Physics notes, chapter 3

Electricity

Textbook Physics Form 5 (DLP), pages 90–133

What this chapter covers

Describe electric fields, electric field strength and the motion of charged particles in a field (3.1)
Define current and potential difference and solve problems using Q = It and V = W/Q (3.1)
Distinguish ohmic and non-ohmic conductors and solve series, parallel and combination circuit problems (3.2)
Explain how length, cross-sectional area and resistivity affect the resistance of a wire, and use R = ρl/A (3.2)
Compare e.m.f. and potential difference and explain internal resistance and voltage drop (3.3)
Solve problems on cells in series and in parallel using ε = I(R + r) (3.3)
Use E = VIt, P = VI, P = I²R and P = V²/R, calculate the cost of electricity in kWh and suggest ways to save energy (3.4)

3.1Current and potential difference

Electric field

An electric field is the region around a charged particle where any other electric charge experiences an electric force.

Everyday signs of an electric field: a charged comb bends a thin stream of tap water, and a charged balloon attracts hair or small pieces of paper. Semolina powder floating on olive oil lines up along the field lines when a high voltage is applied across two electrodes, which is how the shape of a field is shown in the laboratory.

Electric field lines: outward from a positive charge, inward to a negative charge, curved lines from a positive to a negative charge, lines pushing apart for two like charges, and parallel straight lines between two oppositely charged plates — Electric field patterns. The lines give the direction of the force on a positive test charge.

Field lines start on a positive charge and end on a negative charge. They never cross.
Lines are closer together where the field is stronger.
Between two oppositely charged parallel plates the lines are parallel and evenly spaced: a uniform field.

Electric field strength

Electric field strength, E at a point is the electric force acting on a unit positive charge placed at that point.

Situation	Formula	Unit
Any field, force F on charge q	E = F / q	N C⁻¹
Uniform field between plates, voltage V, separation d	E = V / d	V m⁻¹

1 V m⁻¹ is the same as 1 N C⁻¹.

Writing E = V / d for the field around a single point charge. This formula is only for the uniform field between two parallel plates.

Charged particles in an electric field

A metal-coated polystyrene ball touching one of two charged plates picks up that plate's charge, is repelled and is attracted to the other plate. It swings to and fro, carrying charge across, until the supply is switched off.
A candle flame between the plates spreads out because the hot air is ionised. Negative ions go to the positive plate and positive ions go to the negative plate. The flame spreads more towards the negative plate because positive ions are larger and heavier.

Current and potential difference

Current, I is the rate of flow of charge: I = Q / t, so Q = It. Unit: ampere (A) = coulomb per second (C s⁻¹).

Potential difference, V between two points is the work done (energy transferred) in moving one coulomb of charge from one point to the other: V = W / Q. Unit: volt (V) = joule per coulomb (J C⁻¹).

The charge of one electron is e = 1.6 × 10⁻¹⁹ C. So the number of electrons = total charge ÷ (1.6 × 10⁻¹⁹ C).

Worked example: charge and energy

A bulb labelled 3.0 V, 0.2 A is lit for 1 hour. Find the charge and the energy transferred.
Q = It = 0.2 × (1 × 3600) = 720 C.
E = VQ = 3.0 × 720 = 2160 J.

A current of 0.50 A flows for 2 minutes. How many coulombs of charge pass? How many electrons is this?

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Q = It = 0.50 × 120 = 60 C. Electrons = 60 ÷ (1.6 × 10⁻¹⁹) = 3.75 × 10²⁰.

3.2Resistance

Ohmic and non-ohmic conductors

A conductor that obeys Ohm's law (V ∝ I at constant temperature, so the ratio V / I = R stays constant) is ohmic, for example a constantan wire. A filament bulb is non-ohmic: as the current rises the filament gets hotter, its resistance increases and the graph of V against I bends upward.

Two graphs of V against I: a straight line through the origin for a constantan wire and an upward curve through the origin for a filament bulb — Graphs of V against I.

