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 importaat difference between the behaviours of wire P and rod can you
identify from your-results above?

Ask by Gibbs Ortiz. 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 importaat difference between the behaviours of wire P and rod can you
identify from your-results above?

Ask by Gibbs Ortiz. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

**(a) Calculation for wire P**

We use the linear approximation for resistance as a function of temperature with reference temperature $$20^\circ\mathrm{C}$$:  
$$
R = R_{20}\bigl[1+\alpha (T-20)\bigr],
$$
where  
$$
\alpha = 4.3\times10^{-3}\, \mathrm{K}^{-1}.
$$

We are told that at some temperature $$T$$ the resistance is half of that at $$20^\circ\mathrm{C}$$, i.e.,  
$$
R = \frac{1}{2}R_{20}.
$$

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

Dividing both sides by $$R_{20}$$ (assuming $$R_{20}\ne0$$):  
$$
\frac{1}{2} = 1+\alpha (T-20).
$$

Subtract 1 from both sides:  
$$
\alpha (T-20) = -\frac{1}{2}.
$$

Now solve for $$T-20$$:  
$$
T-20 = -\frac{1}{2\alpha} = -\frac{1}{2(4.3\times10^{-3})}.
$$

Calculate the denominator:  
$$
2(4.3\times10^{-3}) = 8.6\times10^{-3}.
$$

Thus,  
$$
T-20 = -\frac{1}{8.6\times10^{-3}} \approx -116.28.
$$

Finally,  
$$
T \approx 20 -116.28 \approx -96.28^\circ\mathrm{C}.
$$

---

**(b) Calculation for rod Q**

Let the resistance at $$20^\circ\mathrm{C}$$ be $$R_{20}$$ and the temperature coefficient be $$\alpha$$. Then the resistance at any temperature $$T$$ is given by:  
$$
R(T)= R_{20}\bigl[1+\alpha (T-20)\bigr].
$$

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

We can express $$R_{20}$$ from the first equation:  
$$
R_{20} = \frac{R}{1+30\alpha}.
$$

Substitute this into the second equation:  
$$
\frac{R}{1+30\alpha}\bigl[1-20\alpha\bigr] = 1.2R.
$$

Cancel $$R$$ (assuming $$R\ne 0$$):  
$$
\frac{1-20\alpha}{1+30\alpha} = 1.2.
$$

Solve for $$\alpha$$:

Multiply both sides by $$(1+30\alpha)$$:  
$$
1-20\alpha = 1.2(1+30\alpha).
$$

Expand the right side:  
$$
1-20\alpha = 1.2 + 36\alpha.
$$

Bring like terms together:  
$$
1 - 1.2 = 36\alpha + 20\alpha,
$$
$$
-0.2 = 56\alpha.
$$

Thus,  
$$
\alpha = -\frac{0.2}{56} \approx -0.00357\, \mathrm{K}^{-1}.
$$

---

**(c) Discussion of the difference between wire P and rod Q**

For wire P, the material has a **positive** temperature coefficient ($$\alpha = 4.3\times10^{-3}\, \mathrm{K}^{-1}$$). This means its resistance increases with temperature. To reduce the resistance to half of its value at $$20^\circ\mathrm{C}$$, the required temperature drop is very large ($$\approx -96^\circ\mathrm{C}$$). Such a dramatic temperature change indicates that for typical temperature changes the resistance of wire P does not vary very much.

In contrast, rod Q exhibits a **negative** temperature coefficient ($$\alpha \approx -0.00357\, \mathrm{K}^{-1}$$). This means its resistance decreases with increasing temperature. Moreover, the resistance changes significantly: from $$1.2R$$ at $$0^\circ\mathrm{C}$$ to $$R$$ at $$50^\circ\mathrm{C}$$. Thus, the behavior of rod Q is opposite to that of wire P and shows a much stronger relative change over a relatively small temperature range.

**(a) Wire P** To find the temperature where the resistance is half of $$20^\circ\mathrm{C}$$, use the formula: $$ R = R_{20}\bigl[1 + \alpha (T - 20)\bigr] $$ Set $$R = \frac{1}{2}R_{20}$$: $$ \frac{1}{2} = 1 + 4.3 \times 10^{-3} (T - 20) $$ Solving for $$T$$: $$ T \approx -96.28^\circ\mathrm{C} $$ --- **(b) Rod Q** Given: - At $$50^\circ\mathrm{C}$$, resistance $$R$$. - At $$0^\circ\mathrm{C}$$, resistance $$1.2R$$. Using the resistance formula: $$ R(50) = R_{20}\bigl[1 + \alpha (50 - 20)\bigr] = R $$ $$ R(0) = R_{20}\bigl[1 + \alpha (0 - 20)\bigr] = 1.2R $$ Solving these equations gives the temperature coefficient $$\alpha \approx -0.00357\, \mathrm{K}^{-1}$$. --- **(c) Difference Between Wire P and Rod Q** - **Wire P** has a **positive** temperature coefficient, meaning its resistance increases with temperature. To reduce resistance to half, a significant temperature drop is needed ($$\approx -96^\circ\mathrm{C}$$), which is impractical for typical temperature changes. - **Rod Q** has a **negative** temperature coefficient, so its resistance decreases with increasing temperature. The resistance changes more significantly over a smaller temperature range compared to wire P. This shows that wire P's resistance changes minimally with temperature, while rod Q's resistance changes more dramatically over a smaller temperature interval.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Physics Questions