Resistance of a resistor at temperature `t^@C is R_t =R_0 (1+alphat + betat^2)`, where `R_0` is the resistance at `0^@C`. The temperature coefficient of resistance at temperature `t^@C` is

Options

A. `((1+alphat+betat^2))/(alpha+2beta t)`
B. `(alpha+2beta t)`
C. `(alpha +2 beta t)/((1 +alphat + beta t^2))`
D. `(alpha +2betat)/(2(1+alpaht +beta r^2))`

Explanation

To find the temperature coefficient of resistance at temperature \( t^\circ C \), we start with the given expression for resistance:

\[ R_t = R_0 (1 + \alpha t + \beta t^2) \]

where \( R_0 \) is the resistance at \( 0^\circ C \), \( \alpha \) is the linear temperature coefficient, and \( \beta \) is the quadratic temperature coefficient.

### Step 1: Understand the definition of the temperature coefficient of resistance
The temperature coefficient of resistance \( \alpha' \) at a temperature \( t \) is defined as:

\[ \alpha' = \frac{1}{R_t} \frac{dR_t}{dt} \]

This means we need to differentiate \( R_t \) with respect to \( t \).

### Step 2: Differentiate \( R_t \) with respect to \( t \)
Let's differentiate the expression for \( R_t \):

\[ \frac{dR_t}{dt} = \frac{d}{dt} \left( R_0 (1 + \alpha t + \beta t^2) \right) \]

Using the constant \( R_0 \):

\[ \frac{dR_t}{dt} = R_0 \left( \alpha + 2\beta t \right) \]

### Step 3: Substitute \( \frac{dR_t}{dt} \) back into the formula for \( \alpha' \)
Now we substitute \( \frac{dR_t}{dt} \) back into the formula for \( \alpha' \):

\[ \alpha' = \frac{1}{R_t} \cdot R_0 \left( \alpha + 2\beta t \right) \]

### Step 4: Substitute \( R_t \) into the equation
Substituting \( R_t = R_0 (1 + \alpha t + \beta t^2) \):

\[ \alpha' = \frac{R_0 (\alpha + 2\beta t)}{R_0 (1 + \alpha t + \beta t^2)} \]

### Step 5: Simplify the expression
The \( R_0 \) cancels out:

\[ \alpha' = \frac{\alpha + 2\beta t}{1 + \alpha t + \beta t^2} \]

### Final Expression
Thus, the temperature coefficient of resistance at temperature \( t^\circ C \) is:

\[ \alpha' = \frac{\alpha + 2\beta t}{1 + \alpha t + \beta t^2} \]

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