Question

A 14.0-m length of wire consists of 9.0 m of copper followed by 5.0 m of aluminum, both of diameter 1.3 mm . A voltage difference of 95 mVis placed across the composite wire. The resistivity of copper is 1.68×10−8Ω⋅m and the resistivity of aluminum is 2.65×10−8Ω⋅m.

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

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

2. What is the current through the wire?

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

3. What is the voltage across the copper part?

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

4. What is the voltage across the aluminum part?

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

Answer 1

Solution:-We can find this as below

=> total Resistance = resistance copper + resistance aluminium
So using the formula

(ρ1 l1 + ρ2 l2)/A

ρ1 (copper)= 1.68×10−8   1

l1 = 9m

ρ2(Al) = 2.82×10−8

l2 = 5 m

A = pi * .0013^2 /4 = 1.32 * 10^-6 m^2

1)total resistance = 0.114 + 0.106 = 0.22 ohm

2)current through the wire= .060 / 0.22 = 0.272 A

3) voltage across copper part = 0.272 * 0.114 = 0.031 V

4) voltage across aluminium part = 0.272 * 0.106 = 0.029 V