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. `(a+2bt)/(1+at+bt^(2))`
B. `(a+2bt)`
C. `(1+at+bt^(2))/(a+2bt)`
D. constant

Answer

To find the temperature coefficient of resistance at temperature \( t^{\circ}C \), we start with the given formula 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 temperature coefficient of resistance
The temperature coefficient of resistance \( \alpha' \) at temperature \( t^{\circ}C \) is defined as:

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

### Step 2: Differentiate \( R_t \) with respect to \( t \)
We need to differentiate \( R_t \) with respect to \( t \):

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

Differentiating \( R_t \):

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

The derivative of the constant \( 1 \) is \( 0 \), the derivative of \( \alpha t \) is \( \alpha \), and the derivative of \( \beta t^2 \) is \( 2\beta t \). Thus,

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

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

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

### 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
Cancel \( R_0 \) from the numerator and denominator:

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

### Final Result
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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