Suppose a 76.9 V battery is connected to a long cylindrical wire, and a current

Question

Suppose a 76.9 V battery is connected to a long cylindrical wire, and a current of 5.92 A flows.

a) Find the number of electrons that pass any given point in the wire every second.
______ electrons.

b) Find the total resistance of the wire.
RTOT = ______

c) If the wire is 93.9 m long, and its diameter is 0.0768 cm, find the resistivity of the wire.
= ________ -m

d) Find the work done by this battery to push one electron through the entire length of the wire.
W = ________ J

e) You will have three chances for this question.
A cylindrical wire of length L has resistance R = 891 . The wire is now passed through a die so it is drawn out to length 4.07L. What is the new resistance of the wire?
HINT: The length is not the only quantity that has changed!
R = ________

Explanation / Answer

a) Current = 5.92 Coulombs per second

5.92C/s / 1.6*10–19 electrons / sec

3.7*1019 electrons per coulomb

b) R = V/A = 76.9 / 5.92 = 12.98

c) R . (Area) / L = resistivity

resistivity = R*r2 / L

r = diametre / 2 = .0.0768*10-2m / 2 = 3.84*10-4m

resistivity = 12.98 * *(3.84*10-4m)2 / 93.9m

resistivity = 6.40*10-8  m

d) 1 electron = 1.6*10–19 C

Work done = Volts * charge = 76.9*1.6*10–19 = 1.23*10–17 J

e)

Volume is a product of length L and cross sectional area A and will remain constant
V= A l=const.
V1=V2
A1L1= A2L2
A2= A1(L1/L2)
L1=L and L2 = 4.07L
A2= A1( L/4.07L)= A1 / 4.07
A2 =  A1 / 4.07
R= p L/A where p is specific material resistance
R1= p (L1/A1)
R2= p(L2/A2)
R2=p(4.07L/ A1 / 4.07)
R2= 16.56 p(L/A1)
R2= 16.56 R1 or

R2= 14759.32 ohms

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