A shunt generator delivers an output voltage of 240 V at a current of 90 A . The armature and field resistances are $$ 0,025 \Omega $$ and $$ 80 \Omega $$ respectively. If the no-load current is $$ 3,09 \mathrm{~A} $$, determine the efficiency of the generator.

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To determine the efficiency of the shunt generator, we need to calculate the input power and the output power. The efficiency can be calculated using the formula:

$$
\text{Efficiency} (\eta) = \frac{\text{Output Power}}{\text{Input Power}} \times 100\%
$$

### Step 1: Calculate Output Power

The output power ($$ P_{\text{out}} $$) can be calculated using the formula:

$$
P_{\text{out}} = V \times I
$$

Where:
- $$ V = 240 \, \text{V} $$ (output voltage)
- $$ I = 90 \, \text{A} $$ (output current)

Substituting the values:

$$
P_{\text{out}} = 240 \, \text{V} \times 90 \, \text{A} = 21600 \, \text{W}
$$

### Step 2: Calculate Input Power

The input power ($$ P_{\text{in}} $$) is the power supplied to the generator, which includes the power consumed by the armature and field windings.

1. **Calculate the field current ($$ I_f $$)**:
   The field current can be calculated using Ohm's law:

$$
   I_f = \frac{V}{R_f}
   $$

Where:
   - $$ R_f = 80 \, \Omega $$ (field resistance)

Substituting the values:

$$
   I_f = \frac{240 \, \text{V}}{80 \, \Omega} = 3 \, \text{A}
   $$

2. **Calculate the total armature current ($$ I_a $$)**:
   The armature current is the sum of the output current and the no-load current:

$$
   I_a = I + I_f
   $$

Where:
   - $$ I = 90 \, \text{A} $$ (output current)
   - $$ I_f = 3 \, \text{A} $$ (field current)

Substituting the values:

$$
   I_a = 90 \, \text{A} + 3 \, \text{A} = 93 \, \text{A}
   $$

3. **Calculate the voltage drop across the armature resistance ($$ R_a $$)**:
   The voltage drop across the armature resistance can be calculated as:

$$
   V_a = I_a \times R_a
   $$

Where:
   - $$ R_a = 0.025 \, \Omega $$ (armature resistance)

Substituting the values:

$$
   V_a = 93 \, \text{A} \times 0.025 \, \Omega = 2.325 \, \text{V}
   $$

4. **Calculate the input voltage ($$ V_{\text{in}} $$)**:
   The input voltage is the sum of the output voltage and the voltage drop across the armature resistance:

$$
   V_{\text{in}} = V + V_a
   $$

Substituting the values:

$$
   V_{\text{in}} = 240 \, \text{V} + 2.325 \, \text{V} = 242.325 \, \text{V}
   $$

5. **Calculate the input power ($$ P_{\text{in}} $$)**:
   The input power can be calculated as:

$$
   P_{\text{in}} = V_{\text{in}} \times I_a
   $$

Substituting the values:

$$
   P_{\text{in}} = 242.325 \, \text{V} \times 93 \, \text{A} \approx 22500.225 \, \text{W}
   $$

### Step 3: Calculate Efficiency

Now we can calculate the efficiency:

$$
\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%
$$

Substituting the values:

$$
\eta = \frac{21600 \, \text{W}}{22500.225 \, \text{W}} \times 100\% \approx 96.00\%
$$

### Conclusion

The efficiency of the shunt generator is approximately **96.00%**.
The efficiency of the shunt generator is approximately 96.00%.

A shunt generator delivers an output voltage of 240 V at a current of 90 A . The armature and field resistances are \( 0,025 \Omega \) and \( 80 \Omega \) respectively. If the no-load current is \( 3,09 \mathrm{~A} \), determine the efficiency of the generator.

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