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GCSE/Combined Science/CCEA

P2.3Electricity: current, p.d., resistance, Ohm’s law; series and parallel circuits

Notes

Electricity — current, voltage, resistance and circuits

Basic electrical quantities

Current (I): the rate of flow of electric charge. Measured in amperes A. Charge flow = current × time: Q = I × t.

Potential difference (p.d.) / Voltage (V): the energy transferred per unit charge. Measured in volts (V). V = E/Q (energy per unit charge).

Resistance (R): how much a component opposes the flow of current. Measured in ohms (Ω).

Ohm's Law

V = I × R (voltage = current × resistance)

Where V is in volts, I is in amperes, A is in ohms.

A component that obeys Ohm's Law at constant temperature is called an ohmic conductor — its I-V graph is a straight line through the origin.

Rearranged: I = V/R; R = V/I.

Example: A bulb with resistance 24 Ω carries a current of 0.5 A. V = 0.5 × 24 = 12 V.

Series circuits

  • Components connected in a single loop.
  • Current is the same everywhere: I_total = I₁ = I₂ = I₃.
  • Voltages add up: V_total = V₁ + V₂ + V₃.
  • Resistance adds up: R_total = R₁ + R₂ + R₃.
  • If one component fails (open circuit), the whole circuit stops.

Parallel circuits

  • Components connected in separate branches.
  • Current splits at junctions: I_total = I₁ + I₂ + I₃.
  • Voltage is the same across each branch: V_total = V₁ = V₂ = V₃.
  • Total resistance is LESS than any individual resistance: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃.
  • If one branch fails, the others continue to work.

Non-ohmic components

Filament bulb: resistance increases as temperature increases (the hotter the filament, the more resistance). I-V graph is a curve (not straight).

Diode: only allows current to flow in one direction. Very high resistance in reverse direction.

LDR (light-dependent resistor): resistance decreases as light intensity increases.

Thermistor: resistance decreases as temperature increases. Used in temperature sensors.

Power in circuits

P = I × V (power = current × voltage) Also: P = I² × R; P = V²/R.

Electrical energy: E = P × t = I × V × t.

Units: power in watts (W); energy in joules (J).

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Practice questions

Try each before peeking at the worked solution.

  1. Question 16 marks

    Ohm's Law calculations

    (a) A resistor has a resistance of 15 Ω. Calculate the current through it when connected to a 9 V battery. (2 marks)
    (b) A current of 2 A flows through a component when connected to a 12 V supply. Calculate the resistance of the component. (2 marks)
    (c) A resistor of 8 Ω carries a current of 3 A. Calculate the voltage across it. (2 marks)

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  2. Question 28 marks

    Series vs parallel circuits

    Two resistors, R₁ = 6 Ω and R₂ = 4 Ω, are connected to a 12 V battery.

    (a) Calculate the total resistance and total current when the resistors are connected in SERIES. (4 marks)
    (b) Calculate the total resistance and total current when the resistors are connected in PARALLEL. (4 marks)

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  3. Question 37 marks

    I-V graphs — non-ohmic components

    (a) Sketch an I-V graph for an ohmic conductor and a filament bulb on the same axes. Label each curve. (4 marks)
    (b) Explain why the filament bulb's I-V graph is not a straight line. (3 marks)

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  4. Question 47 marks

    Power in circuits

    A 40 W light bulb is connected to a 230 V mains supply.

    (a) Calculate the current through the bulb. (2 marks)
    (b) Calculate the resistance of the bulb when operating at this power. (2 marks)
    (c) Calculate the energy used by the bulb in 3 hours. Give your answer in joules. (3 marks)

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Flashcards

P2.3 — Electricity: current, p.d., resistance, Ohm's law; series and parallel circuits

8-card SR deck for CCEA GCSE Double Award Science (GDA2017) topic P2.3

8 cards · spaced repetition (SM-2)