Why a Bulb Draws a Surge of Current the Instant It Is Switched On

Switch on a fan and it starts almost at once, yet the electrons in its wires move slower than an ant crawls. The explanation lies in drift velocity, and the same picture explains resistance, why it changes with temperature, and why a cell never delivers its full emf to a circuit.

This lesson covers drift velocity, Ohm's law and resistivity, and emf, internal resistance and cell combinations.

**Drift velocity is the small average velocity that free electrons gain against an applied field, and it is related to current by $I=neA{v}_d$.

The picture.** Free electrons already dash about randomly at high speed, averaging zero velocity. An electric field E adds a slow drift, and between collisions, separated on average by the relaxation time $\tau$,

$$v_d=\frac{eE\tau }{m}$$

**Deriving $I=neA{v}_d$.** In time t, the electrons within a length $v_{d}t$ of wire pass a cross-section. That volume $A{v}_{d}t$ holds $nA{v}_{d}t$ electrons, carrying charge $neA{v}_{d}t$, so the current is $I=neA{v}_d$.

Worked example. A copper wire of cross-section 1.0 mm${}^2$ carries 1.5 A, with $n=8.5\times 10^{28}$ m${}^{-3}$:

$$v_d=\frac{I}{neA}=\frac{1.5}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 1.0\times 10^{-6}}=1.1\times 10^{-4}{\text{ m s}}^{-1}$$

That is about 0.1 mm each second.

An everyday example. Water in a long, full garden hose flows out the moment the tap opens, because the push travels along the hose even though each drop moves slowly.

The substance. The bulb lights at once because the field spreads through the wire almost at the speed of light — not because electrons race from the switch to the bulb.

**Substituting $v_d=eE\tau /m$ into $I=neA{v}_d$ gives Ohm's law with resistance $R=\rho l/A$, where resistivity $\rho =\frac{m}{n{e}^2\tau }$, and for metals resistance rises with temperature as $R_T=R_0(1+\alpha \Delta T)$.

The derivation.** With $E=V/l$,

$$I=neA\frac{eV\tau }{ml}\Rightarrow V=\left(\frac{m}{n{e}^2\tau }\right)\frac{l}{A}I=RI$$

Worked example 1 — copper's resistivity. With $n=8.5\times 10^{28}$ m${}^{-3}$ and $\tau =2.5\times 10^{-14}$ s:

$$\rho =\frac{9.1\times 10^{-31}}{8.5\times 10^{28}\times (1.6\times 10^{-19}{)}^2\times 2.5\times 10^{-14}}=1.7\times 10^{-8}\Omega \text{ m}$$

Temperature. Hotter ions vibrate more, collisions come sooner, $\tau$ falls and $\rho$ rises. Semiconductors behave the opposite way, because heating frees many more charge carriers.

Worked example 2 — a filament. A tungsten filament has $40\Omega$ at $20^{\circ }$C, with $\alpha =4.5\times 10^{-3}$ K${}^{-1}$. Using the linear formula as an estimate at $2500^{\circ }$C:

$$R=40(1+4.5\times 10^{-3}\times 2480)=486\Omega$$

On 230 V, the cold filament draws $230/40=5.75$ A, but the hot one only $230/486=0.47$ A.

An everyday example. Old filament bulbs usually failed at the moment of switching on, when the cold filament drew its largest current.

The substance. Resistivity is a property of the material; resistance also depends on the wire's shape.

**A cell of emf $\epsilon$ and internal resistance r delivers a terminal voltage $V=\epsilon -Ir$ while supplying current; in series emfs and internal resistances add, and identical cells in parallel keep the same emf but divide r by their number.

Key ideas:**

- emf is the work done per unit charge by the cell, equal to V only when no current flows
- While charging, $V=\epsilon +Ir$
- Series of n cells: $n\epsilon$ and $nr$; parallel of m identical cells: $\epsilon$ and $r/m$

Worked example 1. A cell of 1.5 V and $0.50\Omega$ drives a $2.5\Omega$ resistor:

$$I=\frac{1.5}{2.5+0.50}=0.50\text{ A}V=1.5-0.50\times 0.50=1.25\text{ V}$$

Worked example 2 — combining. Four such cells drive a $4.0\Omega$ load.

$$I_{\text{series}}=\frac{6.0}{4.0+2.0}=1.0\text{ A}{I}_{\text{parallel}}=\frac{1.5}{4.0+0.125}=0.36\text{ A}$$

An everyday example. A torch that takes cells end to end puts them in series to add their emfs for a brighter bulb.

The substance. Series wins when the load resistance is much larger than r, and parallel wins when the load is much smaller — the comparison, not a rule of thumb, decides.