Physics · Lesson 21

Ohm's law & circuits

JAMBWAECNECO

What the syllabus asks

JAMB: Current electricity (e.m.f., p.d., current, internal resistance, Ohm's law, resistivity, series and parallel arrangements, electrical energy and power)

WAEC: Current electricity

NECO: Current electricity

What you will learn

Explain what electric current, potential difference and e.m.f. are, using simple models, then their formal definitions.
Draw and read circuit diagrams, and connect ammeters and voltmeters correctly.
State Ohm's law and its limits, and use V = IR.
Use R = ρL/A for wires.
Find effective resistance for series, parallel and mixed arrangements, and the current and voltage in each part.
Use E = I(R + r) for cells with internal resistance.
Calculate electrical energy, power and the cost of electricity, and choose a suitable fuse.

Key points

I = Q/t (A); V = W/Q (V); R = V/I (Ω).
Ammeter in series (low R); voltmeter in parallel (high R).
Ohm's law: V = IR for metals at constant temperature. Diodes, lamps and thermistors are non-ohmic.
R = ρL/A: longer means more R; thicker means less R.
Series: same current, voltages add, R = R₁ + R₂.
Parallel: same voltage, currents add, 1/R = 1/R₁ + 1/R₂, total smaller than the smallest.
E = I(R + r); terminal p.d. = E − Ir.
P = VI = I²R = V²/R; cost = kW × h × price; fuse slightly above the working current, in the live wire.

Start here: What is electric current?

When you press the switch and the bulb in your room comes on, something has started moving in the wires. What is it? It is charge, carried mainly by those free electrons in the copper wire.

Picture a crowded road in Lagos. Cars are everywhere, but if nobody moves, there is no "traffic flow". Traffic flow is the number of cars passing a point every second. Electric current is the same idea: it is not the amount of charge in the wire, but how much charge passes a point every second.

Definition. Electric current is the rate of flow of electric charge: I = Q/t, where Q is the charge in coulombs (C) and t is the time in seconds (s). The SI unit of current is the ampere (A). One ampere means one coulomb of charge passing a point every second.

How big is a coulomb? Each electron carries a tiny charge, e = 1.6 × 10⁻¹⁹ C. So one coulomb is the charge of about 6.25 × 10¹⁸ electrons. That is over six billion billion electrons! When a 1 A current flows, that many electrons pass every second.

Fig. 21.1: Current is charge per second passing a point.

Which way does current flow?

Here is a historical twist. Scientists agreed on a direction for current long before electrons were discovered. They guessed that current flows from the positive (+) terminal to the negative (−) terminal. We still use that agreed direction, called conventional current. Later we learnt that electrons (being negative) actually drift the other way, from − to +. Both descriptions are correct; in exams, circuit arrows show conventional current unless you are told otherwise.

Worked example 21.1

A charge of 120 C passes through a lamp in 1 minute. Find the current.

Write what we know: Q = 120 C and t = 1 minute.
Convert time to seconds, because the ampere is defined using seconds: t = 60 s.
I = Q/t = 120/60 = 2 A.

2 A

Worked example 21.2

A phone charger delivers 0.5 A for 2 hours. How much charge flows, and how many electrons is that?

t = 2 × 3600 = 7200 s.
Rearrange I = Q/t to get Q = It = 0.5 × 7200 = 3600 C.
Number of electrons = total charge ÷ charge per electron = 3600/(1.6 × 10⁻¹⁹) = 2.25 × 10²².

3600 C; 2.25 × 10²² electrons

⚠️ Common mistake: putting minutes or hours into Q = It. Always convert to seconds first.

Try it yourself. A current of 3 A flows for 5 minutes. What charge passes?

Show solution

t = 300 s, so Q = It = 3 × 300 = 900 C.

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Practice questions

1. A current of 4 mA flows for 5 minutes. Find the charge.

Show worked answer

Q = 4 × 10⁻³ × 300 = 1.2 C.

2. Find the effective resistance of 3 Ω, 6 Ω and 2 Ω in parallel.

Show worked answer

1/R = 2/6 + 1/6 + 3/6 = 1, so R = 1 Ω.

3. A wire has resistance 12 Ω. Another wire of the same material is twice as long with twice the diameter. Find its resistance.

Show worked answer

L × 2 and A × 4, so R × ½: 6 Ω.

4. A cell reads 1.5 V on open circuit and 1.2 V across 4 Ω. Find r.

Show worked answer

I = 0.3 A; lost volts = 0.3 V; r = 1 Ω.

5. A 100 W bulb for 4 h a day and a 1.5 kW kettle for 0.5 h a day: find the energy used in a 30-day month.

Show worked answer

(0.4 + 0.75) × 30 = 34.5 kWh.

Key words

Charge (Q): A property of particles such as electrons; measured in coulombs (C).
Current (I): The rate of flow of charge, Q/t; measured in amperes (A).
Potential difference (V): The energy converted per coulomb between two points; measured in volts (V).
E.m.f. (E): The energy a source gives to each coulomb going round the whole circuit; measured in volts.
Resistance (R): Opposition to current, V/I; measured in ohms (Ω).
Resistivity (ρ): A property of a material: the resistance of a 1 m length of 1 m² cross-section; measured in Ω m.
Ohmic conductor: A conductor that obeys Ohm's law (straight-line I–V graph through the origin).
Internal resistance (r): The resistance inside a cell or battery.
Lost volts: Ir, the part of the e.m.f. used up inside the cell.
Terminal p.d.: The voltage across the cell's terminals when current flows: E − Ir.
kWh: Kilowatt-hour: the energy used by a 1 kW appliance in 1 hour (3.6 × 10⁶ J).
Fuse: A thin wire that melts when the current is too large, protecting the circuit.

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