A simple electric motor has an armature resis | vedclass

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(A) The back $emf$ $(E)$ of a motor is given by $E = V - IR$,where $V$ is the supply voltage,$I$ is the current,and $R$ is the armature resistance.For

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$3$

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(A) The back $emf$ $(E)$ of a motor is given by $E = V - IR$,where $V$ is the supply voltage,$I$ is the current,and $R$ is the armature resistance.For the unloaded motor: $E_1 = 12 - (2 	imes 1) = 10 \,V$.Since the back $emf$ is directly proportional to the speed of the motor $(E \propto \omega)$,a $10 \%$ reduction in speed implies a $10 \%$ reduction in the back $emf$.The new back $emf$ is $E_2 = E_1 - 0.10 	imes E_1 = 10 - 1 = 9 \,V$.Now,for the loaded motor,the current $I_2$ is given by $I_2 = (V - E_2) / R$.Substituting the values: $I_2 = (12 - 9) / 1 = 3 \,A$.

$A$ simple electric motor has an armature resistance of $1 \,\Omega$ and runs from a d.c. source of $12 \,V$. It draws a current of $2 \,A$ 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 ......... $A$

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