A 11.0-m length of wire consists of 6.0 m of copper followed by 5.0 m of aluminu

Question

A 11.0-m length of wire consists of 6.0 m of copper followed by 5.0 m of aluminum, both of diameter 1.3 mm . A voltage difference of 85 mV is placed across the composite wire. The resistivity of copper is 1.68×108m and the resistivity of aluminum is 2.65×108m.

Part A

What is the total resistance (sum) of the two wires?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the current through the wire?

Express your answer to two significant figures and include the appropriate units.

Part C

What is the voltage across the copper part?

Express your answer to two significant figures and include the appropriate units.

Part D

What is the voltage across the aluminum part?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Resistane = resisitivty * L /A

wire are connected in series.
then Req = R1 + R2

R = [ (1.68 x 10^-8 x 6 ) / ( pi (1.3 x 10^-3 / 2)^2 ) ] + [ (2.65 x 10^-8 x 5 ) / ( pi (1.3 x 10^-3 / 2)^2 ) ]

= 0.076 + 0.10 = 0.176 ohm

b) I = V / R = (85 x 10^-3) / (0.176) = 0.483 A

c) V1 = I R1 = 0.483 x 0.076 = 0.0367 V Or 36.7 mV

d) V2 = V - V1 = 85 - 36.7 = 48.3 m V

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