We can solve this problem step-by-step by considering the electrical circuit of the shunt generator.

Given:

• Induced EMF ($E$) = 254 V
• Terminal voltage ($V_t$) = 240 V
• Armature resistance ($R_a$) = 0.04 Ω
• Field circuit resistance ($R_f$) = 24 Ω

Formula:

The total load current ($I_L$) is the sum of the armature current ($I_a$) and the field current ($I_f$): $I_L = I_a + I_f$

Step 1: Calculate the field current ($I_f$):

The field current is calculated using the terminal voltage ($V_t$) and the field circuit resistance ($R_f$): $I_f = \frac{V_t}{R_f}$ $I_f = \frac{240}{24} = 10 \, \text{A}$

Step 2: Calculate the armature current ($I_a$):

The relationship between the induced EMF ($E$), terminal voltage ($V_t$), and armature current ($I_a$) is given by: $E = V_t + I_a R_a$ Rearranging to solve for $I_a$: $I_a = \frac{E - V_t}{R_a}$ $I_a = \frac{254 - 240}{0.04} = \frac{14}{0.04} = 350 \, \text{A}$

Step 3: Calculate the load current ($I_L$):

Substitute $I_a$ and $I_f$ into the equation for $I_L$: $I_L = I_a + I_f$ $I_L = 350 + 10 = 360 \, \text{A}$

Final Answer:

The load current ($I_L$) is 360 A.

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Would you like further clarification or details? Here are five related questions to deepen your understanding:

1. How does armature reaction affect the terminal voltage in practical scenarios?
2. What is the effect of increasing field resistance on the load current?
3. Can this problem be solved if armature reaction were not neglected?
4. How would the load current change if the armature resistance were doubled?
5. What is the efficiency of the generator under these conditions?

Tip: Always check the calculated field and armature currents separately to avoid errors when combining them for load current calculations.