A motor having an armature of resistance 2,Om | vedclass

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(A) The current $I$ in the armature of a $DC$ motor is given by the formula: $I = \frac{V - E_b}{R}$,where $V$ is the supply voltage,$E_b$ is the back

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$5$

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(A) The current $I$ in the armature of a $DC$ motor is given by the formula: $I = \frac{V - E_b}{R}$,where $V$ is the supply voltage,$E_b$ is the back $e.m.f.$,and $R$ is the armature resistance.Given:Supply voltage $V = 220\,V$Back $e.m.f.$ $E_b = 210\,V$Armature resistance $R = 2\,\Omega$Substituting the values:$I = \frac{220 - 210}{2} = \frac{10}{2} = 5\,A$.Therefore,the current in the armature is $5\,A$.

$A$ motor having an armature of resistance $2\,\Omega$ is designed to operate at $220\,V$ mains. At full speed,it develops a back $e.m.f.$ of $210\,V$. When the motor is running at full speed,the current in the armature is: (in $,A$)

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