Question

The armature of the DC motor has $20\Omega$ resistance. It draws current of $1.5A$ when run by $220V$ DC supply. The value of back emf induced in it will be:
A) $150V$
B) $170V$
C) $180V$
D) $190V$

Answer 1

Put the values in the formula $V = {E_b} + IR$ [Where, $V$ is the applied voltage across the armature, ${E_b}$ is the induced emf in the armature by generator action, $I$ is the armature current, and $R$ is the armature resistance].

Complete step by step answer: When the motor armature continues to rotate due to motor action, the armature conductors cut the magnetic flux and, therefore, emfs are induced in them. This induced emf is known as back emf. The direction of this induced emf is such that it opposes the applied force.

Suppose, in a DC motor armature, back emf (${E_b}$) is in series with a resistance, $R$, put across a DC supply mains of $V$ volts. The applied voltage, $V$, must be large enough to balance both the voltage drop in armature resistance and the back emf at all times, i.e., $V = {E_b} + IR$ [Where, $V$ is the applied voltage across the armature, ${E_b}$ is the induced emf in the armature by generator action, $I$ is the armature current, and $R$ is the armature resistance].

Putting all the given values in the equation, we get,
$220 = {E_b} + \left( {20 \times 1.5} \right)$ or, ${E_b} = 220 - \left( {20 \times 1.5} \right) = 220 - 30 = 190$
So, the value of back emf induced in it will be $190V$.

Additional information: Since the back emf is induced due to the generator action, the magnitude of it is, therefore, given by the same expression as that for the generated emf in a generator, i.e., ${E_b} = \dfrac{{\Phi ZN}}{{60}} \times \dfrac{P}{A}$

So, as obvious from the two expressions of the back emf is that it depends among other factors upon the armature speed and armature current depends upon the back emf for a constant applied voltage and armature resistance. If the armature speed is high, the back emf will be large, and therefore, the armature current small. If the speed of the armature is low, then back emf will be less and the armature current more resulting in the development of large torque.

Note: The presence of back emf makes the DC motor a self-regulating machine. So, it makes the DC motor to draw as much armature current as is just sufficient to develop the required load torque.