Question

A copper wire of cross-sectional area of 8.2x10^-7 m^2 has a resistance of 1.11 Ω.

a). How long is the wire?

b). If a potential of 12.0 V is applied across the ends of the wire, what current will flow through the wire?

c). How much electrical power is supplied to the spool?

d). How many electrons (N) pass through this wire in 1.00 minutes?

Answer 1

Part A.

R = rho*L/A

rho = resistivity of copper wire = 1.72*10^-8 ohm-m (Check this value in your reference book)

L = R*A/rho

L = 1.11*8.2*10^-7/(1.72*10^-8)

L = 52.9 m

Part B.

Using ohm's law:

I = V/R = 12.0/1.11

I = 10.8 Amp

Part C.

Power = V^2/R

P = 12.0^2/1.11

P = 129.7 W

Part D.

I = Q/t = N*e/t

N = I*t/e

t = 1 min = 60 sec

N = 10.8*60/(1.6*10^-19)

N = 4.05*10^21 electrons

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