Chapter 3 · Electricity

3.2 Ohm's Law and the Factors Affecting Resistance

Double the push and the current doubles: for a metal wire at steady temperature $V=IR$. The resistance $R=\rho\frac lA$ grows with the length of the wire, falls with its thickness, and depends on the material through its resistivity $\rho$.

Connect a $1.5\ \text{V}$ cell across a short piece of thick copper wire and the wire gets hot and the cell goes flat within minutes; connect the same cell across a torch bulb and it glows quietly for hours. The cell is the same, so the push — the potential difference — is the same. What differs is how strongly each conductor resists the flow of charge. That opposition is called resistance, and it is what lets engineers choose exactly the current each device should take.

In this section we find, by experiment, the simple rule that links current and potential difference for a metal wire — Ohm's law — and use it to define resistance and the ohm. Then we ask what the resistance of a wire depends on: its length, its thickness and what it is made of. That last property, the resistivity, explains why your house is wired with copper, why a toaster's glowing coil is nichrome, and why the handles of tools are covered in rubber.

Ohm's Law

Set up a circuit with a battery, a plug key, a rheostat (a variable resistor that lets you change the current smoothly), an ammeter and a length of nichrome wire, with a voltmeter connected across the wire. Move the rheostat's slider in steps, and at each step write down the ammeter reading $I$ and the voltmeter reading $V$. The ratio $\dfrac VI$ comes out the same every time, as long as the wire does not warm up. Plot $V$ against $I$ and the points lie on a straight line through the origin.

📐 Law 3.1 — Ohm's law

The potential difference $V$ across the ends of a metallic conductor is directly proportional to the current $I$ through it, provided its temperature (and other physical conditions) stays the same:

$$V\propto I,\qquad\text{so}\qquad V=IR,$$

where the constant $R$ is the resistance of the conductor. Equivalently $I=\dfrac VR$: for a given potential difference, the current is inversely proportional to the resistance. A V–I graph of such a conductor is a straight line through the origin.

📘 Definition 3.4 — Resistance and the ohm

The resistance of a conductor is the ratio of the potential difference across it to the current through it,

$$R=\frac VI .$$

It measures how strongly the conductor opposes the flow of charge. The SI unit is the ohm ($\Omega$): a conductor has a resistance of $1\ \Omega$ if a potential difference of $1\ \text{V}$ across it drives a current of $1\ \text{A}$ through it, $1\ \Omega=1\ \text{V/A}$. Larger resistances are given in kilohms ($1\ \text{k}\Omega=10^3\ \Omega$) and megohms ($1\ \text{M}\Omega=10^6\ \Omega$). A component made to have a particular resistance is a resistor.

Reading a V–I graph. With $V$ on the vertical axis and $I$ on the horizontal axis, the slope of the line is $\dfrac VI=R$: a steeper line means a larger resistance. (If the axes are swapped, with $I$ up the side, the slope is $\dfrac1R$ and a steeper line means a smaller resistance — read the axis labels first.) Not everything obeys Ohm's law: the filament of a bulb gets hotter as the current rises, its resistance grows, and its V–I graph bends. Such components are called non-ohmic.

Interactive 1 — The Ohm's Law Lab

A $6.0\ \text{V}$ battery, a rheostat (resistance $R_h$ from $0$ to $50\ \Omega$), an ammeter and a resistor $R$ with a voltmeter across it. Slide the rheostat: the electrons speed up or slow down and both meters change. Press Record to plot the point $(I,\,V)$; the dashed line through the origin has slope $V/I$. Record points for two different resistors and compare the slopes. Play sweeps the rheostat and records automatically.

