An experiment is conducted to measure the electrical resistivity of Nichrome in

Question

An experiment is conducted to measure the electrical resistivity of Nichrome in the form of wires with different lengths and cross-sectional areas. For one set of measurements, a student uses 30 gauge wire, which has a cross-sectional area of 7.30 ? 10?8 m2. The student measures the potential difference across the wire and the current in the wire with a voltmeter and an ammeter, respectively. For each of the measurements given in the following table taken on wires of three different lengths, calculate the resistance of the wires and the corresponding value of the resistivity

L: .54, 1.028, 1.543

V: 5.19, 5.83, 5.95

I: .503, .275, .19

R: 10.3, 21.2, 31.3

resistivity: ?, ?, ?

average resistivity: ?

what percentage it differes from (1.5x10^(-6): ?

Explanation / Answer

apply the relation between resistance R and resistivity rho as

R = rho L/A

where L is length and A is area

also from ohms law as V = iR

where i is current and R is resistance

when I = 0.503 and V = 5.19

R = V/i    = 5.19/0.503 = 10.31 ohms

resisitivity here Rho = RA/L

Rho = 10.31 * 7.3 e -8/0.54

Rho = 1.4 e -6 ohm m

diff % = 1.5-1.4/1.5   = 6.67 %

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when i = 0.275 A

V = 5.83 Volts

R = V/i = 5.83/0.275 = 21.2 ohms

resisitivity here Rho = RA/L

rho = 21.2 * 7.3 e-8/(1.028)

rho = 1.505 e -6 ohm m

% diff = 1.505 -1.5/1.5 = 0.33 %

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when i = 0.19 A

V = 5.95 Volts

R = V/i   = 5.95/0.19 = 31.31 ohms

resisitivity Rho = RA/L

Rho = 31.31 * 7.3 e -8/1.543)

Rho = 1.505 e - ohm m

% diff = 1.505 -1.5/1.5 = 0.33 %

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