Question

An electric motor runs on DC source of emf $200V$ and draws a current of $10A$. If the efficiency be $40$ percent then the resistance of armature is
(A) $2\Omega $
(B) $8\Omega $
(C) $12\Omega $
(D) $16\Omega $

Answer 1

Hint: In order to solve this question, we will first find the input power and output power of the electric motor and then we will find the power loss of the motor and then we will solve for the resistance of the armature.

Formula used:
If P is the power, I is the current and R is the resistance then Power is given by $P = {I^2}R$
Also, $P = VI$ where V is the voltage.

Complete answer:
We have given that, electric motor runs with emf of $200V$ and current $I = 10A$ then, its input power is given by
$
  P = VI \\
  P = 200 \times 10 \\
  P = 2000W \to (i) \\
 $

Now, efficiency of the motor is $40$ percent which means its output power will be $40$ percent of the input power so, output power P’ is given by
$
  P' = 40 \times \dfrac{{2000}}{{100}} \\
  P' = 800W \to (ii) \\
 $

Now, there is a power loss in heating the armature and this loss in power is given by
$P - P'$ on putting the values from equation (i) and (ii) we get,
$
   = 2000 - 800 \\
   = 1200W \\
 $

So, Loss in power say P’’ is $P'' = 1200W$ and the current through the armature is $I = 10A$ and let its Resistance be R then using formula $P = {I^2}R$ we get,
$
  1200 = 100R \\
  R = 12\Omega \\
 $
So, the resistance of the armature is $12\Omega $

Hence, the correct option is (C) $12\Omega $

Note: It should be remembered that, generally in electric motors many coils of wire are mounted to some ferromagnetic material and this part is termed as armature of the motor and its main function is to rotate the shaft to which its attached when electric current passed through these coils.