Topic 4.2 · Unit 4

Electrical quantities

Electric charge and electric fields, current, e.m.f. and p.d., resistance and current–voltage graphs, and the energy, power and cost of using electricity.

In this topic

  1. 4.2.1Electric charge
  2. 4.2.2Electric current
  3. 4.2.3Electromotive force and potential difference
  4. 4.2.4Resistance
  5. 4.2.5Electrical energy and electrical power

Key points

4.2.1 Electric charge

  • There are two kinds of electric charge: positive and negative.
  • Like charges repel. Two positive charges push each other apart. Two negative charges also push each other apart.
  • Unlike charges attract. A positive charge and a negative charge pull towards each other.
  • Charging by friction: rub a polythene rod with a dry cloth. The rod becomes negatively charged.
  • Detecting charge: a charged rod picks up small pieces of paper. Or hang a charged rod from a thread. Bring a second charged rod near it. If they repel, both have the same kind of charge.
  • A gold-leaf electroscope also detects charge. When a charged object is brought near its metal cap, the leaf rises.
  • When two solids are rubbed together, only negative charge (electrons) moves from one to the other. Positive charges do not move.
  • The object that gains electrons becomes negative. The object that loses electrons becomes positive.
  • Conductor or insulator? Make a circuit with a cell, a lamp and two connecting leads with a gap between them. Put the material across the gap. If the lamp lights, the material is an electrical conductor. If it does not light, the material is an insulator.
  • Electron model: in a conductor, some electrons are free to move through the material. These are called free electrons. In an insulator, the electrons are held in their atoms. They are not free to move, so charge cannot flow.
  • Typical conductors: metals such as copper, aluminium and silver, and graphite (carbon).
  • Typical insulators: plastic, rubber, glass and dry wood.
  • Charge is measured in coulombs (C). Extended
  • An electric field is a region where an electric charge feels a force. Extended
  • The direction of an electric field at a point is the direction of the force on a positive charge at that point. Extended
  • Field around a point charge: the field lines are straight lines that spread out from the charge, like the spokes of a wheel (a radial field). They point away from a positive charge and towards a negative charge. Extended
  • Field around a charged conducting sphere: the field lines are straight lines at right angles to the surface. They point away from a positive sphere and towards a negative sphere. The pattern outside the sphere is the same as for a point charge at its centre. Extended
  • Field between two parallel plates with opposite charges: the field lines are straight, parallel and equally spaced. They go from the positive plate to the negative plate, at right angles to the plates. Extended

4.2.2 Electric current

  • An electric current is a flow of electric charge.
  • Conduction in metals: a metal contains free electrons. When a cell is connected, the free electrons move through the metal in one direction. This movement of electrons is the current.
  • An ammeter measures current. It is connected in series, so the current passes through it.
  • Ammeters can be analogue (a needle moves over a scale) or digital (they show a number).
  • Many ammeters have different ranges, for example 0–1 A and 0–5 A. Choose the smallest range that is still larger than the current you expect. This gives the most precise reading.
  • If you do not know the size of the current, start on the largest range. This protects the meter.
  • On an analogue meter, look straight at the needle to read it. Check that it reads zero before the current is switched on.
  • On a digital meter, check the unit shown (A or mA).
  • Direct current (d.c.) flows in one direction only. Cells and batteries give d.c.
  • Alternating current (a.c.) keeps changing direction, many times each second. The mains supply is a.c.
  • Electric current is the charge that passes a point in each unit of time (each second): I = Q / t. Extended
  • Conventional current flows from positive to negative (from the + terminal of the cell, round the circuit, to the − terminal). Extended
  • The free electrons flow the other way: from negative to positive. Extended

4.2.3 Electromotive force and potential difference

  • The electromotive force (e.m.f.) of a source is the electrical work the source does to move a unit charge all the way round a complete circuit.
  • E.m.f. is measured in volts (V).
  • The potential difference (p.d.) across a component is the work done by a unit charge as it passes through that component. This is the energy the charge transfers to the component.
  • P.d. is also measured in volts (V).
  • A voltmeter measures p.d. It is connected in parallel, across the component.
  • Voltmeters can be analogue or digital. Choose the range in the same way as for an ammeter: the smallest range that is larger than the p.d. you expect.
  • Unit charge is one coulomb. So e.m.f. is the work done by the source per coulomb: E = W / Q. Extended
  • P.d. is the work done per coulomb passing through the component: V = W / Q. Extended
  • So 1 V = 1 J/C. A 6 V battery does 6 J of work on each coulomb of charge. Extended

