Electricity

Electric current

I = Q/t: the net charge Q flowing across a cross-section of a conductor divided by the time t taken. Its SI unit is the ampere, where 1 A = 1 C/s; by convention current flows opposite to the direction of flow of electrons.

Electric potential difference

V = W/Q: the work done W to move a charge Q from one point of a current-carrying circuit to another. Its SI unit is the volt: 1 V is the potential difference when 1 joule of work moves 1 coulomb of charge, so 1 V = 1 J/C.

How are an ammeter and a voltmeter connected in a circuit?

An ammeter, which measures current, is always connected in series with the part of the circuit whose current is measured. A voltmeter, which measures potential difference, is always connected in parallel across the two points between which the potential difference is measured.

Ohm's law

V = IR: the potential difference V across a metallic wire is directly proportional to the current I through it, provided its temperature stays the same. R is the wire's resistance, and a V-I graph for the wire is a straight line through the origin.

Resistance

The property of a conductor that resists the flow of charges through it, because the moving electrons are held back by the pull of the surrounding atoms. Its SI unit is the ohm: 1 Ω is the resistance when a potential difference of 1 V drives a current of 1 A.

Rheostat

A variable resistance used to change the resistance in a circuit, and so regulate the current, without changing the voltage source. Since I = V/R, raising the resistance lowers the current.

Factors affecting resistance

R = ρl/A: for a uniform metal conductor, resistance rises in proportion to length l, falls in inverse proportion to cross-sectional area A, and depends on its material through the resistivity ρ. Doubling a wire's length halves the current through it from the same source.

Resistivity

The constant ρ in R = ρl/A, a characteristic property of a material, with SI unit Ω m. Metals and alloys have very low resistivity, from 10^-8 to 10^-6 Ω m, while insulators such as rubber and glass range from 10^12 to 10^17 Ω m; it varies with temperature.

Does the resistivity of a wire change if its length or area of cross-section is changed?

No. Changing the length or area changes the wire's resistance, but resistivity is a characteristic property of the material, so it stays the same. It changes only with a different material or a change in temperature.

How do you work out the new resistance when both the length and the area of cross-section of a wire are changed?

Write R = ρl/A for the original wire, substitute the new length and area, and compare the two, since ρ is unchanged for the same material. In the chapter's example, halving the length and doubling the area gives ρ(l/2)/(2A), one quarter of the original, so a 4 Ω wire becomes 1 Ω.

Why are the coils of electric toasters and irons made of an alloy rather than a pure metal?

An alloy usually has a higher resistivity than the metals it is made from, and it does not readily oxidise, or burn, when very hot. Both properties suit them for electrical heating devices.

Series combination of resistors

Rs = R1 + R2 + R3. The same current flows through every resistor, and the total potential difference equals the sum of those across each resistor (V = V1 + V2 + V3), so the combined resistance is greater than any individual resistance.

How do you find the current and the potential difference across each resistor in a series circuit?

Add the resistances to get Rs, find the current from I = V/Rs, then apply V = IR to each resistor. In the chapter's example, a 20 Ω lamp and a 4 Ω conductor on a 6 V battery give 24 Ω and 0.25 A, so 5 V across the lamp and 1 V across the conductor.

Parallel combination of resistors

1/Rp = 1/R1 + 1/R2 + 1/R3. The potential difference across each resistor is the same, and the total current is the sum of the branch currents (I = I1 + I2 + I3), so joining resistors in parallel decreases the total resistance.

How do you find the current through each resistor and the total resistance of a parallel circuit?

Each branch has the full voltage, so find each current from I = V/R, add them for the total, and get Rp from 1/Rp = 1/R1 + 1/R2 + 1/R3. In the chapter's example, 5 Ω, 10 Ω and 30 Ω on 12 V carry 2.4, 1.2 and 0.4 A: 4 A in all, with Rp = 3 Ω.

How do you find the total resistance of a circuit that combines series and parallel resistors?

Replace each parallel group by its single equivalent resistance, then add the equivalents in series and use Ohm's law for the current. In the chapter's example, 10 Ω and 40 Ω in parallel give 8 Ω, and 30 Ω, 20 Ω and 60 Ω give 10 Ω; in series these make 18 Ω, drawing 0.67 A from 12 V.

Why are household appliances connected in parallel rather than in series?

In series the same current flows through everything, which is impractical for devices like a bulb and a heater that need very different currents, and one failed component breaks the whole circuit. In parallel each gadget draws the current it needs and the total resistance is reduced.

Heating effect of current

In a purely resistive circuit, the energy supplied by the source is dissipated entirely as heat, raising the temperature of the resistor. The effect is used in devices such as electric heaters, irons, toasters, ovens and kettles.

Joule's law of heating

H = I²Rt: the heat produced in a resistor is directly proportional to the square of the current, to the resistance, and to the time for which the current flows. It follows from H = VIt with V = IR.

Why is tungsten used for the filaments of electric bulbs?

The filament must become hot enough to glow without melting. Tungsten is a strong metal with a very high melting point of 3380 °C, and the bulb is filled with chemically inactive nitrogen and argon to prolong the filament's life.

Electric power

P = VI = I²R = V²/R: the rate at which electrical energy is consumed in a circuit. Its SI unit is the watt, the power used by a device carrying 1 A at a potential difference of 1 V, so 1 W = 1 V A.

Kilowatt hour

The commercial unit of electrical energy, commonly called a 'unit': the energy consumed when 1 kW of power is used for 1 hour. 1 kW h = 1000 W × 3600 s = 3.6 × 10^6 J.

How do you calculate the cost of running an appliance over a period?

Multiply the power by the hours of use to get the energy, express it in kilowatt hours, then multiply by the price per unit. In the chapter's example, a 400 W refrigerator used 8 hours a day for 30 days consumes 96 kW h, costing Rs 288 at Rs 3.00 per kW h.

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