Question

A copper wire has resistance of
2.0Ω
at
0oC
and
2.26Ω
at
30oC
. What is its resistance at
50oC
?

Answer 1

Answer on Question #46033 – Physics – Electromagnetism

Question.

A copper wire has resistance of $2.0\Omega$ at 0oC and $2.26\Omega$ at 30oC. What is its resistance at 50oC?

Given:

$$R_1 = 2 \Omega$$ $$T_1 = 0{}^\circ \text{C}$$ $$R_2 = 2.26 \Omega$$ $$T_2 = 30{}^\circ \text{C}$$ $$T_3 = 50{}^\circ \text{C}$$

Find:

$$R_3 = ?$$

Solution.

As we know the resistance's dependence of temperature is expressed the following:

$$R = R_0 [1 + \alpha (T - T_0)]$$

So, we must find the temperature coefficient $\alpha$:

$$\alpha = \frac{1}{R_0} \frac{R - R_0}{T - T_0}$$

We can find the temperature coefficient $\alpha$ for this material, because we know $R_1, T_1, R_2, T_2$:

$$\alpha = \frac{1}{R_1} \frac{R_2 - R_1}{T_2 - T_1}$$

Therefore, we can define the resistance at any temperature:

$$R_3 = R_1 [1 + \alpha (T_3 - T_1)] = R_1 \left[1 + \frac{R_2 - R_1}{R_1} \frac{T_3 - T_1}{T_2 - T_1} \right] = R_1 + (R_2 - R_1) \frac{T_3 - T_1}{T_2 - T_1}$$

Calculate:

$$R_3 = 2 + 0.26 \frac{50}{30} = 2 + 0.433 = 2.433 \Omega$$

Answer.

$$R_3 = R_1 + (R_2 - R_1) \frac{T_3 - T_1}{T_2 - T_1} = 2.433 \Omega$$

http://www.AssignmentExpert.com/