Question

A simple electric motor has an armature resistance of $1\Omega$ and runs from a d.c. source of $12V$. It draws a current of $2A$ when unloaded. When a certain load is connected to it, its speed reduces by $10\%$ of its initial value. The current drawn by the loaded motor is?
A. $3A$
B. $6A$
C. $2A$
D. $1A$

Answer 1

In this question, we need to determine the current drawn by the loaded motor such that the electric motor has an armature resistance of $1\Omega$ and runs from a d.c. source of $12V$. For this, we will first calculate the back E.M.F. and then, calculate the $10\%$ of this E.M.F and then the current drawn by the loaded motor.

Formula Used:
Back E.M.F. of a motor, $E = V - IR$
Where, $V$ is the voltage in volts, $I$ is the current in amperes, and $R$ is the resistance in ohms.

Complete step by step answer:
When a D.C. motor rotates on the action of the driving torque, the conductor through the magnetic field and hence the E.M.F. induce the direction of this E.M.F. opposed to the applied voltage. This is called back E.M.F. Now, let us calculate the back E.M.F. Here, the value of $V$is $12V$, $I$ is $2A$, and $R$ is $1\Omega$. Therefore, by substituting the values of voltage, current, and resistance in the above formula, we get
Back E.M.F is equal to,
$E = 12 - 2(1) \\\ \Rightarrow E = 12 - 2 \\\ \therefore E = 10V$
Now, we have to calculate $10\%$ of this E.M.F. Hence, $10\%$ of this E.M.F. $= 0.1 \times 10 = 1V$.Therefore, the current $= 1 \times 1 = 1A$.Thus, we can calculate that the current drawn by the loaded motor is $2 + 1 = 3A$.

So, option A is the correct answer.

Note: It is worth noting down here that students forget to calculate $10\%$ of the E.M.F. and instead finding the current drawn by the loaded motor by keeping the $100\%$ back E.M.F. Back E.M.F. is the electromotive force which opposes the change in the current which induces it.