I'm studying for an exam, and have come across a question I don't. really understand
The question gives us a back-emf constant and asks for a rated back-emf. Now I know that the back-emf constant is k, but the rated back-emf confused me at first. When the lecture went through the questions, what he said confused me.
The lecturer says that the rated back-emf is E? I don't understand how the back-emf can be both E and k at the same time. What exactly is the difference between E and k? I think the main question I am asking is what exactly is this question asking me to do?
Here is the question:
A DC machine has a back-emf constant of 0.24 V/(rad/s) and a resistance of 0.5 ohms. It is operating from a 12V DC power supply. At its rated speed of 400 rpm, calculate its rated back-emf (that is, the back-emf at rated speed).
Best Answer
There are two key constants with respect to electrical machines
1) \$K_e\$ The BackEMF constant. Units in Volts/rpm (volts/rads ...) -> what voltage will be generated at the terminals if the rotor is spinning.
2) \$K_t\$ The Torque constant. Units in Nm/A - What torque will be generated if current is injected into the stator
(there is a 3rd... \$K_v\$, velocity constant, the reciprocal of \$K_e\$, rpm/Volts) - what speed the rotor could spin at for a given applied voltage
In theory \$K_t\$ = \$K_e\$. In practice, \$K_t\$ < \$K_e\$ as \$K_t\$ is taken at rated current & corresponds to Iron saturation.
So the question:
We thus know \$K_e\$ = 0.24V/rad/s
We also know the rpm = 400rpm
convert rpm to rad/s: \$ \omega = \frac{rpm * 2 * \pi}{60} = 41.8879rad/s \$
There is one more step to get the rated terminal voltage, but I will leave that