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Short shunt cumulative compound generator, 7.5 kW

Short answer

I_L=7500/230≈32.609 A, R_L=230/32.609≈7.053 Ω. V_sh=230+32.609×0.3=239.783 V, I_sh=2.3978 A, I_a=35.007 A. Induced emf E=239.783+35.007×0.4=253.79 V.

A short shunt cumulative 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.

Short shunt: the shunt field is across the armature and the series field is in the load line. Cumulative: the series field aids the shunt field, which does not change the circuit calculation.

Load current and load resistance

IL=PV=7500230=32.609 A,RL=VIL=23032.609=7.053 ΩI_L = \frac{P}{V} = \frac{7500}{230} = 32.609\ \text{A}, \qquad R_L = \frac{V}{I_L} = \frac{230}{32.609} = 7.053\ \Omega

Shunt field and armature currents

Vsh=V+ILRse=230+(32.609)(0.3)=239.783 VV_{sh} = V + I_LR_{se} = 230 + (32.609)(0.3) = 239.783\ \text{V} Ish=239.783100=2.3978 A,Ia=IL+Ish=35.007 AI_{sh} = \frac{239.783}{100} = 2.3978\ \text{A}, \qquad I_a = I_L + I_{sh} = 35.007\ \text{A}

Induced emf

E=Vsh+IaRa=239.783+(35.007)(0.4)=253.79 VE = V_{sh} + I_aR_a = 239.783 + (35.007)(0.4) = 253.79\ \text{V}

Answer: induced emf =253.79 V= 253.79\ \text{V}; load resistance =7.053 Ω= 7.053\ \Omega.

✓ Answer checked by Claude Opus 5.5 (high effort)

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