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What is the resistance of the cube?
the resistivity of Gold is 2.20 × 10-8 Ω·m.
I have a cube of Gold exactly 1cm by 1cm. I apply a current from a power supply of exactly 1 Amp to two diagonal corners. I apply a voltmeter to the two remaining diagonal corners. What Voltage do I see?
3 Answers
- Robert TLv 41 decade agoFavorite Answer
resistivity is 2.2 x 10^-6 ohm-cm.
If I model the cube using 12 resistors, one resistor for each edge, each resistor value would be 4 x (2.2 x 10^-6). This is the case, since if you drive the 4 nets of one face, and ground the 4 nets of the opposite face, the resistance must be 2.2 x 10^-6 ohms, since it is a cube with dimensions of 1cm.
But you are driving the diagonal corners of the cube, so the current of 1A splits 3 ways due to symmetry. Each resistor connected to the current source has 1/3 amp, and the other resistors have 1/6th amp, again, due to symmetry.
The voltage drop across the other diagonal corners is (2/3)x 2.2x10^-6 volts.
Source(s): Verified using above model simulated using Linear Technology's Switcher CAD III. My model used resistors with values of 4 ohms each, which would correspond to a resistivity of 1. Doing this, I got (2/3) volt. Since the voltage scales with resistivity, your answer should be 2/3 x resistivity of Gold. - Anonymous1 decade ago
0.0311126983 x 10^-8 volt?
Just a guess. LOL
1 cm = .01 m and the hypotenuse should be 0.0141421356 m long. E = IR = (1)(2.20 x 10^-8 x 0.0141421356)
Took a good stab at it at least. Should probably have drawn it out on paper.
Ron
- 1 decade ago
if you look at the volt meter it will tell you how many volts
you said you had all that stuff.