Chapter 26
\text{Chapter 26}
Chapter 26
\text{Chapter 26}
Chapter 26
\text{Chapter 26}
Chapter 26
\text{Chapter 26}
ρ−ρ0=ρ0α(T−T0)
\rho-\rho_0=\rho_0\alpha (T-T_0)
R−R0=R0α(T−T0)
R-R_0=R_0\alpha (T-T_0)
Resistivity
\text{Resistivity}
Resistance
\text{Resistance}
J=(ne)vd
\vec{J}=(ne)\vec{v}_d
Drift speed and current density
\text{Drift speed and current density}
E=ρJ
\vec{E}=\rho \vec{J}
R=ρAl
R=\rho\frac{l}{A}
V=Ri
V=R i
P=iV=RV2=Ri2
P=iV=\frac{V^2}{R}=Ri^2
i=∫J⋅dA
i=\int\vec{J}\cdot d{A}
A cylindrical resistor of radius 2.5 mm and length 4.0 cm is made of a material that has
\text{A cylindrical resistor of radius $2.5 \mathrm{~mm}$ and length $4.0 \mathrm{~cm}$ is made of a material that has }
a resistivity of 3.5×10−5Ω. m.
\text{a resistivity of $3.5 \times 10^{-5} \Omega$. $\mathrm{m}$. }
What is the potential difference when the energy dissipation rate in the resistor is 1.0W?
\text{What is the potential difference when the energy dissipation rate in the resistor is $1.0 W$?}
A cylindrical resistor of radius 2.5 mm and length 4.0 cm is made of a material that has
\text{A cylindrical resistor of radius $2.5 \mathrm{~mm}$ and length $4.0 \mathrm{~cm}$ is made of a material that has }
a resistivity of 3.5×10−5Ω. m.
\text{a resistivity of $3.5 \times 10^{-5} \Omega$. $\mathrm{m}$. }
What is the potential difference when the energy dissipation rate in the resistor is 1.0W?
\text{What is the potential difference when the energy dissipation rate in the resistor is $1.0 W$?}
P=RV2
P=\frac{V^2}{R}
⇒
\Rightarrow
V=RP=ρAlP
V=\sqrt{R P}=\sqrt{ \rho \frac{l}{A} P}
=3.5×10−5π2.52×10−60.04×1=0.27V
=\sqrt{ 3.5 \times 10^{-5} \frac{0.04}{\pi 2.5^2 \times 10^{-6}} \times 1}=0.27V
A 1.0m-long wire has a resistance equal to 0.30Ω. A second wire made of identical material
\text{A $1.0\mathrm{m}$-long wire has a resistance equal to $0.30 \Omega$. A second wire made of identical material }
has a length of 2.0 m and a mass equal to the mass of the first wire.
\text{has a length of $2.0 \mathrm{~m}$ and a mass equal to the mass of the first wire.}
What is the resistance of the second wire?
\text{What is the resistance of the second wire?}
A 1.0m-long wire has a resistance equal to 0.30Ω. A second wire made of identical material
\text{A $1.0\mathrm{m}$-long wire has a resistance equal to $0.30 \Omega$. A second wire made of identical material }
has a length of 2.0 m and a mass equal to the mass of the first wire.
\text{has a length of $2.0 \mathrm{~m}$ and a mass equal to the mass of the first wire.}
What is the resistance of the second wire?
\text{What is the resistance of the second wire?}
same mass and density ⇒ same volume
\text{same mass and density $\Rightarrow$ same volume}
V1=A1l1=A2l2⇒A2=A1l2l1=2A1R2=ρA2l2=ρ2A12l1=4ρA1l1=4R1=1.2Ω
\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}
The resistance of a wire at 0∘C is 70.0Ω. If temperature of the wire increases to 100∘C, its
\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 }
resistance increases by 50%. What is resistance of the wire at 120∘C ?
\text{resistance increases by $50 \%$. What is resistance of the wire at $120^{\circ} \mathrm{C}$ ? }
(Ignore changes in the dimensions of the wire)
\text{(Ignore changes in the dimensions of the wire)}
The resistance of a wire at 0∘C is 70.0Ω. If temperature of the wire increases to 100∘C, its
\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 }
resistance increases by 50%. What is resistance of the wire at 120∘C ?
