Question

An electric motor has a back emf of 110 volt and an armature current of 90A. The armature is making 2.5rps. The power developed is –

& \text{A) 110}\times \text{25 Watt} \\\ & \text{B) 110}\times \text{90 Watt} \\\ & \text{C) 90}\times \text{25 Watt} \\\ & \text{D) 90}\times \text{2}\pi \times \text{25 Watt} \\\ \end{aligned}$$

Answer 1

We need to find the power generated in an electric motor which runs with the given quantities. We have to find the relation between the given parameters and the power developed in the armature of the motor to find the solution to this problem.

Complete answer:
We know that electric motors are devices which convert the electrical energy into the mechanical energy to operate an appliance. This conversion of the energy forms requires the utilization of some energy which is wasted in the process by heat mostly. The power developed in the armature or the rotating frame of the motor is due to this energy loss in the form of heat.
The electric motor converts the electric current into mechanical energy by rotating the armature around which the coil is wound. We know that any change in the conditions of the devices tend to develop a resistive force against the change. In electric devices the change in the magnetic field is opposed by developing an electromotive force opposing the change which is equal to the emf in the system. Therefore, the back emf is equal to the potential difference of the electric current supplied to the system.
We know from the Joule’s law of heating that the power dissipated or in this case developed by the armature is related to the voltage and the current as –
$P=VI$
Where. P is the power developed, V is the voltage and I is the current through it.
Now, we can find the power developed in the armature using the above relation and the data given to us as –

& P=VI \\\ & \text{given,} \\\ & V=110V, \\\ & I=90A \\\ & \Rightarrow P=110\times 90 \\\ & \therefore P=9900\text{Watt} \\\ \end{aligned}$$ The power developed in the armature is the product of the voltage and the current. **The correct answer is option B.** **Note:** The back emf is a common phenomenon which we observe in our daily life. Even on our switch boards we can observe a spark when we switch off the circuits of high-power devices like electric motors or the electric iron. This is due to the back emf in these.