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. 3A
B. 6A
C. 2A
D. 1A

Answer

To solve the problem step by step, we need to analyze the given information and apply the relevant formulas.

### Step 1: Identify the given values
- Armature resistance, \( R_a = 1 \, \Omega \)
- Voltage from the DC source, \( V = 12 \, V \)
- Unloaded current, \( I_{\text{unloaded}} = 2 \, A \)

### Step 2: Calculate the back emf when the motor is unloaded
The back emf (\( V_b \)) can be calculated using the formula:
\[ V_b = V - I \cdot R_a \]
Substituting the known values:
\[ V_b = 12 \, V - (2 \, A \cdot 1 \, \Omega) = 12 \, V - 2 \, V = 10 \, V \]

### Step 3: Determine the new speed and back emf when the load is connected
When a load is connected, the speed of the motor becomes half of its unloaded speed. The back emf is proportional to the speed of the motor, so if the speed is halved, the back emf will also be halved:
\[ V_{b,\text{new}} = \frac{1}{2} V_b = \frac{1}{2} \cdot 10 \, V = 5 \, V \]

### Step 4: Calculate the new current drawn with the load
Using the same formula for back emf:
\[ V_b = V - I_{\text{new}} \cdot R_a \]
We can rearrange this to solve for the new current (\( I_{\text{new}} \)):
\[ I_{\text{new}} = \frac{V - V_b}{R_a} \]
Substituting the values we have:
\[ I_{\text{new}} = \frac{12 \, V - 5 \, V}{1 \, \Omega} = \frac{7 \, V}{1 \, \Omega} = 7 \, A \]

### Step 5: Determine the total current drawn from the source
The total current drawn from the source when the load is connected is the sum of the unloaded current and the new current:
\[ I_{\text{total}} = I_{\text{unloaded}} + I_{\text{new}} = 2 \, A + 7 \, A = 9 \, A \]

### Conclusion
The new value of the current drawn by the motor when a load is connected is **9 A**.