أنشئ حسابًا أو سجّل الدخول للانضمام إلى مجتمعك المهني.
Here in this question, the data is not sufficient, so we resort to approximate way.
The Approximate Method:
Full load current ( FLA ) of Circuit Breaker = 1.5 * Full load current of motor But; The full load current of motor ( Approximate ) is equal to ( 2 * rating power )
So that; By Approximate method: FLA of motor = 2 * rating power = 2 * 10 = 20 Amp
Then; The FLA of C.B = 1.5 * FLA of motor =1.5 * 20 = 30 Amp
Note: we must be selected Circuit Breaker; Where it Bears Amperes greater than Ampere motor, and that referred to so as not to occurs Trip for C.B, during process starting. Whereas, While starting operation motor process, the motor consume large Amperes greater than rating current for a few seconds.
in general p= v.i pf since our motor p=10kw then if we assume our motor is a single phase induction motor and line voltage is 240v ac same as well we assume the efficiency of the motor let us say 80% then simply p= v.i .pf which gives I =p/(240*0.8 )=52 amperes
