Electricity and magnetism · GCSE Physics

Current, voltage and resistance

Teacher-written GCSE Physics revision on current, potential difference and resistance: I = Q/t, V = E/Q, R = V/I, Ohm’s law at constant temperature, and the units ampere, volt and ohm.

UNDERSTANDRETRIEVEREMEMBER
THE MEMORY HOOK
Current is the flow of charge. Potential difference is the energy per charge. Resistance is how hard it is to make that charge flow. R = V/I.

The important bits

What you need to know

  1. 1

    Current I = Q / t. Current is in amperes (A), charge in coulombs (C), time in seconds (s). 1 A = 1 C/s.

  2. 2

    Conventional current is from positive to negative. In metals the moving charges are electrons, which drift the other way.

  3. 3

    Potential difference V = E / Q. Potential difference is in volts (V): 1 V = 1 J/C. It is the energy transferred per coulomb of charge.

  4. 4

    Resistance R = V / I, in ohms (Ω). A larger resistance means a smaller current for the same potential difference.

  5. 5

    Ohm’s law holds for an ohmic conductor at constant temperature: I is proportional to V, so a graph of I against V is a straight line through the origin.

  6. 6

    An ammeter is connected in series and has very low resistance. A voltmeter is connected in parallel with the component and has very high resistance.

  7. 7

    Charge is conserved: current is not used up in a resistor. Energy is transferred, so potential difference is dropped across the resistor.

  8. 8

    Rearrange before substituting: Q = It, E = VQ, V = IR. Keep milliamperes as 10⁻³ A and kilohms as 10³ Ω.

Quotations worth analysing

Short evidence. Real method.

Current is the rate of flow of charge.
AQA GCSE Physics, I = Q/t

Rate means divide by time. A current of 2.0 A means 2.0 C passing a point every second, not 2.0 V and not 2.0 Ω.

Potential difference is the work done per coulomb of charge.
AQA GCSE Physics, V = E/Q

A 12 V battery gives 12 J to each coulomb that passes through it. That energy is then transferred in the circuit.

Ohm’s law: current is directly proportional to potential difference at constant temperature.
GCSE Physics ohmic conductors

The constant-temperature clause matters. A filament lamp heats up, so it is not ohmic even though it is a metal wire.

Go deeper

Do not let current get “used up”

Charge is a conserved stuff. If 3.0 A enters a resistor, 3.0 A leaves it. What the resistor does is transfer energy from the electrical pathway to a thermal store, so the potential difference across it tells you how many joules each coulomb loses there. Students draw arrows that shrink along a series circuit and lose the conservation mark. Measure current with an ammeter in the loop; measure potential difference across the component. If the ammeter is in parallel it shorts the component; if the voltmeter is in series it blocks the current because its resistance is huge. Those two wiring rules are as important as the equations. Practise sketching the meters before you calculate R = V/I.

Go deeper

Ohm’s law is a special case, not a personality trait of all circuits

For a resistor at steady temperature, doubling V doubles I, and R stays constant. That is Ohm’s law. The graph of I against V is a straight line; the gradient is 1/R if I is on the y-axis. If the component heats, R rises and the graph curves. Thermistors and LDRs change R because temperature or light changes, not because Ohm’s law failed in a mysterious way: at any instant R is still V/I, but that value is not constant. Write the definition R = V/I for any component, then say whether R is constant. That two-step habit stops you claiming a diode “has no resistance” when the current is almost zero in reverse — it has a very large resistance.

WORKED EXAMPLE

See the idea in action

A charge of 30 C passes a point in 12 s. Current I = Q/t = 30 / 12 = 2.5 A. The same charge transfers 90 J in a resistor. Potential difference V = E/Q = 90 / 30 = 3.0 V. Resistance R = V/I = 3.0 / 2.5 = 1.2 Ω. Checking: E = VIt = 3.0 × 2.5 × 12 = 90 J, which matches. If the 12 s had been left as 12 minutes, I would have been 60 times too small and R 60 times too big.

Exam technique

Turn knowledge into marks

Write I = Q/t, V = E/Q or R = V/I, convert mA and minutes, substitute with units, then calculate. Put the ammeter in series and the voltmeter in parallel. Say “at constant temperature” when you quote Ohm’s law.

Common mistakes

Do not give these marks away

  1. 01

    Putting an ammeter in parallel or a voltmeter in series, or saying current is used up in a resistor.

  2. 02

    Leaving time in minutes in I = Q/t, or mixing mA with A in R = V/I.

  3. 03

    Claiming every component obeys Ohm’s law, including lamps, diodes, LDRs and thermistors.

QUICK RETRIEVAL

A 6.0 V battery sends a current of 0.30 A through a resistor. What is the resistance?

A0.050 Ω

B1.8 Ω

C20 Ω

D18 Ω

Show the answer

20 Ω. R = V/I = 6.0 / 0.30 = 20 Ω. 1.8 Ω is V × I, which is power, not resistance. 0.050 Ω is I/V upside down.

Quick questions

If this is the bit you searched

What is the difference between current and potential difference?

Current is charge flowing per second (A). Potential difference is energy transferred per coulomb (V). You need both to find resistance or power.

Does Ohm’s law always apply?

Only for an ohmic conductor at constant temperature, when I is proportional to V. Lamps, diodes, thermistors and LDRs are non-ohmic.

How should meters be connected?

Ammeter in series with the component. Voltmeter in parallel with the component. Swap them and the readings, and sometimes the component, will be wrong.

Is electron flow the same as conventional current?

They are opposite in direction in a metal. GCSE circuit rules use conventional current, positive to negative, unless a question asks about electrons.