Problem 18.87 Constants Part A Copper wire of diameter 0.269 cm is used to conne

Question

Problem 18.87 Constants Part A Copper wire of diameter 0.269 cm is used to connect a set of appliances at 120 V, which draw 1250 W of power total. The resistivity of copper is 1.68 × 10-8 ·m. What power is wasted in 29.0 m of this wire? Express your answer to three significant figures and include the appropriate units 0] ? Passipated=1 Value | Units Submit Request Answer Part B What is your answer if wire of diameter 0.417 cm is used? Express your answer to three significant figures and include the appropriate units ) ? pated=1 Value ! Units Submit Request Ans Provide Feedback

Explanation / Answer

Part A:

First we determine the current drawn by the source, I.

Write the expression -

P = V*I

=> 1250 = 120*I

=> I = 1250 / 120 = 10.42 A

Now find out the resistance of the copper wire.

cross-section area of the wire A = pi*r^2 = 3.141*(0.00269 / 2)^2 = 5.68 x 10^-6 m^2

write the expression for resistance -

R = k*L / A = (1.68 x 10^-8 x 29) / (5.68 x 10^-6) = 0.0858 ohm

Therefore, power dissipated in the wire, P = I^2 x R = 10.42^2 x 0.0858 = 9.32 W

Part B -

If the diameter of the wire is 0.417 cm

then cross-sectional area A = pi*r^2 = 3.141*(0.00417 / 2)^2 = 13.65 x 10^-6 m^2

so,

R = k*L / A = (1.68 x 10^-8 x 29) / (13.65 x 10^-6) = 0.0357 ohm

Therefore, power dissipated in the wire, P = I^2 x R = 10.42^2 x 0.0357 = 3.88 W

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