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EMF of long shunt compound generator

Short answer

Shunt current is 200/100 = 2 A, so armature and series current is 42 A. Generated emf = V + I_a(R_a + R_se) = 200 + 42×0.9 = 237.8 V.

A dc long shunt compound dc generator has armature winding resistance of 0.4 ohm, series field winding resistance of 0.5 ohm and shunt-field winding resistance of 100 ohms. The generator delivers a current of 40 A to the load at 200 volt. Calculate the emf generated by the armature.

Given: IL=40I_L = 40 A, V=200V = 200 V, Ra=0.4 ΩR_a = 0.4\ \Omega, Rse=0.5 ΩR_{se} = 0.5\ \Omega, Rsh=100 ΩR_{sh} = 100\ \Omega; long-shunt connection.

     +--Ra--Rse--+-----------+
     |           |           |
    (Eg)        Rsh        Load 200 V
     |           |           |
     +-----------+-----------+

In long shunt, the shunt field is across the terminals and the series field carries the armature current.

Step 1: Shunt field current

Ish=VRsh=200100=2 AI_{sh} = \frac{V}{R_{sh}} = \frac{200}{100} = 2\ \text{A}

Step 2: Armature (and series field) current

Ia=Ise=IL+Ish=40+2=42 AI_a = I_{se} = I_L + I_{sh} = 40 + 2 = 42\ \text{A}

Step 3: Generated emf

Eg=V+Ia(Ra+Rse)=200+42×(0.4+0.5)=200+37.8=237.8 V\begin{aligned} E_g &= V + I_a (R_a + R_{se}) \\ &= 200 + 42 \times (0.4 + 0.5) \\ &= 200 + 37.8 = 237.8\ \text{V} \end{aligned}

Brush drop is neglected as it is not given.

Answer: EMF generated Eg=237.8E_g = 237.8 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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