Summary — What This Chapter Covers
- Electric current is a flow of charge carriers; charge is quantised in multiples of e = 1.60 × 10−19 C.
- Q = It and, for a conductor, I = Anvq.
- Potential difference = energy transferred per unit charge: V = W/Q.
- Power P = VI = I2R = V2/R.
- Resistance R = V/I; Ohm's law holds for a metallic conductor at constant temperature.
- I–V characteristics of a metal wire, a filament lamp and a semiconductor diode.
- Resistivity: R = ρL/A. LDR resistance falls as light intensity rises; thermistor resistance falls as temperature rises.
1. Electric Current
Electric current
A flow of charge carriers
An electric current is a flow of charge carriers: electrons in metals, ions in electrolytes. The charge on any carrier is quantised: it is always a whole-number multiple of the elementary charge e = 1.60 × 10−19 C.
Current is the rate of flow of charge, so Q = It; the coulomb is one ampere second.
In a conductor of cross-sectional area A with n charge carriers per unit volume (the number density), each of charge q drifting at average speed v, the current is I = Anvq. Metals have a very large n, so the drift speed is tiny (fractions of a millimetre per second).
Current
Q = It I = Anvq
Worked Example
A current of 2.5 A flows for 3.0 minutes. Find the charge that passes and the number of electrons this represents.
Worked Example
A copper wire of cross-sectional area 1.0 mm2 carries 2.0 A. Copper has 8.5 × 1028 free electrons per m3. Find the drift speed.
Practice Question
The same current flows through a wire that narrows to half its cross-sectional area. What happens to the drift speed in the narrow part?
2. Potential Difference and Power
P.d. and power
Energy per unit charge
The potential difference across a component is the energy transferred (from electrical to other forms) per unit charge passing through it: V = W/Q. One volt is one joule per coulomb.
Power is energy per unit time: P = W/t = VQ/t = VI. Combining with V = IR gives P = I2R and P = V2/R.
P.d. and power
V = W/Q P = VI = I2R = V2/R
Worked Example
A 2.4 kW kettle runs on a 230 V supply. Find the current and its resistance.
Practice Question
A charge of 30 C passes through a lamp with a p.d. of 12 V across it. How much energy is transferred? If this takes 60 s, what is the power?
3. Resistance and Resistivity
Resistance
R = V/I and R = ρL/A
Resistance is defined as the ratio of the p.d. across a component to the current in it: R = V/I (ohm, Ω = V A−1).
Ohm's law: for a metallic conductor at constant temperature, the current is proportional to the p.d. across it.
- Metallic conductor at constant temperature: straight line through the origin (constant R).
- Filament lamp
resistance increases
- Semiconductor diode
Resistivity
LDR
thermistor
Resistance
R = V/I R = ρL/A