A 14.0-m length of wire consists of 9.0 m of copper followed by 5.0 m of aluminu

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 1.68×108m and the resistivity of aluminum is 2.65×108m.

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.

Explanation / Answer

Solution:-We can find this as below

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

(1 l1 + 2 l2)/A

1 (copper)= 1.68×108   1

l1 = 9m

2(Al) = 2.82×108

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

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