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O-Level Physics · Secondary 3 and 4

Current Electricity

Current electricity deals with the flow of charge around circuits. In this topic you learn what current, electromotive force, potential difference and resistance mean, how different components behave on an I-V graph, how resistance depends on the size of a wire, and how current and potential difference share out in series and parallel circuits and in potential dividers. These skills are needed for practical electricity at home and for electromagnetism, where currents produce forces and fields.

What the syllabus expects

  • State that current is the rate of flow of charge and use Q = It
  • Define e.m.f. and potential difference in terms of energy per unit charge and use V = W/Q
  • Define resistance and use R = V/I
  • Sketch and explain the I-V graphs of a metallic conductor at constant temperature, a filament lamp and a semiconductor diode
  • Recall and use R = ρL/A
  • Calculate the combined resistance of resistors in series and in parallel
  • Apply the rules for current and potential difference in series and parallel circuits
  • Describe the use of thermistors and light-dependent resistors in potential divider circuits

Charge, current, e.m.f. and potential difference

Electric charge is measured in coulombs (C). Current is the rate of flow of charge, I = Q/t, measured in amperes (A). Conventional current is taken to flow from the positive terminal round to the negative terminal, although in metal wires the moving charges are electrons travelling the other way. The electromotive force (e.m.f.) of a source is the work done by the source in driving a unit charge around a complete circuit. The potential difference (p.d.) across a component is the work done, or energy converted from electrical energy to other forms, per unit charge passing through it, V = W/Q. Both are measured in volts, where 1 V = 1 J/C. A current of 2.0 A flowing for 30 s carries 2.0 × 30 = 60 C. If the lamp it flows through has a p.d. of 6.0 V, the energy converted is 6.0 × 60 = 360 J.

Resistance and I-V graphs

Resistance is the ratio of the p.d. across a component to the current through it, R = V/I, measured in ohms (Ω). For a metal wire at constant temperature the I-V graph is a straight line through the origin, so its resistance is constant and it obeys Ohm's law. A filament lamp gives a curve that becomes less steep as the voltage rises, because the filament heats up and its resistance increases. A semiconductor diode lets a current flow easily in one direction only. In the reverse direction the current is almost zero. The resistance of a wire depends on its dimensions: R = ρL/A, where ρ is the resistivity of the material, L is the length and A is the cross-sectional area. Doubling the length doubles the resistance, and doubling the diameter makes the area four times larger, so the resistance falls to a quarter.

Series and parallel circuits

In a series circuit the current is the same at every point, and the p.d.s across the components add up to the e.m.f. of the supply. The combined resistance is R = R₁ + R₂ + R₃. In a parallel circuit every branch has the same p.d. across it, and the currents in the branches add up to the current from the supply. The combined resistance is found from 1/R = 1/R₁ + 1/R₂, and it is always smaller than the smallest single resistance. For a 6.0 Ω and a 3.0 Ω resistor in parallel, 1/R = 1/6 + 1/3 = 3/6, so R = 2.0 Ω. An ammeter is connected in series with the component it measures, and a voltmeter is connected in parallel across it.

Potential dividers, thermistors and LDRs

Two resistors in series across a supply share the supply voltage in the ratio of their resistances. The p.d. across R₁ is V₁ = R₁/(R₁ + R₂) × V. With a 12 V supply across 2.0 kΩ and 4.0 kΩ resistors in series, the 2.0 kΩ resistor gets 2/6 × 12 = 4.0 V and the 4.0 kΩ resistor gets 8.0 V. Replacing one resistor with a sensor makes the output change with conditions. The resistance of a thermistor decreases as its temperature rises. The resistance of a light-dependent resistor (LDR) decreases as the light falling on it becomes brighter. In a circuit with an LDR in series with a fixed resistor, the LDR's resistance rises in darkness, so it takes a larger share of the p.d. This output can be used to switch on a lamp automatically at night.

