K323 / 2027
Current of electricity overview

Topic 3 of 4

Resistance and wire dimensions

Resistance relates the potential difference across a component to the current through it. At the same p.d., a greater resistance gives a smaller current.

Current I is measured through a component in amperes. Potential difference V is measured across its terminals in volts. Use readings for the same component when finding its resistance.

R = V/IResistance = potential difference / current. Rearrangements: V = IR and I = V/R. The unit is the ohm, Ω; 1 Ω = 1 V/A.

This defines resistance at the stated operating point: the voltage and current under the conditions of that reading. It does not by itself establish that a component's resistance stays the same when conditions change.

Measure current through and p.d. across the same resistor

A is an ammeter in series. V is a voltmeter connected to the resistor's two terminals. The marked meter polarities suit this d.c. source.

An ammeter in series and a voltmeter in parallel measure the resistorThe cell has its positive plate above its negative plate on the left. The top branch runs through ammeter A and then a resistor before returning to the source along the right and bottom wires. The ammeter's left terminal is positive. Two separate connections descend from the exact left and right resistor terminals to voltmeter V; its left terminal is positive. The supplied ammeter reading is 0.20 ampere and the p.d. across the resistor is 6.0 volts. With ideal meters, the ammeter gives the resistor current, so its resistance is 30 ohm. No wire bypasses the resistor or either meter symbol.+-SourceA+-0.20 AMeasured resistorV+-6.0 VBoth readings concern this resistor

R = V/I = 6.0/0.20 = 30 Ω.

This arrangement uses the ideal meter model: the voltmeter takes negligible current and the ammeter causes negligible p.d. The ammeter therefore measures the resistor current.

The ammeter measures the current in the resistor's path, and the voltmeter connects across its two terminals. With ideal meters, the voltmeter draws negligible current and the ammeter adds negligible resistance. The supplied readings are 6.0 V and 0.20 A.

Use corresponding readings

Determine resistance, then predict current

The p.d. across a resistor is 6.0 V and the current through it is 0.20 A.

R = V/I = 6.0/0.20 = 30 Ω.

If the same resistor remains at 30 Ω when the p.d. becomes 9.0 V, then I = V/R = 9.0/30 = 0.30 A.

The prediction uses the stated constant-resistance assumption. The first measurement alone cannot show that heating or other changes will leave the resistance unchanged.

Optional check A voltmeter reads 6.0 V across a resistor while an ammeter measures 0.20 A through that resistor. What is its resistance at this operating point?
A voltmeter reads 6.0 V across a resistor while an ammeter measures 0.20 A through that resistor. What is its resistance at this operating point?

How wire dimensions affect resistance

Compare wires of the same material at the same temperature. Changing the material or temperature can change the relationship between their dimensions and resistance.

Greater length gives greater resistance
Resistance is proportional to length, R ∝ L. Doubling the length while keeping the cross-section unchanged doubles the resistance. Charge must travel through a longer section of the material.
Greater cross-sectional area gives smaller resistance
Resistance is inversely proportional to cross-sectional area, R ∝ 1/A. Doubling that area while keeping length unchanged halves the resistance. The wider cross-section provides more conducting material alongside the path.

The cross-sectional area is the area you see when looking straight at a cut end. It is not the curved outer surface of the wire. For a circular wire, A = πd2/4, so area changes with the square of the diameter.

Double the length and double the diameter

Compare the same material at the same temperature. Both drawings use the same scale; the labels give ratios, not dimensions to measure from your screen. The area is the circular end section.

Original wire: 8.0 ohm

Original wire: 8.0 ohmThe original wire has length L, circular diameter d and cross-sectional area A. Its supplied resistance is 8.0 ohm. A side view and an end-on circular section show the dimensions. The measured axial length is between the two end faces. Diameter brackets span the wire and the end-on circle. The side-view end faces are drawn as ellipses to suggest their orientation, while the end-on section is a true circle.Side viewLdEnd-on cross-sectionAd

Changed wire: 4.0 ohm

Changed wire: 4.0 ohmThe changed wire is twice as long and twice the diameter. At the same drawing scale, its circular section has twice the radius and four times the area. It is the same material at the same temperature, so its resistance is two divided by four, or one half, of 8.0 ohm: 4.0 ohm. The measured axial length is between the two end faces. Diameter brackets span the wire and the end-on circle. The side-view end faces are drawn as ellipses to suggest their orientation, while the end-on section is a true circle.Side view2L2dEnd-on cross-section4A2d

Twice the diameter gives four times the cross-sectional area. The resistance changes by 2/4 = 1/2, so 8.0 Ω becomes 4.0 Ω.

The second wire has twice the length and twice the diameter. Its end-on area is four times the original area. Material and temperature are kept the same, so the two changes together halve the resistance.

Compare one change at a time

Twice the length and twice the diameter

The original wire has resistance 8.0 Ω.

  1. Doubling its length gives a resistance factor of 2.
  2. Doubling its diameter gives an area factor of 22 = 4, so the resistance factor from area is 1/4.
  3. Combine the factors: new R = 8.0 x 2/4 = 4.0 Ω.

For any comparison at the same material and temperature, Rnew/Rold = (Lnew/Lold) x (Aold/Anew).

Determine resistance and investigate wire length

Use a suitable low-voltage classroom supply, with an ammeter in the current path and a voltmeter across the component. Record the p.d. and current together after the readings settle, and calculate R = V/I. Check meter ranges and polarity before collecting readings.

To investigate length, use the same uniform wire and vary the distance between its electrical contacts. Measure this active length with a ruler. The unused wire beyond the contacts is not part of the measured section.

  1. Keep the wire's material and cross-sectional area fixed. Use several measured active lengths.
  2. At each length, record the current through and p.d. across that section. Calculate its resistance from the paired readings.
  3. Keep the wire temperature as steady as practical: use a suitably low current and switch off between readings where appropriate. Wait for it to return to the comparison conditions if it warms.
  4. Plot resistance vertically against active length horizontally. Proportionality predicts a straight line through the origin for the wire alone at constant cross-section and temperature.

If diameter is needed, use a micrometer or suitable calipers, check for zero error and take readings at several positions and orientations. Avoid squeezing or measuring a damaged part of the wire. Convert the diameter to consistent units before calculating area.

Contact and lead resistance may add to a measured resistance, depending on the voltage lead positions. Wire heating can change readings, and a varying diameter means the cross-section is not uniform. Repeating and averaging readings can reveal scatter, but it does not automatically remove these effects. Explain the specific limitation and how the method addresses it.

Twice the diameter is four times the area. Also keep the voltage and current paired: dividing a voltage across one branch by the current through a different branch does not generally give either branch's resistance.

Optional check A wire has resistance 8.0 ohm. A second wire of the same material and temperature has twice its length and twice its diameter. What is the second resistance?
A wire has resistance 8.0 ohm. A second wire of the same material and temperature has twice its length and twice its diameter. What is the second resistance?