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EMF and load resistance of short shunt compound generator

Short answer

Load current 32.61 A, shunt voltage 239.78 V, armature current 35.01 A. Induced emf is 253.79 V and load resistance 7.05 Ω.

A short shunt compound dc generator supplies 7.5 kW at 230 V. The shunt field, series field and armature resistance are 100, 0.3 and 0.4 ohms respectively. Calculate the induced emf and the load resistance.

Given: P=7.5P = 7.5 kW, V=230V = 230 V, Rsh=100 ΩR_{sh} = 100\ \Omega, Rse=0.3 ΩR_{se} = 0.3\ \Omega, Ra=0.4 ΩR_a = 0.4\ \Omega; short-shunt connection.

     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 230 V
     |      |                 |
     +------+-----------------+

Step 1: Load current

IL=7500230=32.609 AI_L = \frac{7500}{230} = 32.609\ \text{A}

Step 2: Shunt field voltage and current

Vsh=V+ILRse=230+32.609×0.3=239.78 VIsh=239.78100=2.398 A\begin{aligned} V_{sh} &= V + I_L R_{se} = 230 + 32.609 \times 0.3 = 239.78\ \text{V} \\ I_{sh} &= \frac{239.78}{100} = 2.398\ \text{A} \end{aligned}

Step 3: Armature current and induced emf

Ia=IL+Ish=32.609+2.398=35.007 AEg=Vsh+IaRa=239.78+35.007×0.4=239.78+14.00=253.79 V\begin{aligned} I_a &= I_L + I_{sh} = 32.609 + 2.398 = 35.007\ \text{A} \\ E_g &= V_{sh} + I_a R_a = 239.78 + 35.007 \times 0.4 \\ &= 239.78 + 14.00 = 253.79\ \text{V} \end{aligned}

Step 4: Load resistance

RL=VIL=23032.609=7.053 ΩR_L = \frac{V}{I_L} = \frac{230}{32.609} = 7.053\ \Omega

(Or RL=V2/P=2302/7500=7.053 ΩR_L = V^2/P = 230^2/7500 = 7.053\ \Omega.)

Answer: Induced emf Eg≈253.79E_g \approx 253.79 V; load resistance RL≈7.05 ΩR_L \approx 7.05\ \Omega.

✓ Answer checked by Claude Opus 5.5 (high effort)

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