	Ohmic conductor	Non-ohmic conductor
Graph of V against I	Straight line through the origin	Curve through the origin
Gradient (= resistance)	Constant	Increases

Saying a bulb breaks Ohm's law "because of the voltage". The reason is that the temperature of the filament is not constant.

Series and parallel circuits

	Series	Parallel
Current	Same everywhere: I = I₁ = I₂	Splits: I = I₁ + I₂
Potential difference	Shared: V = V₁ + V₂	Same across each branch: V = V₁ = V₂
Effective resistance	R = R₁ + R₂ + R₃	1/R = 1/R₁ + 1/R₂ + 1/R₃

For a combination circuit, first replace each parallel block by its effective resistance, add the series parts, find the total current, then work back outwards to each resistor.

Worked example: combination circuit

6 V cell with a 2 ohm resistor in series with 4 ohm and 12 ohm resistors in parallel

Find the current from the cell and the current through the 4 Ω resistor.
Parallel part: 1/R = 1/4 + 1/12 = 4/12, so R = 3 Ω.
Total resistance = 2 + 3 = 5 Ω. Current I = 6 ÷ 5 = 1.2 A.
V across the 2 Ω resistor = 1.2 × 2 = 2.4 V, so V across the parallel part = 6 − 2.4 = 3.6 V.
Current in the 4 Ω resistor = 3.6 ÷ 4 = 0.9 A (and 3.6 ÷ 12 = 0.3 A in the 12 Ω; 0.9 + 0.3 = 1.2 A).

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the effective resistance?

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1/R = 1/6 + 1/3 = 3/6, so R = 2 Ω. (Parallel resistance is always smaller than the smallest branch.)

Factors that affect the resistance of a wire

Factor	Effect on R (temperature constant)	Graph
Length, l	R is directly proportional to l	R against l: straight line through the origin
Cross-sectional area, A	R is inversely proportional to A; thicker wire, smaller R	R against 1/A: straight line through the origin
Resistivity, ρ (the material)	R is directly proportional to ρ

Resistivity, ρ measures how strongly a material opposes the flow of current. It depends on the material and the temperature. Unit: ohm metre (Ω m).

Putting the three factors together: R = ρl / A, where A = πr² = πd²/4 for a round wire.

s.w.g. (standard wire gauge) is a wire size. The bigger the s.w.g. number, the thinner the wire, and the larger its resistance.

Worked example: wire resistance

A constantan wire (ρ = 49 × 10⁻⁸ Ω m) is 50.0 cm long and 0.6 mm in diameter. Find its resistance.
r = 0.3 mm = 3 × 10⁻⁴ m, so A = π(3 × 10⁻⁴)² = 2.83 × 10⁻⁷ m².
R = ρl / A = (49 × 10⁻⁸ × 0.5) ÷ (2.83 × 10⁻⁷) = 0.87 Ω.

Using the diameter as the radius, or leaving the length in cm. Convert to metres, and halve the diameter before squaring.

Uses of resistivity

Heating elements (rice cooker, kettle, iron): high resistivity and a high melting point, e.g. nichrome, so the wire gets hot and does not melt or oxidise quickly.
Connecting wires: copper has low resistivity, so the wire does not heat up as current flows.
Fuse: a thin wire of low melting point that melts and cuts the circuit when the current is too large.

Material	Resistivity	Examples
Non-conductor (insulator)	Highest	Plastic, wood
Semiconductor	Between conductors and insulators	Silicon, germanium
Conductor	Low	Iron, copper, carbon
Superconductor	Zero below its critical temperature, T_c	Used in coils of MRI scanners

3.3Electromotive force and internal resistance

Electromotive force (e.m.f.), ε is the energy transferred (work done) by an electrical source in moving one coulomb of charge around a complete circuit. Unit: volt (V) = J C⁻¹.

	e.m.f., ε	Potential difference, V
Measured	Across the cell with no current flowing (open circuit)	Across the cell, or between any two points, while current flows
Voltmeter across the cell	Reading is the e.m.f.	Reading is the terminal potential difference, which is less than ε
Meaning	Work done by the source per coulomb in the whole circuit	Work done per coulomb between two points

Internal resistance, r is the resistance of the electrolyte inside the cell. When current flows, energy is lost as heat inside the cell, so the terminal voltage is lower than the e.m.f.

