Electrical Engineering: High-Yield Core for UPPSC AE
What the technical paper tests
The UPPSC Assistant Engineer (Electrical) paper is set at B.E./B.Tech level: a mix of one-line conceptual stems and numericals that must be solved in roughly a minute each. The recurring blocks are circuit theory, electrical machines, power systems, control systems, measurements, power electronics and utilization.
Core rules worth memorising
- Transformer e.m.f.: E = 4.44 f N Φm (the 4.44 is 1.11 × 4, i.e. form factor times 4).
- D.C. machine e.m.f.: E = PΦNZ/60A; developed torque T = 0.159 PΦIZ/A.
- Induction motor: slip s = (Ns − N)/Ns; rotor frequency = sf; rotor copper loss = s × air-gap power; maximum torque is independent of rotor resistance but occurs at s = R2/X2.
- Three-phase circuits: solve per phase; for star V_L = √3 V_ph, for delta I_L = √3 I_ph.
- A transformer reaches maximum efficiency when copper loss equals iron loss.
Worked example
A 50 Hz, 4-pole, 3-phase induction motor runs at 1440 r.p.m. Synchronous speed Ns = 120f/P = 1500 r.p.m.; slip s = (1500 − 1440)/1500 = 0.04, i.e. 4%. Rotor frequency = 0.04 × 50 = 2 Hz. If rotor copper loss is 200 W, air-gap power = 200/0.04 = 5000 W, and gross mechanical power = 5000 − 200 = 4800 W. Note the chain: rotor copper loss is always slip times air-gap power, so one given value fixes the whole power flow.
Traps that cost marks
Mixing line and phase values in three-phase items; using 4fNΦm and dropping the 1.11 factor; substituting slip as a percentage inside formulas; assuming voltage regulation is always positive (it turns negative for a leading power factor); applying Routh's criterion without handling a zero first-column entry; and confusing 'synchronous watt', which expresses torque, with an ordinary power unit.