\text{resistance increases by $50 \%$. What is resistance of the wire at $120^{\circ} \mathrm{C}$ ? }
(Ignore changes in the dimensions of the wire)
\text{(Ignore changes in the dimensions of the wire)}
R−R0=αR0(T−T0)
R-R_0=\alpha R_0\left(T-T_0\right)
R0R100−R0=0.5=α(T100−T0)
\frac{R_{100}-R_0}{R_0}=0.5=\alpha\left(T_{100}-T_0\right)
R0R120−R0=α(T120−T0)
\frac{R_{120}-R_0}{R_0}=\alpha\left(T_{120}-T_0\right)
0.5R0R120−R0=T100−T0T120−T0
\frac{\frac{R_{120}-R_0}{R_0}}{0.5}=\frac{T_{120}-T_0}{T_{100}-T_0}
R120=R0(1+0.5T100−T0T120−T0)=70(1+0.5100120)=70(1.6)=112Ω
\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}
A copper wire of cross-sectional area 2.00×10−6 m2 and length 4.00 m has a current of 2.00 A
\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 $2.00 \mathrm{~A}$ }
uniformly distributed across its area. How much electrical energy is transferred into thermal
\text{uniformly distributed across its area. How much electrical energy is transferred into thermal }
energy in 1.00 hour (resistivity of copper =1.69×10−8Ω.m )
\text{energy in 1.00 hour (resistivity of copper $=1.69 \times 10^{-8} \Omega . \mathrm{m}$ )}
A copper wire of cross-sectional area 2.00×10−6 m2 and length 4.00 m has a current of 2.00 A
\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 $2.00 \mathrm{~A}$ }
uniformly distributed across its area. How much electrical energy is transferred into thermal
\text{uniformly distributed across its area. How much electrical energy is transferred into thermal }
energy in 1.00 hour (resistivity of copper =1.69×10−8Ω.m )
\text{energy in 1.00 hour (resistivity of copper $=1.69 \times 10^{-8} \Omega . \mathrm{m}$ )}
Power = time Electrical energy
\text { Power }=\frac{\text { Electrical energy }}{\text { time }}
Electrical energy =( Power )( time )=(i2R)t=i2Aρlt=(2)22×10−61.69×10−8×4×60×60=487 J
\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}
A light bulb, has a resistance of 15Ω, is connected between the terminals of a 120 V source.
\text{A light bulb, has a resistance of $15 \Omega$, is connected between the terminals of a $120 \mathrm{~V}$ source.}
If the temperature is not ignored, which one of the following answers can be the expected
\text{If the temperature is not ignored, which one of the following answers can be the expected}
output power of the bulb?
\text{output power of the bulb?}
A light bulb, has a resistance of 15Ω, is connected between the terminals of a 120 V source.
\text{A light bulb, has a resistance of $15 \Omega$, is connected between the terminals of a $120 \mathrm{~V}$ source.}
If the temperature is not ignored, which one of the following answers can be the expected
\text{If the temperature is not ignored, which one of the following answers can be the expected}
output power of the bulb?
\text{output power of the bulb?}
Power=RV2=1521202=960W
\text{Power}=\frac{V^2}{R}=\frac{120^2}{15^2 }=960 W
When the light bulb is on, its temperature becomes larger, and its resistance increases
\text{When the light bulb is on, its temperature becomes larger, and its resistance increases }
R′>R⇒Power decreases
R'>R \Rightarrow \text{Power decreases}
The only possible choice is therefore P′=840W
\text{The only possible choice is therefore $P'=840 W$}
At what temperature will aluminum have a resistivity that is three times the resistivity that
\text{At what temperature will aluminum have a resistivity that is three times the resistivity that }
of copper has at 20∘C ? At 20∘C, the resistivity of aluminum is 2.75×10−8Ω.m and the resistivity
\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}
of copper is 1.69×10−8Ω⋅m. The temperature coefficient of aluminum αAl=4.4×10−3 K−1.
\text{
of copper is $1.69 \times 10^{-8} \Omega \cdot \mathrm{m}$. The temperature coefficient of aluminum $\alpha_{\mathrm{Al}}=4.4 \times 10^{-3} \mathrm{~K}^{-1}$.}
At what temperature will aluminum have a resistivity that is three times the resistivity that
\text{At what temperature will aluminum have a resistivity that is three times the resistivity that }
of copper has at 20∘C ? At 20∘C, the resistivity of aluminum is 2.75×10−8Ω.m and the resistivity
\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}
of copper is 1.69×10−8Ω⋅m. The temperature coefficient of aluminum αAl=4.4×10−3 K−1.
\text{
of copper is $1.69 \times 10^{-8} \Omega \cdot \mathrm{m}$. The temperature coefficient of aluminum $\alpha_{\mathrm{Al}}=4.4 \times 10^{-3} \mathrm{~K}^{-1}$.}
ρA−ρA0=ρA0α(T−T0)
\rho_A-\rho_{A_0}=\rho_{A_0} \alpha\left(T-T_0\right)
3ρCo−ρA0=ρA0α(T−T0)
3\rho_{Co}-\rho_{A_0}=\rho_{A_0} \alpha\left(T-T_0\right)
T=T0+ρA0αA3ρCo−ρA0=20+(2.75×10−8)(4.4×10−3)3(1.69×10−8)−2.75×10−8=212∘C
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}
At what T, ρA=3ρA?
\text{At what $T$, $\rho_A=3\rho_{A}$}?
C h a p t e r 2 6 \text{Chapter 26} C h a p t e r 2 6 \text{Chapter 26} C h a p t e r 2 6 \text{Chapter 26} C h a p t e r 2 6 \text{Chapter 26}
Chapter 26
By smstry
Chapter 26
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