Resistance and resistivity
The resistance R of a conductor is the ratio of the potential difference V across it to the current I through it: R = V ÷ I. For a uniform wire the resistance is proportional to its length and inversely proportional to its cross-sectional area.
- resistivity (Ω m)
Measure the current for different voltages across a resistor.
Choose the resistivity, length and diameter of a wire. Find its resistance.
Combining resistors
EMF and internal resistance
The electromotive force (e.m.f.) E of a cell is the energy it gives to each coulomb of charge that passes through it. A real cell has internal resistance r, so when a current I flows, some voltage is lost inside the cell. The terminal p.d. is V = E − Ir. The lost voltage is why a battery’s voltage drops when the current is large.
A cell of e.m.f. 1.5 V and internal resistance 0.5 Ω supplies a current of 0.6 A. What is the voltage at its terminals?
- Voltage lost inside the cell: Ir = 0.6 × 0.5 = 0.3 V.
- V = E − Ir.
Answer: The terminal voltage is 1.2 V.
Electrical energy and power
Choose the power of an appliance, how long it is used and the price of a unit (1 kWh). Find the energy used and the cost.
Domestic wiring and safety
| Wire | Colour | Job |
|---|---|---|
| Live | Brown | Carries current to the appliance; at high voltage |
| Neutral | Blue | Completes the circuit |
| Earth | Green and yellow | Connects the metal case to the ground, so a fault current flows to earth and blows the fuse instead of shocking you |
A fuse is a thin wire that melts and breaks the circuit if the current is too high. Choose a fuse a little above the normal current (3 A, 5 A or 13 A). A circuit breaker does the same job and can be reset. Switches are placed in the live wire. Never touch electrical equipment with wet hands.
Key terms
- Resistivity
- A property of a material: the resistance of a 1 m length with a cross-section of 1 m².
- e.m.f.
- The energy given to each coulomb of charge by a source.
- Internal resistance
- The resistance of the source itself.
- Kilowatt-hour
- The energy used by a 1 kW appliance in 1 hour: 3.6 MJ.
Summary
- R = ρL/A; series: add R; parallel: add reciprocals.
- Terminal voltage V = E − Ir.
- P = VI = I²R = V²/R; energy E = Pt; cost = kWh × tariff.
- Fit a fuse in the live wire and connect the earth wire to metal cases.