4.) (a) At what temperature would the resistance of a wire P made of material with a
temperature coeflicient of resistivity be one-half its resistance at ?
Use as the reference point in your calculation.
(b) A rod Q has a resistance R at and a resistance 1.2 R at . Calculate the
temperature coefficient of resistivity of the material of the rod. Use as the
reference temperature.
© What importapt difference between the behaviours of wire P and rod Q can you
identify from your-results above?

Ask by Sherman Willis. in the United States

Mar 26,2025

Question

4.) (a) At what temperature would the resistance of a wire P made of material with a
temperature coeflicient of resistivity be one-half its resistance at ?
Use as the reference point in your calculation.
(b) A rod Q has a resistance R at and a resistance 1.2 R at . Calculate the
temperature coefficient of resistivity of the material of the rod. Use as the
reference temperature.
© What importapt difference between the behaviours of wire P and rod Q can you
identify from your-results above?

Ask by Sherman Willis. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

(a) For a material with a temperature coefficient of resistivity $$\alpha = 4.3 \times 10^{-3}~\mathrm{K}^{-1}$$ (using $$20^\circ\mathrm{C}$$ as the reference temperature), the usual formula for resistance as a function of temperature is

$$
R = R_{20}\bigl[1+\alpha(T-20)\bigr].
$$

We are told that at some temperature $$T$$ the resistance is half its value at $$20^\circ\mathrm{C}$$; that is,

$$
\frac{1}{2}R_{20} = R_{20}\bigl[1+\alpha(T-20)\bigr].
$$

Dividing both sides by $$R_{20}$$ (assuming $$R_{20}\ne0$$),

$$
\frac{1}{2} = 1+\alpha(T-20).
$$

Solving for $$T$$,

$$
\alpha(T-20) = \frac{1}{2} -1 = -\frac{1}{2},
$$
$$
T-20 = -\frac{1}{2\alpha}.
$$

Substitute the value of $$\alpha$$:

$$
T-20 = -\frac{1}{2(4.3 \times 10^{-3})} \approx -\frac{1}{0.0086} \approx -116.28.
$$
$$
T \approx 20 - 116.28 \approx -96.28^\circ\mathrm{C}.
$$

Thus, the resistance of wire P becomes half its value at about $$-96.3^\circ\mathrm{C}$$.

---

(b) For rod Q, let the resistance at the reference $$20^\circ\mathrm{C}$$ be $$R_{20}$$. Then the resistance at any temperature $$T$$ is given by

$$
R(T)=R_{20}\bigl[1+\alpha(T-20)\bigr].
$$

We are told that:
- At $$50^\circ\mathrm{C}$$ the resistance is $$R$$, so

$$
  R(50)=R_{20}\bigl[1+\alpha(50-20)\bigr]=R_{20}(1+30\alpha)= R.
  $$
  
- At $$0^\circ\mathrm{C}$$ the resistance is $$1.2 R$$, so

$$
  R(0)=R_{20}\bigl[1+\alpha(0-20)\bigr]=R_{20}(1-20\alpha)= 1.2R.
  $$

Dividing the second equation by the first (to eliminate $$R_{20}$$),

$$
\frac{1-20\alpha}{1+30\alpha} = \frac{1.2R}{R} = 1.2.
$$

Now, solve for $$\alpha$$:

$$
1-20\alpha = 1.2(1+30\alpha).
$$

Expanding the right-hand side,

$$
1-20\alpha = 1.2 + 36\alpha.
$$

Bring the $$\alpha$$-terms together:

$$
1 - 1.2 = 36\alpha + 20\alpha,
$$
$$
-0.2 = 56\alpha,
$$
$$
\alpha = -\frac{0.2}{56} \approx -0.00357~\mathrm{K}^{-1}.
$$

Thus, the temperature coefficient of resistivity for rod Q is approximately $$-3.57 \times 10^{-3}~\mathrm{K}^{-1}$$.

---

(c) The results show an important difference in behavior:

- For wire P, the coefficient $$\alpha$$ is positive. This means that as the temperature increases above $$20^\circ\mathrm{C}$$, the resistance increases, and conversely, as the temperature decreases, the resistance decreases.

- For rod Q, the coefficient $$\alpha$$ is negative. This implies that the resistance increases as the temperature decreases (or decreases as the temperature increases).

This contrasting behavior indicates that the two materials have different mechanisms for conduction. Wire P behaves as most metals do (where resistance increases with temperature), while rod Q behaves like a material with a negative temperature coefficient (often seen in semiconductors or other non-metallic materials).
(a) The resistance of wire P becomes half its value at approximately $$-96.3^\circ\mathrm{C}$$. (b) The temperature coefficient of resistivity for rod Q is approximately $$-3.57 \times 10^{-3}~\mathrm{K}^{-1}$$. (c) Wire P has a positive temperature coefficient, meaning its resistance increases with temperature, while rod Q has a negative temperature coefficient, causing its resistance to decrease with temperature.

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