1.1}
]
[
N_2 = 1000 \times 0.9722 \times 0.9091
]
[
N_2 \approx 884.3 \text{ rpm}
]
Final Answer:
The new speed of the motor is approximately 884 rpm.
Explanation:
The speed of a DC series motor is directly proportional to the back EMF and inversely proportional to the flux. When the armature current increases, the back EMF decreases slightly due to the voltage drop in the armature resistance. Simultaneously, since the motor is a series motor, the field flux also increases with the armature current. However, in this case, the field was intentionally increased by 10%, making the flux even larger.
Since the speed is inversely proportional to the flux, the increase in flux results in a lower speed, even though the armature current is higher. This explains why the motor speed decreases from 1000 rpm to about 884 rpm despite the higher armature current.