Q.A silver wire has a resistance of $2.1\ \Omega$ at $27.5\ ^\circ\text{C}$, and a resistance of $2.7\ \Omega$ at $100\ ^\circ\text{C}$. Determine the temperature coefficient of resistivity of silver.

Concept: Temperature Dependence of Resistance — the resistance of a metal changes linearly with temperature over moderate ranges, given by $R_T = R_0 (1 + \alpha \Delta T)$.

Reasoning:

1. The formula relating resistance at two temperatures is:

where $\alpha$ is the temperature coefficient of resistivity.

2. Substitute the given values: $R_1 = 2.1\ \Omega$ at $T_1 = 27.5\ ^\circ\text{C}$, and $R_2 = 2.7\ \Omega$ at $T_2 = 100\ ^\circ\text{C}$.

3. Solve for $\alpha$: …

The key idea is that for most metals over a moderate temperature range, resistance changes linearly with temperature. This is because resistivity itself increases linearly with temperature due to increased lattice vibrations (phonons) scattering electrons. The formula is:

where $R_T$ is resistance at temperature $T$, $R_0$ is resistance at a reference temperature $T_0$, and $\alpha$ is the temperature coefficient of resistivity. The catch: $R_0$ is not given directly — we have two data points, so we must solve for both $R_0$ and $\alpha$.

Let’s work through it step by step.

1. Set up two equations. Let $T_0 = 27.5\ ^\circ\text{C}$ be the reference. Then $R_0 = 2.1\ \Omega$ at $T_0$. At $T_1 = 100\ ^\circ\text{C}$, $\Delta T_1 = 100 - 27.5 = 72.5\ ^\circ\text{C}$, and $R_1 = 2.7\ \Omega$. So:

2. Solve for $\alpha$. Divide both sides by 2.1:

Subtract 1:

$\alpha = \frac{0.2857}{72.5} \approx 0.00394\ ^\circ\text{C}^{-1}$ …

Method: Using the Linear Approximation for Resistance vs. Temperature

This method uses the linear formula for resistance change with temperature, which is valid over moderate temperature ranges (as given in the problem).

Formula

For a conductor, resistance varies approximately linearly with temperature:

Where:

• $R_T$ = resistance at temperature $T$
• $R_0$ = resistance at reference temperature $T_0$
• $\alpha$ = temperature coefficient of resistivity (what we need to find)

Steps

Step 1: Identify the given data

• $R_1 = 2.1\ \Omega$ at $T_1 = 27.5\ ^\circ\text{C}$
• $R_2 = 2.7\ \Omega$ at $T_2 = 100\ ^\circ\text{C}$

Step 2: Choose a reference temperature

Take $T_0 = T_1 = 27.5\ ^\circ\text{C}$ and $R_0 = R_1 = 2.1\ \Omega$.

Step 3: Apply the formula for the second point

Substitute values:

Step 4: Solve for $\alpha$

Divide both sides by 2.1:

Subtract 1:

$\alpha = \frac{0.2857}{72.5}$ …

Here are the common mistakes students make when solving this exact problem, along with how to avoid each one.

1. Using the Wrong Formula (Confusing $\alpha$ for Resistance vs. Resistivity)

The Mistake:

Students often use the formula for the temperature dependence of resistivity ($\rho$) directly on resistance ($R$) without checking if the wire’s dimensions change.

The Correction:

For a metallic wire, if we assume linear expansion is negligible (which is standard in such problems), the temperature coefficient of resistance ($\alpha_R$) is approximately equal to the temperature coefficient of resistivity ($\alpha$).

The correct formula is:

Where:

• $R_t$ = resistance at temperature $t$
• $R_0$ = resistance at reference temperature $t_0$
• $\alpha$ = temperature coefficient of resistivity

How to avoid:

Always write the formula explicitly before plugging numbers. Check whether the problem asks for $\alpha$ of resistivity or resistance — here it’s resistivity, but the formula is the same because dimensions are constant.

2. Using the Wrong Temperature Difference (Celsius vs. Kelvin)

The Mistake:

Some students convert $27.5^\circ\text{C}$ and $100^\circ\text{C}$ to Kelvin, then subtract. Since the coefficient $\alpha$ is defined per degree Celsius, this gives the same numerical difference but can cause confusion if the formula expects Celsius.

The Correction:

The difference $t - t_0$ is the same in Celsius and Kelvin:

So no conversion is needed — but be consistent.

How to avoid:

Stick to Celsius unless the problem explicitly gives a reference temperature in Kelvin. The formula $R_t = R_0[1 + \alpha (t - t_0)]$ uses Celsius differences.

3. Swapping $R_0$ and $R_t$

The Mistake:

Using $R_0 = 2.7\ \Omega$ (the higher temperature) and $R_t = 2.1\ \Omega$ (the lower temperature). This gives a negative $\alpha$, which is wrong for metals.

The Correction:

$R_0$ is the resistance at the lower reference temperature. Here:

• $t_0 = 27.5^\circ\text{C}$, $R_0 = 2.1\ \Omega$
• $t = 100^\circ\text{C}$, $R_t = 2.7\ \Omega$

How to avoid:

Label clearly: “$R_0$ at $t_0$” and “$R_t$ at $t$”. Metals have positive $\alpha$, so if you get a negative number, you swapped them.

4. Forgetting to Subtract 1 After Rearranging

The Mistake:

Plugging into $R_t = R_0[1 + \alpha \Delta t]$ and solving incorrectly — e.g., writing $\alpha = \frac{R_t}{R_0 \Delta t}$ instead of $\alpha = \frac{R_t/R_0 - 1}{\Delta t}$.

The Correction:

From $R_t = R_0(1 + \alpha \Delta t)$:

$\frac{R_t}{R_0} = 1 + \alpha \Delta t$ …