20
ammeter $I$—
voltmeter $V$—
$V/I$—
points recorded—
✏️ Example 1 — Finding a resistance
A potential difference of $6.0\ \text{V}$ across a resistor drives a current of $0.30\ \text{A}$ through it. Find its resistance.
Formula $R=\dfrac VI=\dfrac{6.0\ \text{V}}{0.30\ \text{A}}=20\ \Omega$.
Check $IR=0.30\times20=6.0\ \text{V}$. ✓
✏️ Example 2 — Current drawn by an iron
The heating element of an electric iron has a resistance of $44\ \Omega$ when hot. (a) What current does it draw from a $220\ \text{V}$ supply? (b) What potential difference would drive a current of $2.0\ \text{A}$ through it?
(a) $I=\dfrac VR=\dfrac{220}{44}=5.0\ \text{A}$.
(b) $V=IR=2.0\times44=88\ \text{V}$.
Meaning With the resistance fixed, current and voltage change in the same proportion: $\frac{88}{220}=\frac{2.0}{5.0}=0.40$. ✓
✏️ Example 3 — Testing Ohm's law from data
A student records these readings for a wire:
$I$ (A)0.100.200.300.400.50
$V$ (V)0.801.62.43.24.0
Does the wire obey Ohm's law? Find its resistance, and predict the current when the potential difference is $6.0\ \text{V}$.
Ratios $\dfrac VI=\dfrac{0.80}{0.10}=\dfrac{1.6}{0.20}=\dfrac{2.4}{0.30}=\dfrac{3.2}{0.40}=\dfrac{4.0}{0.50}=8.0\ \Omega$ every time.
Conclusion $V\propto I$: the V–I graph is a straight line through the origin with slope $8.0\ \Omega$, so the wire obeys Ohm's law and $R=8.0\ \Omega$.
Predict $I=\dfrac VR=\dfrac{6.0}{8.0}=0.75\ \text{A}$ (if the wire stays at the same temperature).

Factors Affecting Resistance

Keep the same cell and ammeter, and connect different wires in turn. Experiments of this kind show three things. A longer wire of the same material and thickness gives a smaller current: doubling the length doubles the resistance. A thicker wire gives a larger current: doubling the cross-sectional area halves the resistance. And wires of the same size but different materials — copper, iron, nichrome — have very different resistances. A useful picture: the free electrons must squeeze past the vibrating atoms all the way along the wire, so a longer path means more obstruction, while a wider wire offers more lanes side by side.

📐 Law 3.2 — Resistance of a uniform wire

For a uniform metallic conductor at a given temperature, the resistance is directly proportional to its length $l$ and inversely proportional to its area of cross-section $A$: $R\propto l$ and $R\propto\dfrac1A$. Together,

$$R=\rho\,\frac lA ,$$

where the constant $\rho$ (rho) depends only on the material (and its temperature), not on the size or shape of the wire. For a wire of diameter $d$, $A=\pi r^2=\dfrac{\pi d^2}{4}$.

📘 Definition 3.5 — Resistivity

The resistivity $\rho$ of a material is defined by

$$\rho=\frac{RA}{l} .$$

It is the resistance of a piece of the material of length $1\ \text{m}$ and cross-sectional area $1\ \text{m}^2$. Its SI unit is the ohm metre ($\Omega\ \text{m}$). Resistance belongs to a particular wire; resistivity belongs to the material it is made of — cutting a wire in half halves its resistance but leaves its resistivity unchanged.

MaterialResistivity $\rho$ ($\Omega\ \text{m}$), about $20\,^\circ\text{C}$Typical use
Silver$1.6\times10^{-8}$best conductor; too costly for wiring
Copper$1.7\times10^{-8}$house wiring, connecting wires
Aluminium$2.7\times10^{-8}$overhead power lines (light, cheap)
Tungsten$5.6\times10^{-8}$filaments of bulbs (melts only at about $3400\,^\circ\text{C}$)
Iron$1.0\times10^{-7}$—
Manganin (alloy)$4.4\times10^{-7}$standard resistors
Constantan (alloy)$4.9\times10^{-7}$resistance wires, rheostats
Nichrome (alloy)$1.0\times10^{-6}$heating elements of irons, toasters, heaters
Glass$10^{10}$ to $10^{14}$insulator
Rubber$10^{13}$ to $10^{16}$insulator: cable covers, gloves

Metals have resistivities around $10^{-8}$ to $10^{-6}\ \Omega\ \text{m}$ and are good conductors; insulators such as glass and rubber have resistivities $10^{16}$ or more times larger. Alloys have a higher resistivity than the pure metals they are made from, and they do not oxidise (burn) easily when red-hot — which is why nichrome, not copper, is used for heating elements, while copper and aluminium, with the lowest resistivities, carry current to them with little loss.

