An electric motor runs a `D.C.` source of e.m.f. `200 V` and draws a current of `10A`. If the efficiency is `40%`, then ressistance of the armature is:

Options

A. `2Omega`
B. `8 Omega`
C. `12 Omega`
D. `16 Omega`

Explanation

To find the resistance of the armature of the electric motor, we can follow these steps:

### Step 1: Calculate the Input Power
The input power (P_in) can be calculated using the formula:
\[ P_{\text{in}} = V \times I \]
where:
- \( V = 200 \, \text{V} \) (e.m.f. of the source)
- \( I = 10 \, \text{A} \) (current drawn by the motor)

Substituting the values:
\[ P_{\text{in}} = 200 \, \text{V} \times 10 \, \text{A} = 2000 \, \text{W} \]

### Step 2: Calculate the Output Power
The output power (P_out) can be calculated using the efficiency of the motor:
\[ P_{\text{out}} = \text{Efficiency} \times P_{\text{in}} \]
Given that the efficiency is 40%, we convert it to decimal form:
\[ \text{Efficiency} = \frac{40}{100} = 0.4 \]

Now substituting the values:
\[ P_{\text{out}} = 0.4 \times 2000 \, \text{W} = 800 \, \text{W} \]

### Step 3: Calculate the Power Loss
The power loss (P_loss) in the armature can be calculated as:
\[ P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} \]
Substituting the values:
\[ P_{\text{loss}} = 2000 \, \text{W} - 800 \, \text{W} = 1200 \, \text{W} \]

### Step 4: Relate Power Loss to Resistance
The power loss in the armature can also be expressed using the formula:
\[ P_{\text{loss}} = I^2 \times R \]
where \( R \) is the resistance of the armature.

### Step 5: Solve for Resistance
Rearranging the formula to find \( R \):
\[ R = \frac{P_{\text{loss}}}{I^2} \]
Substituting the known values:
\[ R = \frac{1200 \, \text{W}}{(10 \, \text{A})^2} = \frac{1200}{100} = 12 \, \Omega \]

### Final Answer
The resistance of the armature is:
\[ R = 12 \, \Omega \]