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Circuits overview

Topic 2 of 9

Resistance, material and geometry

Resistance belongs to a component at a stated operating condition. Resistivity describes its material at a stated temperature; length and cross-sectional area then determine a uniform wire's resistance.

For nonzero current, define resistance using the p.d. across the component and the current through it:

R = V/I   or   V = IR
1 Ω = 1 V/A = kg m2 s-3 A-2

The rearrangement V = IR does not establish that R is constant. An ohmic component has proportional I and V under constant physical conditions. On an I-against-V graph its straight-line gradient is 1/R. For a curved non-ohmic characteristic, V/I at a point is not generally the reciprocal tangent gradient.

Use the conducting cross-section

For a uniform wire of electrical length l, cross-sectional area A and resistivity ρ:

R = ρl/A
ρ = RA/l
A = πd2/4   (circular wire of diameter d)

Resistivity has unit Ω m; the symbol ρ here does not mean mass density. A is the area perpendicular to current, not the curved outer surface of the wire. The length is the separation of the relevant electrical contacts. State uniform material and temperature, and assess lead/contact contributions before neglecting them.

A rough p.d. of 0.4 V and current of 0.1 A suggest a few ohms. A diameter of order 4 × 10-4 m gives area of order 10-7 m2. For a roughly metre-long wire, resistivity is then of order 10-7 to 10-6 Ω m. This estimate helps catch powers-of-ten errors before exact calculation.

Worked wire and selected readings

The source voltage is not the wire voltage

A supplied uniform wire has l = 0.800 m and d = 0.400 mm. In this worked model, a voltmeter reads 0.400 V across contacts P/Q and an ammeter reads 0.125 A through the wire. The measured source-terminal supply is 6.0 V; a series control takes the remaining voltage.

Use the wire readings, not the source reading

Use the wire readings, not the source readingA complete low-voltage circuit contains the six-volt source, a 44.8-ohm series control, an ammeter and the chosen wire. Current is 0.125 ampere. The voltmeter taps are exactly at P and Q, the ends of the 0.800-metre electrical length, and read 0.400 volt. The series control takes the other 5.60 volts under negligible meter and lead effects. The source label is its supplied terminal p.d., not a rating used instead of a reading.+-AV44.8 Ω5.60 V0.125 A0.400 VPQ+-6.00 Vat sourceP/Q length: 0.800 mThe voltmeter spans only the wire.

The control and wire share 0.125 A. The wire has 0.400 V across P/Q; using the source's 6.00 V would include the control resistance.

Double length and diameter at fixed resistivity

Double length and diameter at fixed resistivityThe two side views use one length scale: 120 and 240 drawing units correspond to 0.800 and 1.60 metres. Their diameters use one transverse scale, sixteen and thirty-two drawing units, for 0.400 and 0.800 millimetre. The transverse scale is enlarged relative to length. Separately enlarged circular end faces have radii twenty and forty, so the second conducting area is four times the first. The longer, wider wire has half the resistance under the unchanged material and temperature model.0.800 m1.60 md = 0.400 mmd = 0.800 mmEnd faces enlarged togetherArea AArea 4A

Length and diameter use different display scales, each shared by both wires. Length doubles but area quadruples; at unchanged resistivity the resistance is 1.60 ohm, half the original 3.20 ohm.

Resistance falls as the reciprocal of area

Resistance falls as the reciprocal of areaThe positive horizontal variable is area divided by the original area A0. Vertical is resistance divided by its original value R0. For fixed material, temperature and length, R/R0 equals one divided by A/A0. Exact pairs are 0.5 and 2, one and one, two and 0.5, four and 0.25. Only positive areas are shown; the curve is not a straight line.0.5123400.511.52Resistance ratio R/R0Area ratio A/A0

Doubling area halves resistance. Doubling diameter is different: it gives four times the area and one quarter of the resistance if length stays fixed.

The apparatus pairs the wire's P/Q voltage with its current and shows the separate series-control drop. The dimension comparison preserves doubled length and diameter, with enlarged end-face views labelled separately. The final plot holds material and length fixed while comparing area with resistance.
Rwire = 0.400/0.125 = 3.20 Ω
d = 0.400 × 10-3 m
A = π(0.400 × 10-3)2/4
= 1.2566 × 10-7 m2
ρ = (3.20)(1.2566 × 10-7)/0.800
= 5.03 × 10-7 Ω m

Convert diameter to metres before squaring.

Control p.d. = 6.0 - 0.400 = 5.60 V
Rcontrol = 5.60/0.125 = 44.8 Ω

Neglecting meter and lead effects, the control and wire drops add to the 6.0 V terminal supply. Using 6.0/0.125 would give 48 Ω for the combined external path, not the wire's 3.20 Ω.

A second wire of the same material and temperature has twice the length and twice the diameter. Its area is four times as large:

Rnew/Rold = 2/22 = ½
Rnew = 1.60 Ω

Recognise the reciprocal relationship

Hold ρ and l fixed, and call the original area A0 and resistance R0. Then:

R/R0 = 1/(A/A0)

Area ratios 0.5, 1, 2 and 4 give resistance ratios 2, 1, 0.5 and 0.25. The positive-domain graph is a falling reciprocal curve, not a straight line. R against 1/A would instead be linear. R against diameter follows 1/d2, a different horizontal variable.

Determine resistivity from a wire

  1. Use a uniform wire and measure the contact separation l. Connect the voltmeter across those contacts and the ammeter in the wire path, with a suitable low-voltage source and series current control.
  2. Estimate the expected V and I, then choose actual available meter ranges and resolution. A 0-0.1 A range cannot contain the supplied 0.125 A current. Use modest current and check that heating does not appreciably change the wire's resistance.
  3. Use a suitable micrometer to measure diameter at several positions and orientations. Check its zero, use its contact mechanism consistently and retain the raw readings and instrument resolution.
  4. Vary the electrical length, recording matched V/I values to calculate R at each length. Plot R vertically against l; a gradient gives ρ/A, so multiply the gradient by measured A to infer ρ.
  5. Inspect any intercept and departure from a line. A roughly constant contact/lead contribution can produce a nonzero intercept; forcing the line through zero can conceal it. Changing temperature or diameter along the wire challenges the uniform model.

Because area contains d2, a small 1% diameter error produces about a 2% area error and, with R/l fixed, about a 2% resistivity error. Repetition can assess scatter; it cannot correct a persistent micrometer zero, using diameter as radius, or measuring the wrong contact separation.

Optional check A uniform wire has resistance 3.20 ohm. A second wire of the same material at the same temperature has twice the length and twice the diameter. What is its resistance?
A uniform wire has resistance 3.20 ohm. A second wire of the same material at the same temperature has twice the length and twice the diameter. What is its resistance?