A copper wire of length L and cross sectional area A is connected to a 12.0 volt

Question

A copper wire of length L and cross sectional area A is connected to a 12.0 volt DC power supply. The current through the wire is measured at 40A. A platinum wire of the same length and and cross sectional area as the first wire is attached to the first wire and the negative terminal of the battery is connected to the free end of the copper wire, while the positive teriminal is connected to the free end of the platinum wire. The current through the wires is then measured. (a)What is the current through both wires? A (b)The two wires are now separated. The left end of each wire is connected to the negative side of the battery, while the right sides are connected to the positive side of the battery. What is the current supplied by the battery? A

Explanation / Answer

y = resistivity

for copper wire

r = V/i = 12/40 = 0.3 ohms = yl/a

y copper = 1.68 * 10^-8 ohm-m

for platinum wire y = 10.6 * 10^-8 ohm -m = 6.31 times y of copper

so resistance of platiunum = 6.31 * 0.3 = 1.893 ohms

a) in series

total resistance = 2.193

i = v/r = 5.47 amp

b) in parallel

resistance = 0.26 ohms

i = v/r = 46.15 amp

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