You have been given two samples of wire: one nichrome, one aluminum. Describe a procedure you could use to determine which sample was nichrome and which was aluminum
If the wires are the same diameter and the same length then the only difference is the resistivity or ?ρ? of each since: ??Resistance = ρ???L ? A = 4???L ? (π???d)? ??????where??L = wire length ??and???A = cross-sectional area of wire ??or use??d = wire diameter Since the ρ_nichrome is much higher than the ρ_aluminum and given the two wires are the same diameter and length then if you connect the two wires in series across a battery, and measure the voltage across each wire, the wire with the higher voltage will be the nichrome wire because the voltage across the nichrome wire is:?V? = i???(R?) and the voltage across the aluminum wire is: ?Va = i???(Ra)? and the current i is the same through both wires and R? Ra ... or if you can just measure the resistance of each wire, ... the nichrome wire will have a higher value.
first, knowing the resistance of each wire (i dont know it, you need to look it up) i would set up a circuit to put out [X] amt of current (the amt of the lowest resistance in the wires) and run the current thru the wires to a light bulb. if the bulb doesnt light, you now have figured out that this is the wire with the most resistance (again i dont know which has a higher one) for clarification- aluminum resistance = X nichrome resistance = y circuit with current EQUAL to resistance of the lowest wire (for example if X was lowest with 3 ohms, you would set the current to however many amps/volts are required to overcome that) and then connect it to a bulb. when its turned on, if the bulb doesnt light, you have determined that it isnt wire X but wire Y. even more simply, just list the wires by their resistance. if X=3 ohms, Y= 4 ohms, and the bulb doesnt light at 3, you have determined that its wire Y. im assuimng you would have the resistance of the wires? this is how i would do it, but if thats the right way? i dunno. but it is one way.
