Electrical quantities

O Level / IGCSE Physics · Electricity and magnetism. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

There are two kinds of charge, positive and negative, measured in coulombs. Like charges repel and unlike charges attract. Rubbing two insulators together moves electrons from one to the other: the one that gains electrons becomes negative and the one that loses them becomes positive.

Current is the charge passing a point each second. In a metal it is a flow of free electrons from negative to positive, but conventional current is drawn from positive to negative.

The e.m.f. of a source is the work it does per coulomb in driving charge round the whole circuit. The potential difference across a component is the work done per coulomb passing through it. Resistance tells you how much p.d. is needed for each ampere of current.

Rules to remember

Common mistake

Saying that an object becomes positive because it gains positive charge. Only electrons move when solids are rubbed, so a positive object is one that has lost electrons.

Worked example

A charge of 15 C passes through a lamp in 30 s and 180 J of energy is transferred in the lamp. What is the resistance of the lamp?

  1. p.d. = work done ÷ charge = 180 ÷ 15 = 12 V.
  2. current = charge ÷ time = 15 ÷ 30 = 0.50 A.
  3. resistance = p.d. ÷ current = 12 ÷ 0.50 = 24 Ω.

Answer: 24 Ω

Practice questions

Try each one, then open the answer.

1. As the current through a filament lamp increases, its resistance

  1. A
    decreases, because the filament gets hotter
  2. B
    increases, because the filament gets hotter
  3. C
    stays constant, because it obeys Ohm's law
  4. D
    becomes zero
Show answer

Answer: B. The filament heats up; the metal ions vibrate more and impede the electrons, so resistance rises. This is why the I–V graph of a lamp curves.

2. A charge of 60 C passes a point in a circuit in 20 s. What is the current?

  1. A
    0.33 A
  2. B
    3.0 A
  3. C
    40 A
  4. D
    1200 A
Show answer

Answer: B. I = Q ÷ t = 60 ÷ 20 = 3.0 A.

3. A current of 250 mA flows through a lamp for 2.0 minutes. How much charge passes through the lamp?

  1. A
    0.50 C
  2. B
    30 C
  3. C
    500 C
  4. D
    30 000 C
Show answer

Answer: B. 250 mA = 0.25 A and 2.0 min = 120 s. Charge = current × time = 0.25 × 120 = 30 C. Leaving the current in mA gives 30 000 C; leaving the time in minutes gives 0.50 C.

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