Electrical Machines A cumulatively compounded dc motor delivers rated load power at rated speed. If the series field is short-circuited, then Ia increases and ωm decreses both armature current Ia and speed ωm increase both decrese Ia remains constant but ωm increase Ia increases and ωm decreses both armature current Ia and speed ωm increase both decrese Ia remains constant but ωm increase ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines An auto transformer having a transformation ratio of 0.8 supplies a load pf 10 KW. The power transfered inductively from the primary to secondary is 10 KW. 8 KW. 0. 2 KW. 10 KW. 8 KW. 0. 2 KW. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines What is the mechanical power developed by a DC series motor is maximum? Back emf is equal to zero. None of above. Back emf is equal to half the applied voltage. Back emf is equal to applied voltage. Back emf is equal to zero. None of above. Back emf is equal to half the applied voltage. Back emf is equal to applied voltage. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines A slip ring Induction motor runs at 290 rpm at full load, when connected to 50 Hz supply. Determine the number of poles and slip. P = 60, s = 0.96 %. P = 50, s = 4.43 %. P = 20, s = 3.33 %. P = 40, s = 5.79 %. P = 60, s = 0.96 %. P = 50, s = 4.43 %. P = 20, s = 3.33 %. P = 40, s = 5.79 %. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines In salient pole synchronous machine, maximum power occurs at a load angle of 0° 90° less than 90° Greater than 90° 0° 90° less than 90° Greater than 90° ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines A synchronous machine has P number of poles, find the relation between electrical angular momentum (ωe) and mechanical angular momentum (ωm)? ωe = 2/P*ωm ωe = ωm ωe = P/2*ωm ωe = P*ωm ωe = 2/P*ωm ωe = ωm ωe = P/2*ωm ωe = P*ωm ANSWER DOWNLOAD EXAMIANS APP
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