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/tV = I*RR = rho*L/Aseries: R = R1 + R2 + R3parallel: 1/R = 1/R1 + 1/R2, or R = R1*R2/(R1 + R2) for two resistorsE = I*(R + r)terminal p.d. V = E - I*rP = I*V = I^2*R = V^2/RW = I*V*t = I^2*R*t = V^2*t/Renergy in kWh = (power in watts / 1000) * hours1 kWh = 3.6 x 10^6 Jcost = 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.
- Parallel section first: R = (3.0 x 6.0)/(3.0 + 6.0) = 18/9 = 2.0 ohm.
- Total external resistance R = 3.0 + 2.0 = 5.0 ohm.
- Total circuit resistance including internal resistance = 5.0 + 1.0 = 6.0 ohm.
- Current from the cell I = E/(R + r) = 12/6.0 = 2.0 A.
- Terminal p.d. V = E - I r = 12 - (2.0 x 1.0) = 10 V.
- 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.
- 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.
Answer: Current from the cell = 2.0 A; terminal p.d. = 10 V; current in the 6.0 ohm resistor = 0.67 A
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.
- Bulbs: total power = 5 x 60 = 300 W = 0.30 kW. Daily energy = 0.30 x 5 = 1.5 kWh.
- Television: 100 W = 0.10 kW. Daily energy = 0.10 x 4 = 0.40 kWh.
- Electric iron: 1000 W = 1.0 kW. Daily energy = 1.0 x 1 = 1.0 kWh.
- Total energy per day = 1.5 + 0.40 + 1.0 = 2.9 kWh.
- Energy in 30 days = 2.9 x 30 = 87 kWh.
- Cost = 87 x 100 naira = 8700 naira.
Answer: 87 kWh is consumed, giving a bill of 8,700 naira for the month
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.