A simple electric motor has an armature resistance of `1Omega` and runs from a dc source of 12 volt . When running unloaded it draws a current of 2 amp . When a certain load is connected , its speed becomes one-half of its unloaded value . The new value of current drawn

Options

A. `7 A`
B. `2 A`
C. `5 A`
D. `3 A`

Answer

To solve the problem step by step, we will analyze the situation of the electric motor both in unloaded and loaded conditions.

### Step 1: Understand the given data
- Armature resistance (R) = 1 Ω
- Supply voltage (V) = 12 V
- Unloaded current (I_unloaded) = 2 A

### Step 2: Calculate the back EMF when the motor is unloaded
Using Ohm's law, we can find the voltage drop across the armature resistance when the motor is unloaded:
\[ V_{drop} = I_{unloaded} \times R = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \]
Now, the back EMF (E) when the motor is unloaded can be calculated as:
\[ E = V - V_{drop} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V} \]

### Step 3: Analyze the loaded condition
When the motor is loaded, its speed becomes half of its unloaded speed. This implies that the back EMF will also decrease. The back EMF is proportional to the speed of the motor, so if the speed is halved, the back EMF will also be halved:
\[ E_{loaded} = \frac{E_{unloaded}}{2} = \frac{10 \, \text{V}}{2} = 5 \, \text{V} \]

### Step 4: Calculate the new voltage drop across the armature resistance
The voltage drop across the armature resistance when the motor is loaded can be calculated as:
\[ V_{drop, loaded} = V - E_{loaded} = 12 \, \text{V} - 5 \, \text{V} = 7 \, \text{V} \]

### Step 5: Calculate the new current drawn when loaded
Using Ohm's law, we can find the new current drawn (I_loaded) when the motor is loaded:
\[ I_{loaded} = \frac{V_{drop, loaded}}{R} = \frac{7 \, \text{V}}{1 \, \Omega} = 7 \, \text{A} \]

### Step 6: Conclusion
The new value of the current drawn when the motor is loaded is:
\[ \text{I_loaded} = 7 \, \text{A} \]

### Summary of the Solution
- The unloaded current is 2 A.
- The back EMF when unloaded is 10 V.
- When loaded, the back EMF is halved to 5 V.
- The voltage drop across the armature resistance when loaded is 7 V.
- The new current drawn when loaded is 7 A.