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 7.30 108 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.

What is the average value of the resistivity?
· m

By what percent does this differ from the value given in this table (1.50 106 · m)?
%

Explanation / Answer

a) Resistivity = R A / L = (10.3 * 7.3 * 10-8) / (0.54) = 1.4 * 10-6 ohm.m

b) Resistivity = R A / L = (21.3 * 7.3 * 10-8) / (1.028) = 1.5 * 10-6 ohm.m

c) Resistivity = R A / L = (32.1 * 7.3 * 10-8) / (1.543) = 1.52 * 10-6 ohm.m

Average resistivity = [1.4 * 10-6 + 1.5 * 10-6 + 1.52 * 10-6] / 3

= 1.47 * 10-6 ohm.m

Percentage differentiation = (1.5 - 1.47) * 100 / 1.5 = 2 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.