What Is Electrical Quantities? A Visual Guide for IGCSE Physics Students

Your phone, your charger, and the physics behind the glow

You plug your phone into the wall. The screen lights up, the battery icon starts crawling upward, and you go about your day. But something real is happening inside that cable. Tiny charged particles are flowing through the wire, carrying energy from the socket to your phone's battery. The physics behind that flow - charge, current, voltage, resistance, power - is exactly what your IGCSE Physics exam will test you on.

These concepts are called electrical quantities, and they're the foundation for every circuit question you'll face. The good news? Once you understand what each quantity actually means, the equations start to make sense on their own. You won't need to memorise them as disconnected formulas. You'll understand where they come from.

Key facts at a glance

• Charge (Q) is measured in coulombs (C). It's the "stuff" that flows through a circuit.
• Current (I) is the rate of flow of charge, measured in amperes (A). Formula: I = Q / t
• Potential difference (V) is the energy transferred per unit charge, measured in volts (V). Formula: V = W / Q
• Resistance (R) opposes current flow, measured in ohms (Ω). Formula: R = V / I
• E.m.f. is the energy per unit charge supplied by a source such as a battery or cell.
• Power (P) is the rate of energy transfer: P = IV = I²R = V²/R
• Energy (E) transferred in a circuit: E = IVt, measured in joules (J).
• Kilowatt-hour (kWh) is the practical unit of energy used on electricity bills.

The essential equations

You'll use these equations constantly, so it helps to see them together. But don't just memorise the table. Read through the sections below, and you'll see how each formula connects to a real physical idea.

Charge and current: what actually flows

Think of a wire as a narrow pipe, and the charge as water flowing through it. In a metal conductor, the "water" is really free electrons - tiny negatively charged particles that drift through the metal when a circuit is complete. The total charge they carry is measured in coulombs.

Current is simply how fast that charge moves past a given point in the circuit. The equation is:

I = Q / t

If 60 coulombs of charge pass through a wire in 30 seconds, the current is I = 60 / 30 = 2 A. Two amperes. Straightforward enough.

There's one detail that trips students up regularly. Conventional current flows from positive to negative around the external circuit, but electrons actually move the other way, from negative to positive. Both conventions are correct; they describe the same physical process from different perspectives. Your IGCSE exam may ask about either one, so make sure you know both.

Potential difference: the energy behind every circuit

Current tells you how much charge flows. Potential difference tells you how much energy that charge carries. More precisely, potential difference (p.d.) is the energy transferred per coulomb of charge as it passes through a component:

V = W / Q

If a lamp transfers 100 joules of energy when 20 coulombs pass through it, the p.d. across the lamp is V = 100 / 20 = 5 V. Each coulomb gave up 5 joules of energy as light and heat.

A voltmeter measures potential difference. Unlike an ammeter, it goes in parallel across the component you're examining. It samples the energy difference between two points without significantly diverting the current.

Why does this matter practically? Because every component in a circuit uses up some of the energy the battery provides. The p.d. across each component tells you how much energy is converted there. In a series circuit, the individual p.d. values add up to match the total e.m.f. of the supply. That's energy conservation at work, and it's a principle the Cambridge examiners love to test.

Resistance: what slows the flow down

Some components let current through easily. Others put up a fight. Resistance measures how strongly a component opposes the flow of current, and it's measured in ohms (Ω).

R = V / I

This is Ohm's law rearranged. For a conductor at constant temperature, the current through it is directly proportional to the p.d. across it. Double the voltage, double the current. The ratio V / I stays constant, and that constant is the resistance.

Not every component behaves this neatly, though. A filament lamp gets hotter as current increases, which raises its resistance. Its I-V graph curves upward rather than staying straight. A diode only allows current in one direction, blocking reverse flow almost completely. These non-ohmic components come up regularly in IGCSE exam questions, usually asking you to sketch or interpret their characteristic I-V graphs.

