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 current flowing into the fan and the value of back emf e is

Options

A. (a) `200 A, 5 V`
B. (b) `5 A, 200 V`
C. ( c) `5 A, 195 V`
D. (d) `1 A, 0 V`

Explanation

To solve the problem, we need to determine the current flowing into the fan and the value of the back-emf (e) when the fan is running at full speed.

### Step 1: Calculate the Current (I)

We know that the power (P) consumed by the fan is given as 1000 W, and the voltage (V) is 200 V. We can use the formula for power:

\[ P = V \times I \]

Rearranging this formula to find the current (I):

\[ I = \frac{P}{V} \]

Substituting the given values:

\[ I = \frac{1000 \, \text{W}}{200 \, \text{V}} = 5 \, \text{A} \]

### Step 2: Calculate the Power Dissipated in Internal Resistance (P1)

The internal resistance (R) of the fan is given as 1 Ω. The power dissipated in the internal resistance can be calculated using the formula:

\[ P_1 = I^2 \times R \]

Substituting the values we found:

\[ P_1 = (5 \, \text{A})^2 \times 1 \, \Omega = 25 \, \text{W} \]

### Step 3: Calculate the Back-emf (e)

The back-emf (e) can be calculated using the relationship between the source voltage (V), the current flowing through the internal resistance, and the power dissipated. The voltage drop across the internal resistance is equal to the power dissipated divided by the current:

\[ V_{\text{drop}} = I \times R = 5 \, \text{A} \times 1 \, \Omega = 5 \, \text{V} \]

Now, we can find the back-emf (e) using the equation:

\[ e = V - V_{\text{drop}} \]

Substituting the values:

\[ e = 200 \, \text{V} - 5 \, \text{V} = 195 \, \text{V} \]

### Final Answers

- The current flowing into the fan is **5 A**.
- The value of back-emf (e) is **195 V**.

---