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Electric Current and Resistance · Lesson 3 of 6

Ohm's Law and Resistivity

15 min read4 ideas6 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 4
Ohms law
Maths you need

Resistance

Put a potential difference \(V\) across a piece of wire and a current \(I\) flows through it. The ratio \(V/I\) is the resistance \(R\) of that piece. It says how many volts it takes to push one ampere through. Its unit is the ohm, \(1\,\Omega = 1\,\text{V/A}\). The reciprocal \(G = 1/R\) is the conductance, measured in siemens, S.

The ratio \(V/I\) can be worked out for any conductor at any current. The law says more. For a metal wire kept at one temperature, double \(V\) and \(I\) doubles; treble \(V\) and \(I\) trebles. So \(R\) stays the same whatever \(V\) is. This is Ohm's law, and a conductor that obeys it is an ohmic conductor.

law 6.1
Ohm's law

The current through a metallic conductor is proportional to the potential difference across it. This holds while its temperature and other physical conditions stay the same: \(V = IR\), with \(R\) a constant.

RememberOhm's law
\[ V = IR \]
Holds whenan ohmic conductor, such as a metal wire, at a fixed temperature; \(R = V/I\) defines the resistance of any conductor at one current
Symbols\(V\) potential difference across the conductor, \(I\) current through it, \(R\) its resistance
Units\(V\) in V, \(I\) in A, \(R\) in Ω
Potential difference against current for two wiresA graph of the potential difference V across a wire against the current I through it, for two wires A and B. Both graphs are straight lines through the origin, so each wire obeys Ohm's law. Wire A reaches 8 volts at 2 amperes and wire B 4 volts at 4 amperes. The slope V/I is the resistance: 4 ohms for A and 1 ohm for B. The less steep line belongs to the better conductor.12345246810OI (A)V (V) A B
Fig 3.1The potential difference across two wires A and B against the current through them. Both lines are straight and pass through the origin, so both wires obey Ohm's law. The slope of each line is its resistance.

Read Fig 3.1. Wire A has \(8\,\text{V}\) across it at \(2\,\text{A}\), so \(R_A = 8/2 = 4\,\Omega\). Wire B has \(4\,\text{V}\) at \(4\,\text{A}\), so \(R_B = 4/4 = 1\,\Omega\). At any one voltage B carries four times the current. B is the better conductor, and its line is the less steep one.

Checkpoint · numericUnanswered

A heater coil draws \(5\,\text{A}\) from the \(220\,\text{V}\) mains. What is its resistance, in ohms?

Ω± 2 %
CheckpointUnanswered

On a graph of current \(I\) (up) against potential difference \(V\) (across), wire P gives a steeper straight line than wire Q. Which wire has the larger resistance?

Resistivity and the shape of a wire

Material and shape

Two wires of the same copper can have very different resistances. A longer wire has more metal for each electron to drift through, so \(R \propto L\). A thicker wire is like many thin wires side by side, so \(R \propto 1/A\). Together, \(R = \rho L/A\). The constant \(\rho\) is the resistivity of the material.

RememberResistance of a uniform conductor
\[ R = \dfrac{\rho L}{A} \]
Holds whena uniform conductor with the current along its length \(L\), spread evenly over its section \(A\)
Symbols\(R\) resistance, \(\rho\) resistivity of the material, \(L\) length along the current, \(A\) area of the section across the current
Units\(\rho\) in Ω m, \(L\) in m, \(A\) in m²

The resistivity is the material speaking, not the sample. Silver has \(1.6 \times 10^{-8}\,\Omega\,\text{m}\) and copper \(1.7 \times 10^{-8}\,\Omega\,\text{m}\), so house wiring is copper. Nichrome, the alloy of heater coils, has about \(1.0 \times 10^{-6}\,\Omega\,\text{m}\). Silicon has about \(2 \times 10^3\,\Omega\,\text{m}\), and glass more than \(10^{10}\,\Omega\,\text{m}\).

ExampleA hundred metres of copper

A copper wire is \(100\,\text{m}\) long and has a section of \(1\,\text{mm}^2\). Find its resistance, and the current when \(3.4\,\text{V}\) is put across it.

Try it on paper first. The solution is one tap away when you are ready.

For one sample, \(L\) and \(A\) depend on which way the current goes. Take a metal block \(4\,\text{cm}\) long, \(2\,\text{cm}\) deep and \(1\,\text{cm}\) high, of resistivity \(4 \times 10^{-6}\,\Omega\,\text{m}\). Current along its length meets \(L = 4\,\text{cm}\) and \(A = 2\,\text{cm}^2\). Current across its height meets \(L = 1\,\text{cm}\) and \(A = 8\,\text{cm}^2\).

One metal block, two ways through itA metal block 4 centimetres long, 2 centimetres deep and 1 centimetre high, of resistivity 4 times 10 to the minus 6 ohm metre. First chip, the long way: the current enters one 2 by 1 centimetre end face and leaves by the other, so the path is 4 centimetres long and 2 square centimetres wide, and R = 8 times 10 to the minus 4 ohm. Second chip, the short way: the current enters the 4 by 2 centimetre top face and leaves by the bottom, a path 1 centimetre long and 8 square centimetres wide, and R = 5 times 10 to the minus 5 ohm, 16 times less. The shaded faces are where the current enters and leaves. 4 cm 1 cm 2 cm I I I I L = 4 cm, A = 2 cm²: R = 8 × 10⁻⁴ Ω
Fig 3.2One metal block, two paths. The shaded faces are where the current enters and leaves. Switch between the chips and compare the length of the path and the area it crosses.
Checkpoint · numericUnanswered

In Fig 3.2, find the resistance the long way divided by the resistance the short way.

