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EMF of short shunt compound generator (0.5 Ω armature)

Short answer

For short shunt, shunt field voltage is V + I_L R_se = 200 + 50×0.3 = 215 V. Shunt current is 215/200 = 1.075 A, so armature current is 51.075 A. Generated emf = 215 + 51.075×0.5 = 240.54 V.

A dc short shunt compound generator has armature winding resistance of 0.5 ohm, series field winding resistance of 0.3 ohm and shunt field winding resistance of 200 ohms. It supplies a current of 50 Amp to the load at 200 V. Calculate the emf generated by the armature.

Given: IL=50I_L = 50 A, V=200V = 200 V, Ra=0.5 ΩR_a = 0.5\ \Omega, Rse=0.3 ΩR_{se} = 0.3\ \Omega, Rsh=200 ΩR_{sh} = 200\ \Omega; short-shunt connection.

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

Step 1: Voltage across shunt field

Series field carries load current:

Vsh=V+ILRse=200+50×0.3=215 VV_{sh} = V + I_L R_{se} = 200 + 50 \times 0.3 = 215\ \text{V}

Step 2: Shunt field and armature current

Ish=215200=1.075 AIa=IL+Ish=50+1.075=51.075 A\begin{aligned} I_{sh} &= \frac{215}{200} = 1.075\ \text{A} \\ I_a &= I_L + I_{sh} = 50 + 1.075 = 51.075\ \text{A} \end{aligned}

Step 3: Generated emf

Eg=Vsh+IaRa=215+51.075×0.5=215+25.54=240.54 V\begin{aligned} E_g &= V_{sh} + I_a R_a \\ &= 215 + 51.075 \times 0.5 \\ &= 215 + 25.54 = 240.54\ \text{V} \end{aligned}

Brush drop is neglected as it is not given.

Answer: EMF generated by the armature Eg≈240.54E_g \approx 240.54 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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