4.2.4 Resistance

  • Resistance is the p.d. across a component divided by the current in it: R = V / I. The unit is the ohm (Ω).
  • A larger resistance means a smaller current for the same p.d.
  • Experiment to find a resistance:
    1. Connect the resistor (or wire) in series with an ammeter, a variable resistor and a power supply.
    2. Connect a voltmeter in parallel across the resistor.
    3. Read V and I.
    4. Change the variable resistor and take more pairs of readings.
    5. Work out R = V / I for each pair and find the mean.
    6. Switch off between readings, so the resistor does not get hot.
  • For a metal wire made of one material:
    • a longer wire has a larger resistance
    • a thicker wire (larger cross-sectional area) has a smaller resistance.
  • For a metallic wire at constant temperature, resistance is directly proportional to length: R ∝ l. Double the length and the resistance doubles. Extended
  • Resistance is inversely proportional to cross-sectional area: R ∝ 1 / A. Double the area and the resistance halves. Extended
  • Current–voltage (I–V) graphs: current I goes on the vertical axis and p.d. V on the horizontal axis. Negative values mean the p.d. is reversed. Extended
  • Resistor of constant resistance: a straight line through the origin. It continues the same way for negative values. This is because R stays the same, so I is directly proportional to V. Extended
  • Filament lamp: a curve through the origin. It is steep at first, then it gets less steep as V increases. The shape is the same for negative values. Extended
  • Why the lamp curve bends: a larger current makes the filament hotter. A hotter filament has a larger resistance. So each extra volt gives a smaller increase in current. Extended
  • Diode: for negative values of V (reverse direction), the current is zero or almost zero. For positive values, the current stays almost zero until V reaches a small value. Then the current rises steeply. Extended
  • Why: a diode has a very high resistance in one direction and a low resistance in the other. So current flows through it in one direction only. Extended

4.2.5 Electrical energy and electrical power

  • An electric circuit transfers energy. The energy goes from the source (for example a cell or the mains supply) to the components. Then it goes from the components into the surroundings.
  • Example: a battery lights a torch lamp. Energy goes from the battery to the lamp. The lamp transfers it to the surroundings by light and by heating.
  • Electrical power is current × p.d.: P = IV. Power is measured in watts (W).
  • Electrical energy transferred is current × p.d. × time: E = IVt. Energy is measured in joules (J).
  • The kilowatt-hour (kW h) is a unit of energy. One kilowatt-hour is the energy transferred by an appliance with a power of 1 kW that is used for 1 hour.
  • 1 kW h = 1000 W × 3600 s = 3.6 × 106 J.
  • Electricity companies charge for each kW h used. To find a cost:
    1. Energy in kW h = power in kW × time in hours.
    2. Cost = energy in kW h × price of 1 kW h.

Model

Current, voltage and resistance: compare the direction of electron flow with the direction of conventional current. Then change the length and the thickness of the wire and watch the ammeter reading.

3D modelCurrent, voltage and resistance
Current, voltage and resistanceOpen full screen: Current, voltage and resistance

Equations

  • CurrentExtended

    I = Q / t

    I = current (A); Q = charge (C); t = time (s)

  • Electromotive forceExtended

    E = W / Q

    E = e.m.f. (V); W = work done by the source (J); Q = charge (C)

  • Potential differenceExtended

    V = W / Q

    V = p.d. (V); W = work done (J); Q = charge (C)

  • Resistance

    R = V / I

    R = resistance (Ω); V = p.d. (V); I = current (A)

  • Resistance of a metallic wireExtended

    R ∝ l and R ∝ 1 / A

    R = resistance (Ω); l = length of the wire (m); A = cross-sectional area of the wire (m2); same material at the same temperature

  • Electrical power

    P = IV

    P = power (W); I = current (A); V = p.d. (V)

  • Electrical energy

    E = IVt

    E = energy transferred (J); I = current (A); V = p.d. (V); t = time (s)

  • Cost of electricity

    cost = energy used in kW h × price of 1 kW h

    energy in kW h = power (kW) × time (h); 1 kW h = 3.6 × 106 J

Current Extended

A charge of 90 C passes a point in a wire in 1.0 minute. Find the current.

  • Given: Q = 90 C, t = 1.0 minute = 60 s
  • I = Q / t
  • I = 90 ÷ 60
  • I = 1.5 A

How long does it take for 450 C to pass the same point?

  • Rearrange: t = Q / I
  • t = 450 ÷ 1.5
  • t = 300 s (5.0 minutes)

Electromotive force Extended

A battery does 54 J of work to move 6.0 C of charge round a circuit. Find its e.m.f.

  • Given: W = 54 J, Q = 6.0 C
  • E = W / Q
  • E = 54 ÷ 6.0
  • E = 9.0 V

Potential difference Extended

When 2.5 C of charge passes through a lamp, it does 15 J of work. Find the p.d. across the lamp.