Interactive 2 — The Wire Resistance Lab

A $1.5\ \text{V}$ cell and an ammeter are connected across a wire. Choose the material and set the length and diameter: the resistance $R=\rho\frac lA$, the current and the speed of the electrons update. The graph below plots $R$ against $l$ for your wire (solid) and for a reference wire — nichrome, $0.50\ \text{mm}$ diameter (dashed): straight lines through the origin, and a thinner wire gives a steeper line. Play stretches the wire to show $R\propto l$.

1.000.50
$A=\pi d^2/4$—
$\rho$—
$R=\rho l/A$—
$I=1.5\ \text{V}/R$—
$R\div R_{\text{ref}}$—
✏️ Example 4 — Resistance of a heater wire
Find the resistance of a nichrome wire $2.0\ \text{m}$ long and $0.40\ \text{mm}$ in diameter. Take $\rho=1.0\times10^{-6}\ \Omega\ \text{m}$.
Area $r=0.20\ \text{mm}=2.0\times10^{-4}\ \text{m}$, so $A=\pi r^2=\pi\times(2.0\times10^{-4})^2=1.26\times10^{-7}\ \text{m}^2$.
Formula $$R=\rho\frac lA=\frac{1.0\times10^{-6}\times2.0}{1.26\times10^{-7}}=15.9\ \Omega\approx16\ \Omega.$$
Check Units: $\dfrac{\Omega\ \text{m}\times\text{m}}{\text{m}^2}=\Omega$. ✓ Convert millimetres to metres before squaring.
✏️ Example 5 — Cutting and stretching a wire
A uniform wire has a resistance of $10\ \Omega$. (a) It is cut into two equal halves. What is the resistance of each half? (b) Another identical wire is stretched uniformly to twice its length. What is its new resistance?
(a) Same material and area, half the length: $R\propto l$, so each half has $5.0\ \Omega$.
(b) Idea Stretching does not change the volume $Al$ of the metal. If the length doubles, $l'=2l$, the area must halve, $A'=\tfrac A2$.
(b) Working $$R'=\rho\frac{l'}{A'}=\rho\frac{2l}{A/2}=4\,\rho\frac lA=4\times10=40\ \Omega.$$
Meaning The length doubling alone would double $R$; the thinner wire doubles it again. The resistivity is unchanged in both cases.
✏️ Example 6 — Identify the material
A wire $1.5\ \text{m}$ long with a cross-sectional area of $0.50\ \text{mm}^2$ has a resistance of $1.32\ \Omega$. Find the resistivity of its material and identify it from the table.
Units $1\ \text{mm}^2=(10^{-3}\ \text{m})^2=10^{-6}\ \text{m}^2$, so $A=0.50\times10^{-6}=5.0\times10^{-7}\ \text{m}^2$.
Formula $$\rho=\frac{RA}{l}=\frac{1.32\times5.0\times10^{-7}}{1.5}=4.4\times10^{-7}\ \Omega\ \text{m}.$$
Answer The table gives $4.4\times10^{-7}\ \Omega\ \text{m}$ for manganin, the alloy used in standard resistors.
⚠️ Common mistakes

1. Forgetting "constant temperature". Ohm's law holds only while the conductor's temperature does not change. A glowing bulb filament does not obey it.

2. Reading the slope the wrong way. On a $V$ (up) – $I$ (across) graph the slope is $R$; on an $I$ (up) – $V$ (across) graph it is $\frac1R$. Check the axes before saying which wire has more resistance.

3. Square millimetres. $1\ \text{mm}^2=10^{-6}\ \text{m}^2$, not $10^{-3}\ \text{m}^2$. And use the radius, $r=\frac d2$, in $A=\pi r^2$.

