Coaching Centre
Voltage, current and resistance
Electric charge and current in amps, voltage as energy per coulomb, using ammeters and voltmeters, resistance and what affects it, Ohm's law V = IR, I–V graphs for resistors and filament lamps, and the resistor colour code.
When you flick a light switch, the light comes on instantly, yet the electrons in the wire crawl along at less than a millimetre per second. They don't have to race from the power station: the wires are already full of electrons, and the switch just sets them all moving at once, like pushing marbles into a tube that is already full.
Current: the flow of charge
Metal wires contain huge numbers of free electrons, each with a tiny negative electric charge. Charge is measured in coulombs (C); one coulomb is the charge of about 6 billion billion (6.2 × 10¹⁸) electrons. A cell pushes these electrons around a closed circuit.
Current (I) is the amount of charge passing a point each second, measured in amperes or amps (A). One amp is one coulomb per second.
current (A) = charge (C) ÷ time (s)
I = Q ÷ t
Small currents are measured in milliamps (mA): 1000 mA = 1 A. A phone charging draws about 1 to 3 A; an LED might use 20 mA.
Electrons actually flow from the negative terminal round to the positive. Before electrons were discovered, scientists had agreed that current flows from positive to negative. This conventional current is still used on circuit diagrams, and it gives the right answers for everything you'll calculate.
Voltage: energy per charge
A cell or battery gives energy to every coulomb of charge that passes through it. The charge carries that energy around the circuit and hands it over to the components (heating a wire, lighting a lamp, turning a motor). Voltage (V), also called potential difference, is the energy transferred per coulomb, measured in volts (V).
voltage (V) = energy (J) ÷ charge (C)
A 1.5 V AA cell gives 1.5 J to each coulomb. A 9 V battery gives 9 J. Australian mains power is 230 V (the standard was changed from 240 V in 2000, though you may still see 240 V written on older appliances).
A useful picture is water in pipes: the voltage is like the pressure from a pump, the current is how much water flows past each second, and a narrow section of pipe is like a resistance. Like any model, it isn't perfect: water can leak out of a broken pipe, but charge can't leak out of a broken circuit; it just stops.
Measuring current and voltage
- An ammeter measures current. It goes in series, in the loop, so all the current flows through it. A good ammeter has almost no resistance, so it doesn't change the current it is measuring.
- A voltmeter measures the voltage across a component. It goes in parallel, connected to each end of the component, to compare the energy carried by the charge before and after it. A good voltmeter has a very high resistance so almost no current goes through it.
Connect the red (+) terminal of a meter towards the positive side of the battery. A multimeter can be switched to measure current, voltage or resistance.
Resistance
As electrons move through a wire they collide with the vibrating atoms, which slows them down and transfers energy as heat. This opposition to current is resistance (R), measured in ohms (Ω). The resistance of a wire is greater if it is:
- longer: there are more atoms to collide with
- thinner: there is less room for the electrons to flow side by side
- made of a poorer conductor, such as nichrome (used in toaster and heater elements) rather than copper
- hotter (for metals), because the atoms vibrate more and get in the way more often.
Resistors are components built to have a set resistance, used to control the current in electronic circuits.
Ohm's law
In 1827 the German teacher Georg Ohm found that, for a metal wire kept at the same temperature, the current is directly proportional to the voltage: double the voltage and the current doubles. The voltage divided by the current is the resistance.
voltage (V) = current (A) × resistance (Ω)
V = I R so I = V ÷ R and R = V ÷ I
A 12 V battery is connected to a 48 Ω resistor. I = V ÷ R = 12 ÷ 48 = 0.25 A (250 mA).
A torch globe draws 300 mA from a 4.5 V battery. First convert: 300 mA = 0.3 A. R = V ÷ I = 4.5 ÷ 0.3 = 15 Ω.
A current of 50 mA flows through a 220 Ω resistor. V = I × R = 0.05 × 220 = 11 V.
Always change mA to A and kΩ to Ω before you substitute. 1 kΩ = 1000 Ω.
An I–V graph shows how the current changes as the voltage is increased.
- A fixed resistor at constant temperature gives a straight line through the origin. It obeys Ohm's law and is called ohmic. A steeper line means more current for the same voltage, so a lower resistance.
- A filament lamp gives a curve that flattens out. As the current rises, the filament heats up to over 2000 °C, its atoms vibrate more, and its resistance increases. It is non-ohmic.
- Choose the 10 Ω resistor. Set the supply to several voltages from 0 to 12 V and press Record reading each time. What shape is the graph?
- Repeat with 47 Ω. How does the gradient compare? Explain using R = V ÷ I.
- Choose the Mystery resistor. Record at least four readings and use the graph to find its resistance. Then check your answer against the colour bands.
- Choose the 12 V lamp and record readings. Why does the graph curve? Does the lamp have a single resistance?
- Watch the power readout as you increase the voltage on the 10 Ω resistor. Why might a real resistor get hot or burn out?
Reading a resistor
Small resistors are too tiny to print numbers on, so they use coloured bands. The first two bands are digits, the third tells you how many zeros to add, and a fourth band (gold or silver) shows the tolerance (±5% or ±10%).
| Colour | Black | Brown | Red | Orange | Yellow | Green | Blue | Violet | Grey | White |
|---|---|---|---|---|---|---|---|---|---|---|
| Value | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
A resistor has bands yellow, violet, red, gold.
Yellow = 4, violet = 7, so the digits are 47.
Red = 2, so add two zeros: 4700 Ω = 4.7 kΩ.
Gold = ±5%, so the real value is between 4465 Ω and 4935 Ω.
Test yourself
A charge of 30 C passes through a lamp in 60 s. The current is
I = Q ÷ t = 30 ÷ 60 = 0.5 A.
A 9 V battery gives each coulomb of charge
1 volt = 1 joule per coulomb.
The current through a 20 Ω resistor is 0.6 A. The voltage across it is
V = I R = 0.6 × 20 = 12 V.
Which change would decrease the resistance of a copper wire?
A thicker wire gives more room for current, so less resistance.
The I–V graph for a filament lamp curves because
Hotter metal atoms vibrate more, so the resistance rises and the current increases less for each extra volt.
Label the parts of the circuit.
Sort the changes to a metal wire.
Current is measured in with an connected in . Voltage is the energy carried per of charge. Resistance is measured in .
A student measures the current through a component at different voltages: 2 V, 0.10 A; 4 V, 0.20 A; 6 V, 0.30 A; 8 V, 0.40 A. Calculate the resistance, state whether the component obeys Ohm's law, and describe how the results would be different if the component were a filament lamp.
For every reading, R = V ÷ I gives 20 Ω (for example 2 ÷ 0.10 = 20 Ω and 8 ÷ 0.40 = 20 Ω). Because the resistance stays the same, the current is directly proportional to the voltage, and an I–V graph would be a straight line through the origin, so the component obeys Ohm's law (it is ohmic). If it were a filament lamp, the current would increase by smaller and smaller amounts as the voltage rose, giving a curve that flattens out. This is because a bigger current heats the filament; its metal atoms vibrate more and collide with electrons more often, so the lamp's resistance increases and V ÷ I would be larger at higher voltages.