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Problem 4In a long shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A . Determine (i) induced emf, (ii)total power generated. Given that shunt field, series field, divertor and armature resistances are $92 \Omega, 0.015 \Omega, 0.03 \Omega$ and $0.032 \Omega$ respectively.

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## Solution 1

#### Solution By Steps***Step 1: Calculate Voltage Drop in Series Field and Divertor***The total resistance in the series path (series field + divertor) is \(0.015 \Omega + 0.03 \Omega = 0.045 \Omega\). The voltage drop across this path is \(150 A imes 0.045 \Omega = 6.75 V\).***Step 2: Calculate Voltage Drop in Armature***The voltage drop across the armature is \(150 A imes 0.032 \Omega = 4.8 V\).***Step 3: Calculate Total Voltage Drop***The total voltage drop in the series path and armature is \(6.75 V + 4.8 V = 11.55 V\).***Step 4: Calculate Induced EMF***The induced EMF is the terminal voltage plus the total voltage drop, which is \(230 V + 11.55 V = 241.55 V\).***Step 5: Calculate Total Power Generated***The total power generated is the product of the induced EMF and the current delivered, which is \(241.55 V imes 150 A = 36232.5 W\).#### Final Answer(i) Induced EMF: \(241.55 V\)(ii) Total Power Generated: \(36232.5 W\)This problem involves calculating the induced electromotive force (EMF) and the total power generated in a long shunt compound generator, considering various resistances and current flows. Understanding these calculations is crucial for the design and operation of electrical generators.

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