4. Resistance versus resistivity. Changing the length or thickness changes $R$, never $\rho$. Only a change of material (or temperature) changes $\rho$.

5. Stretching. When a wire is stretched, its area shrinks too: doubling the length makes the resistance four times, not twice, as large.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional: drawing wires through a die

In JEE Main, NEET and JEE Advanced 2026; not in the CBSE Class 10 syllabus.

When a wire is drawn out, its volume $V_0=Al$ stays fixed, so $A=\dfrac{V_0}{l}$ and $R=\rho\dfrac lA=\dfrac{\rho\,l^2}{V_0}$: at constant volume $R\propto l^2$. Stretching to $n$ times the length multiplies the resistance by $n^2$. In terms of the diameter, $l=\dfrac{V_0}{A}\propto\dfrac1{d^2}$ and $A\propto d^2$, so $R\propto\dfrac{1}{d^4}$: drawing a wire to half its diameter makes it $16$ times as resistive. For example, a $2.0\ \Omega$ wire drawn to a third of its diameter has resistance $2.0\times3^4=162\ \Omega$. Entrance-exam problems often hide these ratios behind "same mass" or "same volume".

Exercises

I
Ohm's law. Assume the temperature stays constant.
  • 1 A potential difference of $12\ \text{V}$ drives $0.40\ \text{A}$ through a resistor. Find its resistance.
  • 2 Find the current through a $25\ \Omega$ resistor connected across $5.0\ \text{V}$.
  • 3 What potential difference is needed to drive $0.60\ \text{A}$ through a $15\ \Omega$ resistor?
  • 4 A bulb connected to $220\ \text{V}$ draws $0.50\ \text{A}$. Find the resistance of its hot filament.
  • 5 An $8.0\ \Omega$ resistor is connected to a $2.0\ \text{V}$ cell, and then to a battery of four such cells. Find the current each time.
  • 6 The potential difference across a resistor is doubled. What happens to the current through it?
  • 7 Define $1$ ohm.
  • 8 A current of $2.0\ \text{mA}$ flows through a resistor when the potential difference across it is $5.0\ \text{V}$. Find its resistance in ohms and in kilohms.
Check: enter any two of $V$, $I$, $R$ (leave the third blank) and compare with your answer
$V$—
$I$—
$R$—
1 $R=\tfrac VI=\tfrac{12}{0.40}=30\ \Omega$.
2 $I=\tfrac VR=\tfrac{5.0}{25}=0.20\ \text{A}$.
3 $V=IR=0.60\times15=9.0\ \text{V}$.
4 $R=\tfrac{220}{0.50}=440\ \Omega$.
5 $I=\tfrac{2.0}{8.0}=0.25\ \text{A}$; with four cells $V=8.0\ \text{V}$ and $I=\tfrac{8.0}{8.0}=1.0\ \text{A}$ — four times the voltage, four times the current.
6 It doubles too, since $I=\tfrac VR$ with $R$ constant.
7 A conductor has a resistance of $1\ \Omega$ if a potential difference of $1\ \text{V}$ across its ends makes a current of $1\ \text{A}$ flow through it.
8 $R=\tfrac{5.0}{2.0\times10^{-3}}=2500\ \Omega=2.5\ \text{k}\Omega$.
II
V–I graphs and the experiment.
  • 1 For a wire, $V=1.0$, $2.0$, $3.0$, $4.0\ \text{V}$ give $I=0.25$, $0.50$, $0.75$, $1.00\ \text{A}$. Does it obey Ohm's law? Find $R$.
  • 2 On a graph with $V$ on the vertical axis and $I$ on the horizontal axis, the line for wire P is steeper than the line for wire Q. Which has the larger resistance?
  • 3 The same two wires are plotted with $I$ on the vertical axis and $V$ on the horizontal axis. Which line is now steeper, and why?
  • 4 The straight V–I line of a resistor passes through the origin and the point ($0.40\ \text{A}$, $2.0\ \text{V}$). Find $R$ and the current when $V=3.5\ \text{V}$.
  • 5 The V–I graph of a component is a curve. What does this tell you?
  • 6 Why must the current be switched off between readings, or kept small, when verifying Ohm's law?
  • 7 What is the job of the rheostat in the Ohm's law experiment?
1 $\tfrac VI=4.0\ \Omega$ for every pair: yes, $V\propto I$, and $R=4.0\ \Omega$.
2 P: the slope $\tfrac VI$ is the resistance, so the steeper line has the larger $R$.
3 Q's line is now steeper: the slope is $\tfrac IV=\tfrac1R$, which is larger for the smaller resistance.
4 $R=\tfrac{2.0}{0.40}=5.0\ \Omega$; $I=\tfrac{3.5}{5.0}=0.70\ \text{A}$.
