I = dQ/dt -= nqv_d A J = 1/A = nqv_d rho = E/J rho = rho_0 [1 + alpha (T - T_0)]
ID: 1628866 • Letter: I
Question
I = dQ/dt -= nqv_d A J = 1/A = nqv_d rho = E/J rho = rho_0 [1 + alpha (T - T_0)] R = rho L/A V = IR E = V/L (Conductor Length L): R = R_0 [1 + alpha (T - T_0]] Household wiring uses copper wire with a 1.5 times 10^-5 m^2 cross sectional area, (a) Find the resistance of a 1610 m length of this wire at 20 degree C. The resistivity of copper wire at 20 degree C is 1.72 times 10^-8 Ohm middot m. (b) Find the resistance of this wire if its temperature is raised to 150 degree C. The temperature coefficient of resistance of copper is 0.00393 (degree C)^-1. A 3.5 A current flows through a typical automobile headlight. How many coulombs of charge will flow through the headlight during a nonstop 2.5 hour nighttime drive? (3600 s = 1 hour) A steady current of 48 A flows through a long aluminum conductor with cross sectional area 1.6 times 10^-3 m^2. The resistivity of aluminum is 2.75 times 10^-8 Ohm middot m. For a 50 m section of this conductor find: (a) current density, (b) resistance per meter (c) electric field within the conductor, and (d) potential drop across this section.Explanation / Answer
1(a)
area A=1.5*10^-5 m^2
length L=1610 m
resistivity p=1.72*10^-8 ohm-m
now we find the resistance
resistance R=pL/A=1.72*10^-8*1610/1.5*10^-5=1.846 ohms
(b)
the resistance R=1.846[1+0.00393(150-20)]=2.79 ohms
2).
I = dQ/dT
dQ = I*dT
Q = I*T
Q = (3.5)*(2.5)*(3600)
Q = 3.15*10^4 C
3).
current I=48 A
area of cross sectional A=1.6*10^-3 m^2
length L=50 m
resistivity p=2.75*10^-8 ohm-m
(a) current density J=i/A=48/1.6*10^-3=3*10^4 A/m^2
(b) resistance R=pL/A=2.75*10^-8*50/1.6*10^-3=8.6*10^-4 ohms
(c) electric field E=pJ=2.75*10^-8*3*10^4=8.3*10^-4 N/m
(d) the potential differeence V=iR=48*8.6*10^-4=0.04128 V
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