\text{Chapter 26}

\rho-\rho_0=\rho_0\alpha (T-T_0)

R-R_0=R_0\alpha (T-T_0)

\text{Resistivity}

\text{Resistance}

\vec{J}=(ne)\vec{v}_d

\text{Drift speed and current density}

\vec{E}=\rho \vec{J}

R=\rho\frac{l}{A}

V=R i

P=iV=\frac{V^2}{R}=Ri^2

i=\int\vec{J}\cdot d{A}

\text{A cylindrical resistor of radius $2.5 \mathrm{~mm}$ and length $4.0 \mathrm{~cm}$ is made of a material that has }

\text{a resistivity of $3.5 \times 10^{-5} \Omega$. $\mathrm{m}$. }

\text{What is the potential difference when the energy dissipation rate in the resistor is $1.0 W$?}

\text{A cylindrical resistor of radius $2.5 \mathrm{~mm}$ and length $4.0 \mathrm{~cm}$ is made of a material that has }

\text{a resistivity of $3.5 \times 10^{-5} \Omega$. $\mathrm{m}$. }

\text{What is the potential difference when the energy dissipation rate in the resistor is $1.0 W$?}

P=\frac{V^2}{R}

\Rightarrow

V=\sqrt{R P}=\sqrt{ \rho \frac{l}{A} P}

=\sqrt{ 3.5 \times 10^{-5} \frac{0.04}{\pi 2.5^2 \times 10^{-6}} \times 1}=0.27V

\text{A $1.0~\mathrm{m}$-long wire has a resistance equal to $0.30 ~\Omega$. A second wire made of identical material }

\text{has a length of $2.0 \mathrm{~m}$ and a mass equal to the mass of the first wire.}

\text{What is the resistance of the second wire?}

\text{A $1.0~\mathrm{m}$-long wire has a resistance equal to $0.30 ~\Omega$. A second wire made of identical material }

\text{has a length of $2.0 \mathrm{~m}$ and a mass equal to the mass of the first wire.}

\text{What is the resistance of the second wire?}

\text{same mass and density $\Rightarrow$ same volume}

\begin{aligned} & V_1=A_1 l_1=A_2 l_2 \Rightarrow A_2=A_1 \frac{l_1}{l_2}=\frac{A_1}{2} \\ & R_2=\rho \frac{l_2}{A_2}=\rho \frac{2 l_1}{\frac{A_1}{2}}=4 \rho \frac{l_1}{A_1}=4 R_1=1.2 \Omega \end{aligned}

\text{The resistance of a wire at $0^{\circ} \mathrm{C}$ is $70.0 ~\Omega$. If temperature of the wire increases to $100^{\circ} \mathrm{C}$, its }

\text{resistance increases by $50 \%$. What is resistance of the wire at $120^{\circ} \mathrm{C}$ ? }

\text{(Ignore changes in the dimensions of the wire)}

\text{The resistance of a wire at $0^{\circ} \mathrm{C}$ is $70.0 ~\Omega$. If temperature of the wire increases to $100^{\circ} \mathrm{C}$, its }

\text{resistance increases by $50 \%$. What is resistance of the wire at $120^{\circ} \mathrm{C}$ ? }

\text{(Ignore changes in the dimensions of the wire)}

R-R_0=\alpha R_0\left(T-T_0\right)

\frac{R_{100}-R_0}{R_0}=0.5=\alpha\left(T_{100}-T_0\right)

\frac{R_{120}-R_0}{R_0}=\alpha\left(T_{120}-T_0\right)

\frac{\frac{R_{120}-R_0}{R_0}}{0.5}=\frac{T_{120}-T_0}{T_{100}-T_0}

\begin{aligned} R_{120} & =R_0\left(1+0.5 \frac{T_{120}-T_0}{T_{100}-T_0}\right) \\ & =70\left(1+0.5 \frac{120}{100}\right)=70(1.6)=112 \Omega \end{aligned}