5 $V$ is not proportional to $I$: the component's resistance changes with the current (for example a filament that heats up), so it does not obey Ohm's law.
6 A current heats the wire, and a hotter wire has a higher resistance; Ohm's law is tested at constant temperature.
7 It changes the total resistance of the circuit, and so lets you vary the current (and the potential difference across the wire) in steps without changing the battery.
III
Resistivity. Use the table of resistivities where needed.
  • 1 Find the resistance of a copper wire $10\ \text{m}$ long with a cross-sectional area of $1.0\ \text{mm}^2$.
  • 2 Find the resistance of a nichrome wire $1.0\ \text{m}$ long and $0.20\ \text{mm}$ in diameter.
  • 3 A wire has a resistance of $4.0\ \Omega$. What is the resistance of a wire of the same material and thickness but twice as long?
  • 4 A wire has a resistance of $4.0\ \Omega$. What is the resistance of a wire of the same material and length but twice the diameter?
  • 5 A $6.0\ \Omega$ wire is cut into three equal pieces. Find the resistance and the resistivity of each piece compared with the original.
  • 6 A wire $2.0\ \text{m}$ long with cross-sectional area $2.0\times10^{-7}\ \text{m}^2$ has a resistance of $10\ \Omega$. Find the resistivity and identify the material.
  • 7 What length of constantan wire of cross-sectional area $0.49\ \text{mm}^2$ has a resistance of $5.0\ \Omega$?
  • 8 What is the SI unit of resistivity? Does the resistivity of a wire depend on its length?
1 $R=\tfrac{1.7\times10^{-8}\times10}{1.0\times10^{-6}}=0.17\ \Omega$.
2 $A=\pi(1.0\times10^{-4})^2=3.14\times10^{-8}\ \text{m}^2$; $R=\tfrac{1.0\times10^{-6}\times1.0}{3.14\times10^{-8}}=32\ \Omega$.
3 $R\propto l$: $8.0\ \Omega$.
4 Twice the diameter gives four times the area, so $R=\tfrac{4.0}{4}=1.0\ \Omega$.
5 Each piece is a third as long: $2.0\ \Omega$. The resistivity is the same as before — same material.
6 $\rho=\tfrac{RA}{l}=\tfrac{10\times2.0\times10^{-7}}{2.0}=1.0\times10^{-6}\ \Omega\ \text{m}$: nichrome.
7 $l=\tfrac{RA}{\rho}=\tfrac{5.0\times0.49\times10^{-6}}{4.9\times10^{-7}}=5.0\ \text{m}$.
8 The ohm metre ($\Omega\ \text{m}$). No — resistivity depends only on the material (and its temperature), not on the length or thickness.
IV
Materials and applications.
  • 1 Why are the heating elements of toasters and irons made of an alloy rather than a pure metal?
  • 2 Why is tungsten used for the filaments of bulbs?
  • 3 Why are copper and aluminium used for connecting wires and power cables?
  • 4 From copper, glass, nichrome and iron, name the best conductor and the insulator.
  • 5 Power line. An aluminium cable is $10\ \text{km}$ long with a cross-sectional area of $3.0\ \text{cm}^2$. Find its resistance.
  • 6 Challenge. A $5.0\ \Omega$ wire is stretched uniformly to three times its length. Find its new resistance.
  • 7 Challenge. Wires A and B are made of the same metal. A is twice as long as B and has twice its diameter. Find $\tfrac{R_A}{R_B}$.
1 Alloys have a much higher resistivity, so a short coil has the resistance needed to produce heat, and they do not oxidise (burn) readily when red-hot.
2 Tungsten has a very high melting point (about $3400\,^\circ\text{C}$), so it can glow white-hot without melting.
3 They have very low resistivity, so long wires have little resistance and waste little energy as heat; aluminium is also light and cheap for overhead lines.
4 Best conductor: copper (lowest resistivity). Insulator: glass.
5 $l=1.0\times10^4\ \text{m}$, $A=3.0\times10^{-4}\ \text{m}^2$: $R=\tfrac{2.7\times10^{-8}\times1.0\times10^4}{3.0\times10^{-4}}=0.90\ \Omega$.
6 Volume constant: $l'=3l$, $A'=\tfrac A3$, so $R'=9R=45\ \Omega$.
7 $R\propto\tfrac{l}{d^2}$: $\tfrac{R_A}{R_B}=\tfrac{2}{2^2}=\tfrac12$.
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board · 2 marks State Ohm's law. Sketch the V–I graph of a conductor that obeys it and say what its slope represents.

