A fan operates at 200 volt `(DC)` consuming `1000W` when running at full speed . It's internal wiring has resistance `1 Omega` . When the fan runs at full speed , its speed becomes constant. This is because the torque due to magnetic field inside tha fan is balanced by the torque due to air resistance on the blades of the fan and torque due to friction between the fixed part and the shaft of the fan. The electrical power going into the fan is spent `(i)` in the internal resistance as heat, call it `P_(1)(ii)` in doing work against internal friction and air resistance producing heat, sound etc. call it `P_(2)`. When the coil of fan rotates, an emf is also induced in the coil. This opposes the external emf applied to snd the current into the fan. This emf is called back-emf,call it `'e'`.
Answer the following questions when the fan is running at full speed.
The value of power `P_(1)` is

Options

A. (a) `1000 W` s
B. (b) `975 W`
C. ( c) `25W`
D. (d) `200 W`

Explanation

To find the value of power \( P_1 \) dissipated in the internal resistance of the fan, we can follow these steps:

### Step 1: Identify the given values
- Voltage \( V = 200 \, \text{V} \)
- Power \( P = 1000 \, \text{W} \)
- Internal resistance \( R = 1 \, \Omega \)

### Step 2: Calculate the current \( I \)
Using the formula for power:
\[ P = V \times I \]
We can rearrange this to find the current \( I \):
\[ I = \frac{P}{V} \]
Substituting the known values:
\[ I = \frac{1000 \, \text{W}}{200 \, \text{V}} = 5 \, \text{A} \]

### Step 3: Calculate the power \( P_1 \) dissipated in the internal resistance
The power dissipated in the internal resistance can be calculated using the formula:
\[ P_1 = I^2 \times R \]
Substituting the values of \( I \) and \( R \):
\[ P_1 = (5 \, \text{A})^2 \times 1 \, \Omega = 25 \, \text{W} \]

### Final Answer
The value of power \( P_1 \) is \( 25 \, \text{W} \).

---