Worked example

A 12 V supply is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a 12 Ω resistor. Calculate (a) the total resistance of the circuit, (b) the current from the supply, (c) the p.d. across the 4.0 Ω resistor and (d) the current in the 6.0 Ω resistor.

  1. Parallel part: 1/R = 1/6.0 + 1/12 = 2/12 + 1/12 = 3/12, so R = 12 ÷ 3 = 4.0 Ω.
  2. Total resistance in series: 4.0 + 4.0 = 8.0 Ω.
  3. Supply current: I = V/R = 12 ÷ 8.0 = 1.5 A.
  4. P.d. across the 4.0 Ω resistor: V = IR = 1.5 × 4.0 = 6.0 V.
  5. The rest of the 12 V is across the parallel part: 12 − 6.0 = 6.0 V, and both branches share this p.d.
  6. Current in the 6.0 Ω resistor: I = 6.0 ÷ 6.0 = 1.0 A. Check: the 12 Ω branch carries 6.0 ÷ 12 = 0.5 A, and 1.0 + 0.5 = 1.5 A.

Answer: (a) 8.0 Ω (b) 1.5 A (c) 6.0 V (d) 1.0 A

Common mistakes

  • Thinking current is used up as it passes through a lamp. The current is the same before and after a component in series. It is energy that is converted, not charge.
  • Adding parallel resistances directly, or finding 1/R and forgetting to invert it at the end. The combined parallel resistance must be smaller than the smallest branch resistance.
  • Saying a filament lamp obeys Ohm's law. Its resistance increases as it heats up, so its I-V graph is a curve, not a straight line.
  • Connecting a voltmeter in series or an ammeter in parallel when drawing circuits. Ammeters go in series, voltmeters go across the component.
  • Using the diameter instead of the cross-sectional area in R = ρL/A, or converting mm² to m² wrongly. Use A = πr² and remember 1 mm² = 10⁻⁶ m².
  • Getting the direction wrong for sensors. A thermistor's resistance falls when it gets hotter, and an LDR's resistance falls when it gets brighter.

Exam tips

  • Redraw complicated circuits neatly, replace each parallel group with its combined resistance first, then work outwards to the supply.
  • Check every answer with the circuit rules: currents into a junction equal currents out, and p.d.s around a series loop add up to the e.m.f.
  • When describing an I-V graph, state what happens to the resistance and give the physical reason, such as the filament getting hotter.

Practice questions

1. Easy: A current of 0.25 A flows through a lamp for 2.0 minutes. How much charge passes through the lamp?

Show answer

Q = It = 0.25 × (2.0 × 60) = 0.25 × 120 = 30 C.

2. Medium: A wire is 2.0 m long with a cross-sectional area of 0.50 mm². The resistivity of its material is 5.0 × 10⁻⁷ Ω m. Calculate its resistance and the current through it when a p.d. of 3.0 V is applied.

Show answer

Area = 0.50 mm² = 0.50 × 10⁻⁶ = 5.0 × 10⁻⁷ m². R = ρL/A = 5.0 × 10⁻⁷ × 2.0 ÷ 5.0 × 10⁻⁷ = 2.0 Ω. Current I = V/R = 3.0 ÷ 2.0 = 1.5 A.

3. Harder: A 9.0 V supply is connected across a fixed 3.0 kΩ resistor in series with a thermistor. At 20 °C the thermistor has a resistance of 6.0 kΩ, and at 60 °C its resistance is 1.5 kΩ. Find the p.d. across the thermistor at each temperature.

Show answer

At 20 °C: total = 3.0 + 6.0 = 9.0 kΩ, so the thermistor gets 6.0/9.0 × 9.0 = 6.0 V. At 60 °C: total = 3.0 + 1.5 = 4.5 kΩ, so it gets 1.5/4.5 × 9.0 = 3.0 V. The p.d. across the thermistor falls as it warms because its resistance falls.

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