Law The potential difference across a metallic conductor is directly proportional to the current through it, provided its temperature remains constant: $V=IR$.
Graph $V$ on the vertical axis, $I$ on the horizontal axis: a straight line through the origin. Its slope $\frac VI$ is the resistance $R$.
Exam tip "At constant temperature" is a marked phrase — never leave it out.

E2. Board · 3 marks A wire $3.0\ \text{m}$ long and $0.40\ \text{mm}$ in diameter has a resistance of $12\ \Omega$. (a) Find the resistivity of its material. (b) What would be the resistance of a wire of the same material and length but $0.80\ \text{mm}$ in diameter?

Area $r=2.0\times10^{-4}\ \text{m}$, $A=\pi r^2=1.26\times10^{-7}\ \text{m}^2$.
(a) $\rho=\dfrac{RA}{l}=\dfrac{12\times1.26\times10^{-7}}{3.0}=5.0\times10^{-7}\ \Omega\ \text{m}$.
(b) Doubling the diameter makes the area four times as large, so $R=\dfrac{12}{4}=3.0\ \Omega$. Exam tip: for (b) use ratios — no need to recalculate with $\rho$.

E3. Board · Assertion–Reason Assertion (A): The heating elements of electric irons are made of nichrome. Reason (R): Nichrome has a high resistivity and does not oxidise readily at high temperatures. (A) Both A and R are true and R explains A. (B) Both are true but R does not explain A. (C) A is true but R is false. (D) A is false but R is true.

Idea Both statements are true, and the high resistivity and resistance to burning are exactly why nichrome is chosen.
Answer (A). Exam tip: in assertion–reason items, ask "does R answer why A happens?" — here it does.

E4. Board · Case study In an Ohm's law experiment a student obtains: $V=0.5$, $1.0$, $1.5$, $2.0$, $2.5\ \text{V}$ and $I=0.10$, $0.20$, $0.30$, $0.40$, $0.50\ \text{A}$. (i) Find the resistance of the wire from the graph or the readings. [2 marks] (ii) What current would flow at $4.0\ \text{V}$? [1 mark] (iii) The wire is replaced by one of the same material and thickness but twice as long. What current flows at $2.5\ \text{V}$? [1 mark]

(i) The points lie on a straight line through the origin; slope $=\dfrac VI=\dfrac{2.5}{0.50}=5.0\ \Omega$.
(ii) $I=\dfrac{4.0}{5.0}=0.80\ \text{A}$.
(iii) Twice the length: $R=10\ \Omega$, so $I=\dfrac{2.5}{10}=0.25\ \text{A}$. Exam tip: find the slope from two well-separated points on the line, not from one reading.

