An electric heater consumes `1` kilowatt power when connected across a `100` volt. D.C. supply. If this heater is to be used with `200V`, `50Hz` A.C. supply, the value of inductance to be connected in series with it is

Options

A. `5.5H`
B. `0.55H`
C. `0.055H`
D. `1.1H`

Answer

To solve the problem, we need to find the value of inductance that should be connected in series with an electric heater when it is used with a 200V, 50Hz AC supply. The heater consumes 1 kW of power when connected to a 100V DC supply.

### Step-by-Step Solution:

1. **Determine the Resistance of the Heater:**
- The power consumed by the heater is given as \( P = 1 \text{ kW} = 1000 \text{ W} \).
- The voltage across the heater when connected to the DC supply is \( V = 100 \text{ V} \).
- Using the formula for power:
\[ P = \frac{V^2}{R} \]
We can rearrange this to find the resistance \( R \):
\[ R = \frac{V^2}{P} = \frac{100^2}{1000} = \frac{10000}{1000} = 10 \, \Omega \]

2. **Calculate the Current through the Heater:**
- The current \( I \) flowing through the heater can be calculated using Ohm's law:
\[ I = \frac{V}{R} = \frac{100}{10} = 10 \, \text{A} \]

3. **Determine the Impedance Required for AC Supply:**
- The AC supply voltage is \( V_{AC} = 200 \text{ V} \).
- The RMS current \( I_{RMS} \) through the heater remains the same, which is \( 10 \text{ A} \).
- The impedance \( Z \) required for the AC supply can be calculated using:
\[ Z = \frac{V_{AC}}{I_{RMS}} = \frac{200}{10} = 20 \, \Omega \]

4. **Relate Impedance to Resistance and Inductive Reactance:**
- The total impedance \( Z \) in an inductive circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Here, \( R = 10 \, \Omega \) and we need to find \( X_L \) (the inductive reactance).
- Setting up the equation:
\[ 20 = \sqrt{10^2 + X_L^2} \]
- Squaring both sides:
\[ 400 = 100 + X_L^2 \]
- Rearranging gives:
\[ X_L^2 = 400 - 100 = 300 \]
- Taking the square root:
\[ X_L = \sqrt{300} \approx 17.32 \, \Omega \]

5. **Calculate the Inductance:**
- The inductive reactance \( X_L \) is related to the inductance \( L \) by the formula:
\[ X_L = 2 \pi f L \]
- Where \( f = 50 \text{ Hz} \). Rearranging for \( L \):
\[ L = \frac{X_L}{2 \pi f} = \frac{17.32}{2 \pi \times 50} \]
- Calculating:
\[ L = \frac{17.32}{100 \pi} \approx \frac{17.32}{314.16} \approx 0.055 \, \text{H} \]

### Final Answer:
The value of inductance to be connected in series with the heater is approximately \( 0.055 \, \text{H} \).

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