A heating element has a resistance of 100 `Omega` at room temperature. When it is connected to a supply of 220 V, a steady current of 2A passes in it and temperature is `500^(@)C` more than room temperature. What is the temperature coefficient of resistance of the heating element ?

Options

A. `2xx10^(-4).^(@)C^(-1)`
B. `0.5xx10^(-4).^(@)C^(-1)`
C. `1xx10^(-4).^(@)C^(-1)`
D. `5xx10^(-4).^(@)C^(-1)`

Answer

To find the temperature coefficient of resistance (α) of the heating element, we can follow these steps:

### Step 1: Identify the given values
- Resistance at room temperature (R₀) = 100 Ω
- Voltage (V) = 220 V
- Current (I) = 2 A
- Increase in temperature (ΔT) = 500 °C

### Step 2: Calculate the resistance at the operating temperature (R')
Using Ohm's Law, we can find the resistance when the heating element is connected to the supply:
\[ R' = \frac{V}{I} \]
Substituting the values:
\[ R' = \frac{220 \, \text{V}}{2 \, \text{A}} = 110 \, \Omega \]

### Step 3: Use the formula for resistance change with temperature
The relationship between resistance and temperature is given by:
\[ R' = R_0 (1 + \alpha \Delta T) \]
Where:
- R' = resistance at the elevated temperature
- R₀ = resistance at room temperature
- α = temperature coefficient of resistance
- ΔT = change in temperature

### Step 4: Substitute known values into the equation
We know R' = 110 Ω, R₀ = 100 Ω, and ΔT = 500 °C. Substituting these into the equation:
\[ 110 = 100 (1 + \alpha \cdot 500) \]

### Step 5: Solve for α
First, divide both sides by 100:
\[ 1.1 = 1 + 500\alpha \]

Now, isolate α:
\[ 1.1 - 1 = 500\alpha \]
\[ 0.1 = 500\alpha \]
\[ \alpha = \frac{0.1}{500} \]
\[ \alpha = \frac{1}{5000} \]
\[ \alpha = 2 \times 10^{-4} \, \text{°C}^{-1} \]

### Final Answer
The temperature coefficient of resistance (α) of the heating element is:
\[ \alpha = 2 \times 10^{-4} \, \text{°C}^{-1} \]

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