To solve for the generated electromotive force (EMF) and the power developed by the generator, we can proceed with the following steps:

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Step 1: Identify given data

• Load current ($I_L$): $100 \, \text{A}$
• Terminal voltage ($V$): $220 \, \text{V}$
• Shunt field resistance ($R_f$): $50 \, \Omega$
• Series field resistance ($R_s$): $0.025 \, \Omega$
• Armature resistance ($R_a$): $0.05 \, \Omega$
• Brush drop: $1 \, \text{V}$ per brush ($2 \times 1 = 2 \, \text{V}$ total)

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Step 2: Calculate shunt field current

The current through the shunt field can be calculated as: $I_f = \frac{V}{R_f}$ Substitute the values: $I_f = \frac{220}{50} = 4.4 \, \text{A}$

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Step 3: Calculate armature current

The armature current is the sum of the load current and the shunt field current: $I_a = I_L + I_f$ Substitute the values: $I_a = 100 + 4.4 = 104.4 \, \text{A}$

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Step 4: Calculate total resistance drop

The voltage drop across the armature and series field resistances is: $V_{\text{drop}} = I_a (R_a + R_s)$ Substitute the values: $V_{\text{drop}} = 104.4 (0.05 + 0.025) = 104.4 \times 0.075 = 7.83 \, \text{V}$

Adding the brush drop ($2 \, \text{V}$): $V_{\text{total drop}} = V_{\text{drop}} + \text{Brush Drop} = 7.83 + 2 = 9.83 \, \text{V}$

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Step 5: Calculate generated EMF

The generated EMF ($E$) is the sum of the terminal voltage and the total resistance drop: $E = V + V_{\text{total drop}}$ Substitute the values: $E = 220 + 9.83 = 229.83 \, \text{V}$

Rounded to the nearest whole number: $E = 230 \, \text{V}$

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Step 6: Calculate power developed

The power developed in the armature is given by: $P = E \times I_a$ Substitute the values: $P = 230 \times 104.4 = 24,012 \, \text{W}$

Rounded to the nearest whole number: $P = 24,012 \, \text{W}$

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Final Answers:

• Generated EMF: $230 \, \text{V}$
• Power Developed: $24,012 \, \text{W}$

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Do you need further details or have any questions?

Related Questions:

1. How does brush drop affect the performance of a DC generator?
2. What are the advantages of using a compound generator over a simple DC generator?
3. Explain the role of the shunt and series field windings in this generator.
4. How would increasing the load current affect the generated EMF?
5. What is the efficiency of this generator if the input mechanical power is given?

Tip:

Always consider brush drop in DC generators as it directly affects the generated voltage and output performance.