  • Given: W = 15 J, Q = 2.5 C
  • V = W / Q
  • V = 15 ÷ 2.5
  • V = 6.0 V

Resistance

The p.d. across a resistor is 6.0 V. The current in it is 0.40 A. Find its resistance.

  • Given: V = 6.0 V, I = 0.40 A
  • R = V / I
  • R = 6.0 ÷ 0.40
  • R = 15 Ω

The same resistor is connected to a 4.5 V supply. Find the current.

  • Rearrange: I = V / R
  • I = 4.5 ÷ 15
  • I = 0.30 A

Resistance of a metallic wire Extended

A wire is 2.0 m long. Its resistance is 6.0 Ω.

(a) Find the resistance of a 5.0 m length of the same wire.

  • Same material and same cross-sectional area, so R ∝ l.
  • The length is 5.0 ÷ 2.0 = 2.5 times larger.
  • R = 6.0 × 2.5 = 15 Ω

(b) A second wire is made of the same material. It is 2.0 m long, but its cross-sectional area is 3 times larger. Find its resistance.

  • Same length, so R ∝ 1 / A.
  • The area is 3 times larger, so the resistance is 3 times smaller.
  • R = 6.0 ÷ 3 = 2.0 Ω

Electrical power

A ceiling fan works on the 230 V mains. The current in it is 0.30 A. Find its power.

  • Given: V = 230 V, I = 0.30 A
  • P = IV
  • P = 0.30 × 230
  • P = 69 W

Electrical energy

A lamp is connected to a 230 V supply. The current is 0.40 A. It is switched on for 5.0 minutes. Find the energy transferred.

  • Given: I = 0.40 A, V = 230 V, t = 5.0 minutes = 300 s
  • E = IVt
  • E = 0.40 × 230 × 300
  • E = 27 600 J = 2.8 × 104 J (28 kJ) to 2 s.f.

Cost of electricity

An air conditioner has a power of 1.5 kW. It is used for 8.0 hours each day. Electricity costs MVR 3.00 per kW h (an example price). Find the cost for one day.

  • Energy = power × time = 1.5 kW × 8.0 h = 12 kW h
  • Cost = 12 × 3.00
  • Cost = MVR 36.00
  • The same energy in joules: 12 × 3.6 × 106 = 4.3 × 107 J (2 s.f.)

Common mistakes

  • Students write that the positive charges move when a rod is rubbed. / The mark scheme wants: only electrons (negative charge) move. The object that gains electrons becomes negative.
  • Students connect the voltmeter in series, or the ammeter in parallel. / The mark scheme wants: ammeter in series; voltmeter in parallel across the component.
  • Students say a kilowatt-hour is a unit of power or of time. / The mark scheme wants: the kW h is a unit of energy, the energy used by a 1 kW appliance in 1 hour.
  • Students put the time in seconds or minutes when they work out kW h. / The mark scheme wants power in kW and time in hours.
  • Students say that conventional current and electron flow go the same way. / The mark scheme wants: conventional current from + to −; electrons from − to +. Extended
  • Students think a wire with double the diameter has half the resistance. / The mark scheme wants the area: double the diameter gives 4 times the area, so the resistance is 4 times smaller. Extended

Exam tips

  • Define e.m.f. and p.d. carefully. Both are work done per unit charge. E.m.f. is for the source and the whole circuit. P.d. is for one component.
  • Describe an experiment to find resistance: draw or name the circuit (ammeter in series, voltmeter in parallel), say what you measure and how you calculate R.
  • Explain charging by friction in steps: rubbing transfers electrons → one object gains electrons and becomes negative → the other loses electrons and becomes positive.
  • Calculate: write the equation, substitute with units converted (minutes to seconds, kW to W), then give the answer with its unit.
  • Sketch I–V graphs with labelled axes. Show the diode’s current as zero for negative V. Show the lamp curve bending towards the V axis. Extended
  • Explain the lamp graph with the chain: more current → filament hotter → resistance increases → graph less steep. Extended
  • Draw electric field lines with arrows. Arrows point away from positive charges and towards negative charges. Between plates, draw straight, parallel, equally spaced lines. Extended
  • A typical 1-mark answer: “Direct current flows in one direction only.”
  • A typical 2-mark answer to “Why does a metal conduct but plastic does not?”: “A metal has free electrons that can move through it (1). In plastic, the electrons are held in their atoms and cannot move (1).”
  • A typical 1-mark answer: “An electric field is a region where a charge feels a force.” Extended
  • A typical 3-mark answer to “Explain the shape of the I–V graph for a filament lamp”: “As the p.d. increases, the current increases (1). The filament gets hotter (1). Its resistance increases, so the graph gets less steep (1).” Extended