Electromotive force and internal resistance

Your battery doesn't just push charge around the circuit. It also has to push charge through itself. The total energy per coulomb that the battery supplies is called the electromotive force (e.m.f.). Don't let the name mislead you: e.m.f. isn't actually a force. It's a voltage, measured in volts, just like p.d.

Here's the key distinction. E.m.f. is the energy supplied per coulomb by the source. P.d. is the energy converted per coulomb in an external component. If the battery were perfect, with zero internal resistance, the e.m.f. would equal the terminal p.d. exactly. But real batteries aren't perfect.

If you're sitting the Extended paper, you could be asked to calculate e.m.f. and internal resistance from experimental data. A common setup involves measuring the terminal p.d. at several different currents and plotting V against I. The y-intercept gives the e.m.f., and the negative gradient gives the internal resistance. It's a neat experiment that ties the theory to real measurements.

Power: how fast energy gets used up

Power measures the rate at which energy is transferred. A 60 W light bulb converts 60 joules of electrical energy into light and heat every second. Simple as that.

The base equation is:

P = IV

But you can substitute Ohm's law to get two more versions that are just as useful:

• P = I²R - useful when you know current and resistance but not voltage
• P = V²/R - useful when you know voltage and resistance but not current

These three forms appear in different problem types. The trick is choosing the right one for the information you've been given. If the question hands you V and I, use P = IV. If it gives you I and R, go with P = I²R. If you have V and R, pick P = V²/R. Choosing the direct formula saves you an extra calculation step and reduces the chance of a slip.

Worked example 1: Choosing the right power equation

A 12 V car headlamp has a resistance of 3 Ω. Calculate the power dissipated.

1. You know V = 12 V and R = 3 Ω, but not I.
2. Use P = V²/R = 12² / 3 = 144 / 3 = 48 W.

You could also find I first (I = V/R = 12/3 = 4 A) and then use P = IV = 4 x 12 = 48 W. Both routes give the same answer, but the direct route is quicker.

Electrical energy and the kilowatt-hour

Energy is the total amount of electrical work done, while power is the rate. They're connected by time:

E = IVt

Or equivalently, E = Pt. Energy is measured in joules, and one joule is one watt for one second.

Worked example 2: Calculating energy in joules

An electric heater rated at 2000 W runs for 45 minutes. How much energy does it transfer?

1. Convert time to seconds: 45 minutes = 2700 s
2. E = Pt = 2000 x 2700 = 5,400,000 J = 5400 kJ

That's a huge number. And this is exactly why electricity companies don't bill you in joules. They use the kilowatt-hour (kWh) instead. One kilowatt-hour is the energy transferred when a 1 kW appliance runs for 1 hour:

1 kWh = 1000 W x 3600 s = 3,600,000 J = 3.6 MJ

For the heater above: 2 kW x 0.75 hours = 1.5 kWh. Much easier to work with than millions of joules, right?

Worked example 3: Cost of electricity

An oven rated at 3 kW is used for 2 hours. If electricity costs 15p per kWh, what is the cost?

1. Energy = 3 kW x 2 h = 6 kWh
2. Cost = 6 x 15p = 90p

These cost calculations are tested frequently. They're not difficult, but you need to be careful with your units. Stay in kW and hours throughout, and you won't go wrong.

Common pitfalls and how to dodge them

Test yourself

Try these questions before looking at the answers. They reflect the kind of calculations you'll meet on your IGCSE Physics paper, and working through them yourself is the single best way to check whether you've truly understood the concepts.

1. A current of 0.5 A flows through a resistor for 4 minutes. How much charge passes through it?
2. A battery has an e.m.f. of 9 V. When connected to a circuit, the current is 2 A and the terminal p.d. is 7 V. What is the internal resistance of the battery? (Extended)
3. A 920 W microwave runs for 5 minutes. Calculate the energy used in (a) joules and (b) kilowatt-hours.
4. A 6 V battery is connected to a 4 Ω resistor. Find the current, the power dissipated, and the energy transferred in 2 minutes.
5. Electricity costs 14p per kWh. How much does it cost to run a 3 kW oven for 1.5 hours?