To solve this problem, we will calculate each parameter step by step.

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, $V_b = 2 \, \text{V}$
• Iron and friction losses, $P_{\text{iron+friction}} = 1 \, \text{kW}$

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a. Generated E.M.F ($E_g$)

The total current through the armature is the sum of the load current and the current through the shunt field:

$I_f = \frac{V}{R_f} = \frac{220}{50} = 4.4 \, \text{A}$ $I_a = I_L + I_f = 100 + 4.4 = 104.4 \, \text{A}$

The generated E.M.F is calculated as:

$E_g = V + I_a (R_a + R_s) + V_b$ $E_g = 220 + 104.4 \times (0.05 + 0.025) + 2$ $E_g = 220 + 104.4 \times 0.075 + 2$ $E_g = 220 + 7.83 + 2 = 229.83 \, \text{V}$

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b. Copper Loss

Copper losses occur in the armature and field windings:

$P_{\text{armature loss}} = I_a^2 \cdot R_a$ $P_{\text{armature loss}} = (104.4)^2 \cdot 0.05 = 545.78 \, \text{W}$

$P_{\text{series loss}} = I_L^2 \cdot R_s$ $P_{\text{series loss}} = (100)^2 \cdot 0.025 = 250 \, \text{W}$

$P_{\text{shunt loss}} = \frac{V^2}{R_f}$ $P_{\text{shunt loss}} = \frac{220^2}{50} = 968 \, \text{W}$

Total copper loss: $P_{\text{copper loss}} = P_{\text{armature loss}} + P_{\text{series loss}} + P_{\text{shunt loss}}$ $P_{\text{copper loss}} = 545.78 + 250 + 968 = 1763.78 \, \text{W}$

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c. Output Power of the Prime Mover

The total power supplied by the prime mover must include:

1. Electrical output power (to the load)
2. Copper losses
3. Iron and friction losses

The electrical output power is: $P_{\text{electrical output}} = V \cdot I_L$ $P_{\text{electrical output}} = 220 \cdot 100 = 22,000 \, \text{W}$

Total power supplied by the prime mover: $P_{\text{prime mover}} = P_{\text{electrical output}} + P_{\text{copper loss}} + P_{\text{iron+friction}}$ $P_{\text{prime mover}} = 22,000 + 1763.78 + 1000 = 24,763.78 \, \text{W}$

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d. Generator Efficiency

The efficiency of the generator is given by: $\eta = \frac{\text{Output Power}}{\text{Input Power}} \times 100$ $\eta = \frac{P_{\text{electrical output}}}{P_{\text{prime mover}}} \times 100$ $\eta = \frac{22,000}{24,763.78} \times 100 = 88.85\%$

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Final Results:

a. Generated E.M.F ($E_g$): 229.83 V
b. Copper Loss: 1763.78 W
c. Output Power of the Prime Mover: 24.763.78 W
d. Generator Efficiency: 88.85%

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Do you need more detailed explanations, or have further questions?
Here are 5 related questions to explore:

1. How do changes in field resistance affect generator performance?
2. What is the significance of brush drop in DC machines?
3. How are copper losses minimized in practical applications?
4. Why is the shunt field current so much smaller than the armature current?
5. What factors contribute to iron and friction losses?

Tip: Always ensure proper maintenance of generator components to reduce friction and improve efficiency.