Coefficient of linear expnsion of material of resistor is `alpha` . Its temperature coefficient of resistivity and resistance are `alpha_(p)` and `alpha_(R)` respectively , the correct relation is .

Options

A. `alpha_(R) = alpha_(rho) - alpha`
B. `alpha_(R) = alpha_(rho) + alpha`
C. `alpha_(R) = alpha_(rho) + 3 alpha`
D. `alpha_(R) = alpha_(rho) - 3alpha`

Answer

To derive the correct relation between the coefficient of linear expansion of the material of a resistor (α), the temperature coefficient of resistivity (αₚ), and the temperature coefficient of resistance (αᵣ), we can follow these steps:

### Step 1: Understand the basic relationships
The resistance \( R \) of a resistor is given by the formula:
\[ R = \rho \frac{L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

### Step 2: Define initial conditions
At an initial temperature \( T_0 \):
- Resistance \( R_0 = \rho_0 \frac{L_0}{A_0} \)

### Step 3: Consider changes due to temperature increase
When the temperature increases by \( \Delta T \):
- The new resistance \( R \) can be expressed as:
\[ R = R_0 (1 + \alpha_R \Delta T) \]
- The resistivity changes as:
\[ \rho = \rho_0 (1 + \alpha_\rho \Delta T) \]
- The length changes as:
\[ L = L_0 (1 + \alpha \Delta T) \]
- The area changes as:
\[ A = A_0 (1 + 2\alpha \Delta T) \]

### Step 4: Substitute the new values into the resistance formula
Substituting the new values into the resistance formula gives:
\[ R = \rho (1 + \alpha \Delta T) \frac{L}{A} \]
Substituting for \( \rho \), \( L \), and \( A \):
\[ R = \left( \rho_0 (1 + \alpha_\rho \Delta T) \right) \frac{L_0 (1 + \alpha \Delta T)}{A_0 (1 + 2\alpha \Delta T)} \]

### Step 5: Simplify the expression
Using the initial condition \( R_0 = \rho_0 \frac{L_0}{A_0} \):
\[ R = R_0 \frac{(1 + \alpha_\rho \Delta T)(1 + \alpha \Delta T)}{(1 + 2\alpha \Delta T)} \]

### Step 6: Use binomial expansion for small \( \Delta T \)
For small \( \Delta T \), we can use the binomial expansion:
\[ (1 + x) \approx 1 + x \quad \text{for small } x \]
Thus, we can approximate:
\[ R \approx R_0 \left(1 + \alpha_\rho \Delta T + \alpha \Delta T - 2\alpha \Delta T\right) \]
This simplifies to:
\[ R \approx R_0 (1 + (\alpha_\rho - \alpha) \Delta T) \]

### Step 7: Equate the expressions for \( R \)
From the earlier expression for \( R \):
\[ R \approx R_0 (1 + \alpha_R \Delta T) \]
Equating the two expressions for \( R \):
\[ 1 + \alpha_R \Delta T = 1 + (\alpha_\rho - \alpha) \Delta T \]

### Step 8: Solve for the relation
Cancelling the common terms:
\[ \alpha_R = \alpha_\rho - \alpha \]

### Final Relation
Thus, the correct relation is:
\[ \alpha_R = \alpha_\rho - \alpha \]