E5. NEET A wire of resistance $R$ is stretched uniformly until its length is $n$ times the original. Its new resistance is (A) $nR$ (B) $n^2R$ (C) $\dfrac Rn$ (D) $\dfrac R{n^2}$

Working Volume constant: $l'=nl$, $A'=\dfrac An$, so $R'=\rho\dfrac{nl}{A/n}=n^2R$.
Answer (B). Exam tip: (A) forgets that the wire gets thinner — the commonest wrong answer.

E6. JEE Main Two wires of the same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. The ratio of their resistances is (A) $1:8$ (B) $8:1$ (C) $1:2$ (D) $2:1$

Working $R\propto\dfrac{l}{d^2}$: $\dfrac{R_1}{R_2}=\dfrac{l_1}{l_2}\times\left(\dfrac{d_2}{d_1}\right)^2=\dfrac12\times\dfrac14=\dfrac18$.
Answer (A). Exam tip: the diameter enters squared — write the proportionality before substituting ratios.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. Does a line drawn with a pencil obey $R=\rho\frac lA$ — and can you use it to estimate the resistivity of graphite?

Why it matters. Printed and "drawn" electronics — carbon tracks on paper, flexible sensors in sports clothing, touch pads, low-cost medical test strips — rely on thin conducting films whose resistance is set by their length, width and thickness. Researchers around the world are developing paper-based sensors for clinics with no laboratory, and the first design check is always $R=\rho\frac lA$.

What to do.

  1. With a soft pencil (6B or 8B), shade solid strips $1.0\ \text{cm}$ wide and $2$, $4$, $6$, $8$ and $10\ \text{cm}$ long on plain paper, pressing evenly. Shade each strip three times over.
  2. Measure the resistance of each strip end to end with a digital multimeter on its resistance (Ω) range, pressing the probes firmly on the ends. Plot $R$ against $l$.
  3. Now draw strips $10\ \text{cm}$ long and $0.5$, $1.0$, $1.5$ and $2.0\ \text{cm}$ wide. Plot $R$ against $\frac1{\text{width}}$.
  4. Are your graphs straight lines through the origin? Discuss what spoils the fit (uneven shading, contact resistance at the probes).
  5. Estimate the film thickness as about $10\ \mu\text{m}$ and calculate a resistivity for pencil graphite from $\rho=\frac{RA}{l}$.

Tools: soft pencils, ruler, digital multimeter, graph paper or a spreadsheet.

Hand in: Both graphs, your estimate of $\rho$ with a comment on its uncertainty, and a comparison with the metals in the table above.

Going further. Draw two strips side by side and join them at one end. Predict their combined resistance before measuring — Section 3.3 explains the rule.

Instructor note: typical readings are tens to a few hundred kilohms for a $10\ \text{cm}\times1\ \text{cm}$ strip, depending strongly on pencil grade and pressure. With $R\approx100\ \text{k}\Omega$, $l=0.10\ \text{m}$, width $0.010\ \text{m}$ and thickness $1\times10^{-5}\ \text{m}$, $\rho\approx\frac{10^5\times10^{-7}}{0.10}=0.1\ \Omega\ \text{m}$ — millions of times that of copper; pure graphite is much lower (around $10^{-5}\ \Omega\ \text{m}$) because pencil "lead" is mixed with clay and the film is porous. Linear $R$–$l$ graphs with a small positive intercept (probe contact resistance) are the usual result and make a good discussion of systematic error.

🧪 Self-check quiz

0/5 correct

Q1. A resistor carries $0.20\ \text{A}$ when the potential difference across it is $6.0\ \text{V}$. Its resistance is

Q2. The V–I graph of a conductor that obeys Ohm's law is

Q3. The SI unit of resistivity is

Q4. A wire's length is doubled and its cross-sectional area is halved. Its resistance becomes

Q5. The material best suited to the heating element of a toaster is

← 3.1 Electric Charge, Current and Potential Difference3.3 Resistors in Series and in Parallel →