Question

A silver wire has a resistance of 2.1 Ohm at 27.5 degree C , and a resistance of 2.7 Ohm at 100 degree C . Determine the temperature coefficient of resistivity of silver.

Answer 1

1. Understand the Principle

The relationship between resistance and temperature for most metals (like silver) over a moderate temperature range is approximately linear and can be described by the formula:

$R_2 = R_1 [1 + \alpha(T_2 - T_1)]$

Where:
$R_1$ is the resistance at the initial temperature $T_1$.
$R_2$ is the resistance at the final temperature $T_2$.
* $\alpha$ (alpha) is the temperature coefficient of resistivity, which is the value we need to find.

2. Identify the Given Information

• Initial Resistance ($R_1$) = 2.1 Ω
• Initial Temperature ($T_1$) = 27.5 °C
• Final Resistance ($R_2$) = 2.7 Ω
• Final Temperature ($T_2$) = 100 °C

3. Rearrange the Formula to Solve for α

We need to isolate $\alpha$ in the equation.

1. Start with the formula:
   $R_2 = R_1 [1 + \alpha(T_2 - T_1)]$

2. Divide both sides by $R_1$:
   $\frac{R_2}{R_1} = 1 + \alpha(T_2 - T_1)$

3. Subtract 1 from both sides:
   $\frac{R_2}{R_1} - 1 = \alpha(T_2 - T_1)$

4. Divide by the change in temperature $(T_2 - T_1)$:
   $\alpha = \frac{\frac{R_2}{R_1} - 1}{T_2 - T_1}$

A simpler arrangement for calculation is:
$R_2 - R_1 = R_1 \alpha (T_2 - T_1)$
$\alpha = \frac{R_2 - R_1}{R_1 (T_2 - T_1)}$

4. Substitute the Values and Calculate

Using the second, simpler formula:

• Change in Resistance: $R_2 - R_1 = 2.7 \, \Omega - 2.1 \, \Omega = 0.6 \, \Omega$
• Change in Temperature: $T_2 - T_1 = 100 \, °C - 27.5 \, °C = 72.5 \, °C$

Now, plug these into the rearranged formula for $\alpha$:

$\alpha = \frac{0.6 \, \Omega}{2.1 \, \Omega \times (72.5 \, °C)}$

$\alpha = \frac{0.6}{152.25}$

$\alpha \approx 0.00394088... \, °C^{-1}$

5. Final Answer

Rounding to three significant figures, the temperature coefficient of resistivity for silver is:

α = 0.00394 °C⁻¹

Alternatively, in scientific notation:

α = 3.94 x 10⁻³ °C⁻¹