± 2 %
Ohms law inside the metal

Ohm's law inside the metal

The last lesson gave two results. The drift speed is \(v_d = eE\tau/m\), and the current density is \(J = nev_d\). Put the first into the second. The current density at a point is proportional to the field at that point. This is Ohm's law, derived from the drift of electrons.

Deriving \(V = I\,\dfrac{\rho L}{A}\) step 1 of 4
1

\[ J = nev_d = \dfrac{ne^2\tau}{m}\,E = \sigma E \]

Why. put \(v_d = eE\tau/m\) into \(J = nev_d\), and name the constant the conductivity \(\sigma\).

RememberOhm's law inside a conductor
\[ \begin{gathered} \vec J = \sigma \vec E \\ \rho = \dfrac{1}{\sigma} = \dfrac{m}{ne^2\tau} \end{gathered} \]
Holds whenan ohmic material at a fixed temperature; \(\tau\) is the relaxation time
Symbols\(\vec J\) current density, \(\vec E\) field at the same point, \(\sigma\) conductivity, \(\rho\) resistivity, \(n\) free electrons per unit volume, \(e\) and \(m\) charge and mass of the electron
Units\(\sigma\) in S/m, \(\rho\) in Ω m

The formula says what makes a good conductor: many free electrons, and a long time between collisions. Put in copper's numbers. \(ne^2\tau\) \(= 8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2\) \(\times 2.5 \times 10^{-14}\) \(= 5.44 \times 10^{-23}\). So \(\rho = 9.1 \times 10^{-31}/(5.44 \times 10^{-23})\) \(= 1.7 \times 10^{-8}\,\Omega\,\text{m}\), the measured value.

Checkpoint · numericUnanswered

A metal has conductivity \(\sigma = 5 \times 10^7\,\text{S/m}\). A field of \(0.1\,\text{V/m}\) acts inside a wire of this metal with section \(1\,\text{mm}^2\). Find the current, in amperes.

A± 2 %
Stretching a wire
Maths you need

Stretching a wire

Draw a wire out to a greater length and no metal is added or lost. So its volume \(AL\) stays the same, and the section shrinks as the length grows. If the length becomes \(k\) times as long, the section becomes \(1/k\) of what it was. Both changes raise the resistance: \(R = \rho (kL)/(A/k) = k^2 \rho L/A\).

Stretching a wireAbove, a wire of length L_0 and resistance 8 ohms. Below, the same wire stretched by a slider to k times its length. Its volume stays the same, so its cross-section falls to 1/k of the original and it is drawn thinner. The resistance grows twice over: once because the path is k times longer and once because it is k times narrower, so R = k squared times 8 ohms: 18 ohms at k = 1.5 and 32 ohms at k = 2. L0, A0, 8 Ω L L/L0 = 1.50, A/A0 = 0.67 R = 1.50² × 8 Ω = 18.0 Ω
SimAn \(8\,\Omega\) wire, stretched by the slider. Watch the wire grow longer and thinner at once. The resistance grows as the square of the stretch factor.

An \(8\,\Omega\) wire stretched to twice its length has \(2^2 \times 8 = 32\,\Omega\). Stretched to one and a half times, it has \(1.5^2 \times 8 = 18\,\Omega\). A stretch of only 10 % raises the resistance by 21 %, since \(1.1^2 = 1.21\).

Checkpoint · numericUnanswered

A wire of resistance \(5\,\Omega\) is drawn out until its length is three times the original. Find the new resistance, in ohms.

Ω± 2 %
CheckpointUnanswered

A wire is drawn out until its radius is half the original. By what factor does its resistance change?

Cutting a wire into pieces and folding it double are the other shape changes; the next chapter, on circuits, treats them with resistors joined together.

Story · Georg Simon Ohm, 1827
A law nobody wanted

Georg Simon Ohm was a school teacher of mathematics in Cologne. He made his own wires of different lengths and thicknesses and measured the current each one passed. In 1827 he published the law in a book in Berlin. It was received coldly, and Ohm spent years in poorly paid teaching. In 1841 the Royal Society of London gave him its Copley Medal, and the unit of resistance now carries his name.

Exam traps
  1. Taking the slope of every graph as \(R\).
    On a V-I graph the slope is \(R\); on an I-V graph it is \(1/R\). Read the axes first.
  2. Saying that \(R = V/I\) is Ohm's law.
    \(R = V/I\) defines resistance for any conductor. Ohm's law is the claim that this ratio stays constant.
  3. Doubling the resistance when a wire is stretched to twice its length.
    The section halves as well, so the resistance becomes four times as large.
  4. Treating resistivity as a property of the wire.
    \(\rho\) belongs to the material; \(R = \rho L/A\) belongs to the sample.