Electric current, resistance and Ohm's law

The second half of Physics II is about circuits, and all circuits rest on the four formulas of this week: $I = dQ/dt$, $J = I/A$, $\Delta V = IR$ and $R = \rho L/A$. The questions on this topic are usually short calculations: the resistance of a wire, the number of electrons from a charge, ratio questions like “by what factor does the resistance change if the wire is stretched”. These are easy points; almost all the lost points come from unit conversion and confusing diameter with radius.

Golden rules:

1. Current = charge passing per unit time. $1\ \text{A} = 1\ \text{C/s}$. Convert minutes to seconds.
2. The direction of current is the direction of motion of positive charge. In a metal the electrons move opposite to the current.
3. $R = \rho L/A$: proportional to length, inversely proportional to cross-sectional area. Area grows with the square of the diameter.
4. $\Delta V = IR$ can always be written as a definition; a material is “ohmic” only if $R$ is constant.
5. To convert mm² to m², multiply by $10^{-6}$. Not $10^{-3}$.

Electric current

The net charge passing through a surface per unit time is called the electric current:

$I_{\text{avg}} = \frac{\Delta Q}{\Delta t}, \qquad I = \frac{dQ}{dt}$

Its unit is the ampere: $1\ \text{A} = 1\ \text{C/s}$. Current is a scalar quantity but it has a direction: by convention, the direction positive charges would flow. In a copper wire the moving charges are electrons, so the electrons move opposite to the direction of the current. The result does not change: a current of electrons going right is the same current as positive charge going left.

If the current changes with time, there are two connections:

• If the charge function is given, differentiate: $I(t) = dq/dt$.
• If the current function is given, integrate: $Q = \int I\,dt$.

PITFALL · average or instantaneous? In the question ”$q(t)$ is given, what is the current at $t = 2$ s?”, if you compute $q(2)/2$ you get the average current. The instantaneous current is the derivative. If the question says “instantaneous”, differentiate.

How many electrons?

Each electron carries $e = 1.60 \times 10^{-19}$ C. If a charge $Q$ is made of $N$ electrons:

$N = \frac{Q}{e} = \frac{I\,\Delta t}{e}$

Example. A current of 0.50 A flows through a wire for 10 minutes. How many electrons pass?

Δt = 10 min = 600 s
Q  = I Δt = 0.50 × 600 = 300 C
N  = Q / e = 300 / 1.60e-19 = 1.9 × 10^21 electrons

Example (varying current). The discharge current of a capacitor is $I(t) = 4.0\,e^{-t/2.0}$ A ($t$ in seconds). How much charge flows in total?

$Q = \int_0^{\infty} 4.0\,e^{-t/2.0}\,dt = 4.0 \times 2.0 = 8.0\ \text{C}$

Knowing that $\int_0^\infty e^{-t/\tau}dt = \tau$ reduces this type of question to one line. You will see the same thing in RC circuits in week 14.

Current density and drift

Current density is the current through a unit cross-sectional area:

$J = \frac{I}{A} \qquad [\text{A/m}^2]$

$J$ is a vector, and its direction is the direction of the current. If there is an electric field inside the wire, the charges slowly drift under its influence. You will see the drift speed and its microscopic model in detail in week 12. What matters for this week is this: if there is a current inside a conductor, the electric field is not zero. The rule ”$E = 0$ in a conductor” holds only in electrostatic equilibrium (static charges); a wire carrying current is not in equilibrium.

In most metals the current density is proportional to the field:

$\vec{J} = \sigma \vec{E}$

$\sigma$ is the conductivity, and its inverse $\rho = 1/\sigma$ is the resistivity. This expression is called microscopic Ohm’s law.

Resistance and Ohm’s law

Consider a uniform wire of length $L$ and cross-section $A$. If the potential difference between its ends is $\Delta V$, the field inside is $E = \Delta V/L$. Put this into $J = \sigma E$:

$\frac{I}{A} = \sigma \frac{\Delta V}{L} \quad\Rightarrow\quad \Delta V = \left(\frac{L}{\sigma A}\right) I = R\,I$

What is inside the parentheses is the resistance of the wire:

$R = \frac{\Delta V}{I}, \qquad R = \rho\frac{L}{A}$

Its unit is the ohm: $1\ \Omega = 1\ \text{V/A}$. The unit of resistivity is $\Omega\cdot\text{m}$.

Ohm’s law is really not a law of nature but a behavior of some materials: if $R$ stays constant, independent of the applied voltage, the material is ohmic. The $I$-$\Delta V$ graph is a straight line through the origin with slope $1/R$. Elements such as the diode are nonohmic: the graph is curved. You can still write the definition $R = \Delta V/I$, but this ratio changes with voltage.

The values are orders of magnitude. The $\rho$ you will use on the exam is given in the question.

Example: ohmic or not, from a table

The measurements of two elements are given. Element X: 0.10 A at 2.0 V, 0.20 A at 4.0 V, 0.30 A at 6.0 V. Element Y (a lamp filament): 0.20 A at 2.0 V, 0.40 A at 6.0 V.

X:  2.0/0.10 = 20 Ω   4.0/0.20 = 20 Ω   6.0/0.30 = 20 Ω   → R constant, ohmic
Y:  2.0/0.20 = 10 Ω   6.0/0.40 = 15 Ω                    → R increasing, not ohmic

The resistance of Y increases with voltage because the filament heats up (you will see the change of resistance with temperature in week 12). On the graph, X is a straight line through the origin and Y is a curve whose slope keeps decreasing. Be careful: on an $I$-$\Delta V$ graph the slope is $1/R$, not $R$. If the slope is decreasing, the resistance is increasing.

Example: resistance of a wire

An aluminum wire ($\rho = 2.8\times10^{-8}\ \Omega\cdot$m) with a diameter of 2.0 mm and a length of 100 m. What is its resistance?

r = d/2 = 1.0 mm = 1.0e-3 m
A = π r² = π (1.0e-3)² = 3.14e-6 m²
R = ρ L / A = (2.8e-8)(100) / (3.14e-6) = 0.89 Ω

PITFALL · the diameter is given. If the question says “diameter”, first divide by two. If you put the diameter in place of the radius, the area comes out 4 times larger and the resistance 4 times smaller. This wrong answer is always among the choices.

Ratio questions

A favorite exam type: “Two wires of the same material, one twice as long and twice as thick as the other, what is the ratio of the resistances?” Write the formula as a ratio and insert the numbers:

$\frac{R_A}{R_B} = \frac{L_A}{L_B}\cdot\frac{A_B}{A_A} = \frac{L_A}{L_B}\cdot\left(\frac{d_B}{d_A}\right)^2$

A stretched wire. When a wire is drawn out, material does not disappear: the volume $V = LA$ stays constant. If the length becomes $n$ times larger, the area becomes $1/n$ of its value:

$R' = \rho\frac{nL}{A/n} = n^2 R$

If the length doubles, the resistance becomes four times larger.

TIP · unit-value model. In a ratio question take $\rho = L = A = 1$, apply the change, and compute the new $R$. No calculator needed, and less room for error.

Example: current density and field

5.0 A flows through a copper wire with a cross-section of 2.0 mm². What are $J$ and $E$?

A = 2.0 mm² = 2.0e-6 m²
J = I/A = 5.0 / 2.0e-6 = 2.5e6 A/m²
E = ρ J = (1.7e-8)(2.5e6) = 4.3e-2 V/m

The field is very small: because copper is such a good conductor, a small field is enough even for a large current. The voltage across a 20 m piece of this wire is $EL = 0.85$ V.

Common mistakes