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Current Electricity, Ohm's Law and Energy Cost

Physics · WAEC and JAMB · SS2 and SS3

Electricity is the highest-scoring topic for a prepared candidate because the questions are formulaic: combine resistors, apply Ohm's law, then cost the energy. Nigerian papers favour a NEPA-style billing calculation, so learn the kilowatt-hour cold.

What you need to know

  • Electric current is the rate of flow of charge, I = Q/t, measured in amperes. One coulomb is the charge that passes when a current of one ampere flows for one second. Conventional current flows from positive to negative, opposite to the actual drift of electrons.
  • Potential difference between two points is the work done in moving one coulomb of charge between them, V = W/Q, measured in volts. Electromotive force is the total energy supplied by a source per coulomb, including the energy wasted inside the source itself.
  • Ohm's law states that the current through a conductor is directly proportional to the potential difference across its ends, provided temperature and other physical conditions remain constant. Hence V = IR, and R is the resistance in ohms.
  • A conductor obeying Ohm's law gives a straight-line V against I graph through the origin, and the slope of that line is the resistance. A filament lamp, a thermistor and a diode are non-ohmic, so their graphs are curved.
  • Resistance of a uniform wire R = rho L / A: it increases with length, decreases with cross-sectional area, and depends on the resistivity of the material. For most metals resistance rises with temperature; for semiconductors and thermistors it falls.
  • Resistors in series carry the same current and their resistances add: R = R1 + R2 + R3. The potential difference divides between them in proportion to their resistances.
  • Resistors in parallel have the same potential difference across each and their reciprocals add: 1/R = 1/R1 + 1/R2. The combined resistance of a parallel arrangement is always smaller than the smallest individual resistor, and for just two resistors R = R1R2/(R1 + R2).
  • A real cell has internal resistance r, so E = I(R + r) and the terminal potential difference V = E - Ir. The terminal p.d. falls as the current drawn rises, which is why car headlamps dim briefly when the starter motor is engaged.
  • Cells in series give E(total) = sum of the emfs and r(total) = sum of the internal resistances, which is used when a large voltage is wanted. Identical cells in parallel keep the emf of one cell but reduce the internal resistance to r/n, which is used when a large current is wanted.
  • Electrical energy W = IVt = I^2Rt = V^2t/R, in joules, and electrical power P = IV = I^2R = V^2/R, in watts. A device rated 240 V, 1000 W draws I = P/V = 1000/240 = 4.17 A.
  • The commercial unit of electrical energy is the kilowatt-hour, the energy used by a 1 kW appliance running for 1 hour. One kWh = 1000 W x 3600 s = 3.6 x 10^6 J. The energy meter in a house counts kilowatt-hours, and the bill is kWh used multiplied by the tariff per kWh.
  • The ammeter is connected in series and should have very low resistance; the voltmeter is connected in parallel across the component and should have very high resistance. Connecting an ammeter in parallel across a supply short-circuits it and can destroy the meter.
  • A fuse is a thin wire of low melting point placed in the live wire; it melts and breaks the circuit when the current exceeds its rating. A fuse is rated slightly above the normal working current of the appliance. Earthing carries fault current safely to ground, and switches must always be fitted in the live wire.
  • A primary cell such as the Leclanche or dry cell cannot be recharged, while a secondary cell such as the lead-acid accumulator or a lithium-ion battery stores charge chemically and can be recharged. Defects of the simple voltaic cell are polarisation, cured by a depolariser such as manganese dioxide, and local action, cured by amalgamating the zinc with mercury.

Key terms

Electric current
The rate of flow of electric charge through a conductor, measured in amperes.
Potential difference
The work done in moving one coulomb of charge from one point to another in an electric circuit.
Electromotive force
The total work done by a source in driving one coulomb of charge round a complete circuit, including through the source itself.
Resistance
The opposition a conductor offers to the flow of current through it, equal to the ratio of potential difference to current.
Resistivity
The resistance of a specimen of a material of unit length and unit cross-sectional area, measured in ohm metres.
Internal resistance
The resistance of the source itself to the current it drives, causing the terminal voltage to fall below the emf when current flows.
Kilowatt-hour
The electrical energy consumed by an appliance of power one kilowatt operating for one hour, equal to 3.6 x 10^6 J.

Formulae

  • I = Q/t
  • V = I*R
  • R = rho*L/A
  • series: R = R1 + R2 + R3
  • parallel: 1/R = 1/R1 + 1/R2, or R = R1*R2/(R1 + R2) for two resistors
  • E = I*(R + r)
  • terminal p.d. V = E - I*r
  • P = I*V = I^2*R = V^2/R
  • W = I*V*t = I^2*R*t = V^2*t/R
  • energy in kWh = (power in watts / 1000) * hours
  • 1 kWh = 3.6 x 10^6 J
  • cost = energy in kWh * tariff per kWh

Worked examples

A cell of emf 12 V and internal resistance 1.0 ohm is connected to a 3.0 ohm resistor in series with a parallel combination of a 3.0 ohm and a 6.0 ohm resistor. Calculate the current drawn from the cell, the terminal potential difference and the current in the 6.0 ohm resistor.

  1. Parallel section first: R = (3.0 x 6.0)/(3.0 + 6.0) = 18/9 = 2.0 ohm.
  2. Total external resistance R = 3.0 + 2.0 = 5.0 ohm.
  3. Total circuit resistance including internal resistance = 5.0 + 1.0 = 6.0 ohm.
  4. Current from the cell I = E/(R + r) = 12/6.0 = 2.0 A.
  5. Terminal p.d. V = E - I r = 12 - (2.0 x 1.0) = 10 V.
  6. P.d. across the parallel section = I x 2.0 = 2.0 x 2.0 = 4.0 V, so current in the 6.0 ohm resistor = 4.0/6.0 = 0.67 A.
  7. Check: current in the 3.0 ohm branch = 4.0/3.0 = 1.33 A, and 0.67 + 1.33 = 2.0 A, which matches the main current. Also 6.0 V across the series resistor plus 4.0 V across the parallel section gives 10 V, the terminal p.d.

A household uses five 60 W bulbs for 5 hours a day, a 100 W television for 4 hours a day and a 1000 W electric iron for 1 hour a day. If electricity is charged at 100 naira per kilowatt-hour, calculate the bill for 30 days.

  1. Bulbs: total power = 5 x 60 = 300 W = 0.30 kW. Daily energy = 0.30 x 5 = 1.5 kWh.
  2. Television: 100 W = 0.10 kW. Daily energy = 0.10 x 4 = 0.40 kWh.
  3. Electric iron: 1000 W = 1.0 kW. Daily energy = 1.0 x 1 = 1.0 kWh.
  4. Total energy per day = 1.5 + 0.40 + 1.0 = 2.9 kWh.
  5. Energy in 30 days = 2.9 x 30 = 87 kWh.
  6. Cost = 87 x 100 naira = 8700 naira.

The mistake to avoid

Candidates add resistors in parallel the way they add them in series, so three 3-ohm resistors in parallel come out as 9 ohms instead of 1 ohm. Check every parallel answer against this rule: the combined resistance must be smaller than the smallest branch. In billing questions the other killer is leaving power in watts and multiplying straight by the tariff; convert watts to kilowatts first, because the tariff is per kilowatt-hour.

In the exam

Redraw any circuit given in words, mark the current direction, and reduce parallel sections to a single resistor before doing anything else. Show the formula, the substitution and the answer with its unit on three separate lines; WAEC gives method marks for the substitution even when the arithmetic slips. For cost questions, end with the naira figure clearly stated, not just the kWh.