The armature of a generator of resistance `1Omega` is rotated at its rated speed and produces 125 V without load and 115 V with full load. The current in the armature coil is

Options

A. 240 A
B. 10 A
C. 1 A
D. 2A

Answer

To find the current in the armature coil of the generator, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance of the armature (R) = 1 Ω
- Voltage without load (E) = 125 V
- Voltage with full load (V) = 115 V

2. **Understand the Concept of Back EMF:**
- When the generator is under full load, the voltage across the armature is less than the no-load voltage due to the back EMF (E_b) generated. The back EMF opposes the applied voltage.

3. **Calculate the Back EMF:**
- The back EMF can be calculated as:
\[ E_b = E - V \]
- Substituting the values:
\[ E_b = 125 \, \text{V} - 115 \, \text{V} = 10 \, \text{V} \]

4. **Apply Ohm's Law to Find the Current:**
- The current (I) in the armature can be calculated using Ohm’s Law:
\[ I = \frac{V}{R} \]
- Here, V is the voltage across the armature when it is under load, which is 115 V, and R is the resistance of the armature:
\[ I = \frac{115 \, \text{V}}{1 \, \Omega} = 115 \, \text{A} \]

5. **Calculate the Current Considering Back EMF:**
- The effective voltage that drives the current through the armature is the difference between the voltage with no load and the back EMF:
\[ I = \frac{E_b}{R} = \frac{10 \, \text{V}}{1 \, \Omega} = 10 \, \text{A} \]

6. **Final Answer:**
- The current in the armature coil is **10 A**.