A 250 V, 50 kW, short-shunt compound DC generator has the following data : armature resistance = 0:05 Ω series field resistance = 0:05 Ω, shunt field resistance = 130 Ω and 2 V is the total brush contact drop. What is the value of the total current supplied by the generator?

In short shunt compound generator:

Here, V_t = Terminal Voltage (V)

IL = Line Current (A)

Rse = Series field resistance (Ω)

Ra = Armature resistance (Ω)

Rsh = Shunt field resistance (Ω)

Ish = Shunt current (A)

Ia = Armature current (A)

Vb = Brush voltage drop (V)

Va = Armature terminal voltage (V)

Formula:

Va = Vt + ILRsh

Ia = Ish + Il

Eg = Va + IaRa + 2Vb

Calculation:

Given:

Vt = 250 V

IL = P / Vt  = 50 k / 250 = 200 A

Rse = 0.05 Ω

Ra = 0.05 Ω

Rsh = 130 Ω

Total brush contact drop = 2Vb = 2 V

Va = Vt + IlRsf

Va = 250 + 200 × 0.05

Va = 260 V

Ia = Ish + Il

Ia = 2 + 200 = 202 A

Eg = Va + IaRa + 2Vb

Eg = 260 + 202 ×  0.05 + 2 

Eg = 272.1 V

∴ Total current supplied by the generator = I_a = 202 A = 0.202 kA.

Important Points

In long shunt compound generator:

Formula:

Ia = Ish + Il

Eg = Vt + I_a(R_a + Rsf) + 2Vb