1 Calculate the resistance of an aluminum cylinder that has

1. Calculate the resistance of an aluminum cylinder that has a length of 10.0 cm and a cross-sectional area of 2.00e-4 m^2. Repeat the calculation for a cylinder of the same dimensions and made of glass having a resistivity of 3.0e10 ohm m.(Answers below.)

resistance of aluminum cylinder = 0.0000141 ohm

resistance of glass = 1.5e13.

Using this same problem, what if the same amount of aluminum is used to make a cylinder with 4.0 times the original length? What will be the resistance of the resulting cylinder?(in units of u ohm)?

---------- u ohm

2. Calculate the ressitance per unit length of a thin 22-gauge Nichrome wire which has a radius of 0.375 mm. If a potential difference of 15V is maintained across a 1.0 m length of the Nichrome wire, what is the current of the wire?(Answers below)

cross-sectional area of the wire = 0.000000442 m^2

resistance per unit length = 3.39 ohm/m.

I=4.424

Using the same problem, calculate the electric field in the wire assuming that the wire carries a current of 7.9 A.

E= --------- N/C

Solution

1. Resistance = resistivity x length / cross section area

R = rho L / A

for aluminium, rho = 2.82 x 10^-8 ohm m

R = (2.82 x 10^-8 x 0.10 ) / (2 x 10^-4 ) = 1.41 x 10^-5 ohm

For glass:

rho = 3. x 10^10 ohm m

R = (3. x 10^10 x 0.10 ) / (2 x 10^-4 ) = 1.5 x 10^13 ohm

volume = AL this will be same as same amount is used.

L\' = 4L

volume = A\'L\' = A L

A\' (4L) = AL

A\' = A/4

R\' = rho L\' / A\' = rho 4L / (A/4) = 16 rho L / A

and rho L / A = 1.41 x 10^-5 ohm

R\' = 16 x 1.41 x 10^-5 = 2.256 x 10^-4 ohm = 225.6 u ohm