\text{A copper wire of cross-sectional area $2.00 \times 10^{-6} \mathrm{~m}^2$ and length $4.00 \mathrm{~m}$ has a current of $2.00 \mathrm{~A}$ }

\text{uniformly distributed across its area. How much electrical energy is transferred into thermal }

\text{energy in 1.00 hour (resistivity of copper $=1.69 \times 10^{-8} \Omega . \mathrm{m}$ )}

\text{A copper wire of cross-sectional area $2.00 \times 10^{-6} \mathrm{~m}^2$ and length $4.00 \mathrm{~m}$ has a current of $2.00 \mathrm{~A}$ }

\text{uniformly distributed across its area. How much electrical energy is transferred into thermal }

\text{energy in 1.00 hour (resistivity of copper $=1.69 \times 10^{-8} \Omega . \mathrm{m}$ )}

\text { Power }=\frac{\text { Electrical energy }}{\text { time }}

\begin{aligned} \text { Electrical energy } & =(\text { Power })(\text { time })=\left(i^2 R\right) t=i^2 \frac{\rho l}{A} t \\ & =(2)^2 \frac{1.69 \times 10^{-8} \times 4}{2 \times 10^{-6}} \times 60 \times 60=487 \mathrm{~J} \end{aligned}

\text{A light bulb, has a resistance of $15 ~\Omega$, is connected between the terminals of a $120 \mathrm{~V}$ source.}

\text{If the temperature is not ignored, which one of the following answers can be the expected}

\text{output power of the bulb?}

\text{A light bulb, has a resistance of $15 ~\Omega$, is connected between the terminals of a $120 \mathrm{~V}$ source.}

\text{If the temperature is not ignored, which one of the following answers can be the expected}

\text{output power of the bulb?}

\text{Power}=\frac{V^2}{R}=\frac{120^2}{15^2 }=960 W

\text{When the light bulb is on, its temperature becomes larger, and its resistance increases }

R'>R \Rightarrow \text{Power decreases}

\text{The only possible choice is therefore $P'=840 W$}

\text{At what temperature will aluminum have a resistivity that is three times the resistivity that }

\text{of copper has at $20^{\circ} \mathrm{C}$ ? At $20^{\circ} \mathrm{C}$, the resistivity of aluminum is $2.75 \times 10^{-8} \Omega$.$\mathrm{m}$ and the resistivity}

\text{ of copper is $1.69 \times 10^{-8} \Omega \cdot \mathrm{m}$. The temperature coefficient of aluminum $\alpha_{\mathrm{Al}}=4.4 \times 10^{-3} \mathrm{~K}^{-1}$.}

\text{At what temperature will aluminum have a resistivity that is three times the resistivity that }

\text{of copper has at $20^{\circ} \mathrm{C}$ ? At $20^{\circ} \mathrm{C}$, the resistivity of aluminum is $2.75 \times 10^{-8} \Omega$.$\mathrm{m}$ and the resistivity}

\text{ of copper is $1.69 \times 10^{-8} \Omega \cdot \mathrm{m}$. The temperature coefficient of aluminum $\alpha_{\mathrm{Al}}=4.4 \times 10^{-3} \mathrm{~K}^{-1}$.}

\rho_A-\rho_{A_0}=\rho_{A_0} \alpha\left(T-T_0\right)

3\rho_{Co}-\rho_{A_0}=\rho_{A_0} \alpha\left(T-T_0\right)

T=T_0+\frac{3 \rho_{Co}-\rho_{A_0}}{\rho_{A_0} \alpha_A}=20+\frac{3\left(1.69 \times 10^{-8}\right)-2.75 \times 10^{-8}}{\left(2.75 \times 10^{-8}\right)\left(4.4 \times 10^{-3}\right)}=212^{\circ} \mathrm{C}

\text{At what $T$, $\rho_A=3\rho_{A}$}?