Voltage drop (lost volts) = Ir = ε − V. With an external resistance R: ε = I(R + r), or ε = V + Ir.

Straight line graph of V against I that starts at the e.m.f. on the vertical axis and slopes downwards — V = ε − Ir: the intercept on the V axis is the e.m.f. and the gradient is −r.

From a V against I graph of a cell: vertical intercept = ε, gradient = −r, so r = −gradient.

Cells in series and in parallel

n identical cells (ε, r each)	Series	Parallel
Effective e.m.f.	nε	ε (same as one cell)
Effective internal resistance	nr	r / n
Main advantage	Higher voltage	Lower internal resistance and the cells last longer

Worked example: two cells

Two cells, each ε = 1.5 V and r = 0.5 Ω, supply a 10 Ω resistor.
Series: ε = 3.0 V, r = 1.0 Ω, so I = 3.0 ÷ (10 + 1.0) = 0.27 A.
Parallel: ε = 1.5 V, r = 0.25 Ω, so I = 1.5 ÷ (10 + 0.25) = 0.146 A.
Series gives the larger current because its effective e.m.f. is larger.

Adding the e.m.f. of cells connected in parallel. A parallel bank has the e.m.f. of one cell; only its internal resistance falls.

A cell of e.m.f. 1.5 V gives a terminal voltage of 1.2 V when it supplies 0.6 A. Find its internal resistance.

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Ir = ε − V = 1.5 − 1.2 = 0.3 V, so r = 0.3 ÷ 0.6 = 0.5 Ω.

Electric and hybrid vehicles use batteries (and solar panels at charging stations) arranged in series and parallel to give the high voltage and high current the motor needs, and so reduce the use of fossil fuel and air pollution.

3.4Electrical energy and power

Quantity	Formula	Unit
Electrical energy	E = VQ = VIt	joule (J)
Electrical power	P = E / t = VI	watt (W) = J s⁻¹
Using V = IR	P = I²R = V² / R	watt (W)

The label on an appliance, such as 240 V, 700 W, gives its working voltage and power rating. At 240 V it uses 700 J of electrical energy every second.

Worked example: lamp

A fluorescent lamp is labelled 240 V, 32 W. Under normal use:
Current I = P / V = 32 ÷ 240 = 0.133 A.
Resistance R = V² / P = 240² ÷ 32 = 1800 Ω.
Energy in 3 hours = 32 × (3 × 3600) = 345 600 J = 346 kJ.

Cost of electricity

1 kilowatt-hour (kWh) is the energy used by a 1 kW appliance in 1 hour, and equals one "unit" of electricity. 1 kWh = 3.6 × 10⁶ J.

Cost = number of units (kWh) × cost per unit, where units = power in kW × time in hours.

Worked example: comparing lamps

A 40 W CFL and a 12 W LED give the same brightness and are on 12 h a day for 30 days at RM0.218 per unit.
CFL: 0.040 × 12 × 30 = 14.4 kWh, cost RM3.14.
LED: 0.012 × 12 × 30 = 4.32 kWh, cost RM0.94.
The LED uses less energy for the same light, so it is more efficient.

Putting watts straight into the cost formula. Change W to kW (divide by 1000) and use hours, not seconds, when working in kWh.

Saving electricity at home

Switch off appliances, lights and fans when not in use.
Use energy-saving lamps such as LEDs.
Wash a full load of laundry; keep air-conditioner filters clean and doors and windows closed.
Choose appliances with a high efficiency or energy-saving label.

An electric kettle rated 240 V, 2000 W is used for 15 minutes. Find the current and the energy used in kWh.

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I = P / V = 2000 ÷ 240 = 8.3 A. Energy = 2.0 kW × 0.25 h = 0.50 kWh.