A 4-pole, long shunt lap wound generator supplies 20 kW at a terminal voltage of 400 V. The armature resistance is 0.02 Ω, series field resistance is 0.04 Ω and shunt field resistance is 100 Ω. The brush drop may be taken as 1.0 V. Determine the induced EMF.

Concept:

In a long-shunt compound generator:

Here, Vt = Terminal Voltage (V)

Il = Line Current (A)

Rsf = Series field resistance (Ω)

Ra = Armature resistance (Ω)

Rshf = Shunt field resistance (Ω)

Ish = Shunt current (A)

Ia = Armature current (A)

Eg = Generated EMF (V)

V_b = Brush voltage drop ( = 2 V_b1 = 2 × voltage drop per brush)

Formula: 

\({I_{sh}}=\frac{{{V_t}}}{{{R_{shf}}}}\)

Ia = Ish + Il

Eg = Vt + Ia(Ra + Rsf) + 2 Vb1

Calculation: 

Given:

Output Power = P = 20 kW

Vt = 400 V

Ra = 0.02 Ω

Rsf = 0.04 Ω

Rshf = 100 Ω 

Vb1 = 1 V

I_sh = V_t / R_shf = 400 / 100 = 4 A

P = V_t × I_l ⇒ 20000 = 400 × Il

Il = 50 A

Ia = Ish + Il

Ia = 4 + 50 = 54 A

Eg = Vt + Ia(Ra + Rsf) + 2 Vb1

Eg = 400 + 54(0.02 + 0.04) + 2 × 1

Eg = 405.24 V

Important Points

In a short-shunt compound generator:

Va = Vt + IlRsf

Ia = Ish + Il

\({I_{sh}}=\frac{{{V_t}}}{{{R_{shf}}}}\)

Eg = Va + IaRa