If the carrier density of Copper is 8.5 times 10^28 m^-3 then what would be the

Question

If the carrier density of Copper is 8.5 times 10^28 m^-3 then what would be the electron drift speed in the copper wire? The copper wire is stretched so that its new length is 1.6 times its original length. What would be its new resistance? The resistivity of copper is 1.7 times 10^ -8 ohm.m. What is the resistance of a 1.4 m -long copper wire that 0.40mm in diameter? if this copper wire is connected in series with a 2 Ohm resistor and a 3 V battery, how much current will flow through the copper wire? How much power is dissipated by the 2 Ohm resistor?

Explanation / Answer

1) given

rho = 1.7*10^8 ohm.m

a) R = rho*L/A

= rho*L/(pi*d^2/4)

= 1.7*10^-8*1.4/(pi*0.0004^2/4)

= 0.1894 ohms

b) I = V/Rnet

= 3/(2+0.1894)

= 1.37 A

c) P(2) = 1.37^2*2

= 3.76 Watts

d) Vd = I/(A*n*q)

= I/(pi*d^2/4*n*q)

= 1.37/(pi*0.0004^2/4*8.5*10^28*1.6*10^-19)

= 8.02*10^-4 m/s or 0.802 mm/s

e) new resistance = n^2*R

= 1.6^2*0.1894

= 0.4848 ohms

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