Current Electricity | ICSE Class 10 Physics Notes

This note covers electric charge and current, potential difference, resistance and Ohm's law, the experimental verification of Ohm's law and the V-I graph, ohmic and non-ohmic conductors, the factors that affect resistance, specific resistance, conductors, semiconductors, insulators and superconductors, the emf, terminal voltage and internal resistance of a cell, and resistors in series, in parallel and in simple networks, with worked numerical problems.

What are electric charge and electric current?

In a metal conductor, some electrons are practically free to move, and they carry the current among fixed positive ions. In electrolytic solutions, positive and negative ions both move.

What is electric charge?

Electric charge (Q) is the quantity whose flow makes up a current. The SI unit of electric charge is the coulomb (C). An electron carries a negative charge of 1.6 × 10⁻¹⁹ C, so one coulomb is the charge on nearly 6 × 10¹⁸ electrons.

How is electric current defined?

Electric current (I) is the rate of flow of electric charge through a cross-section of a conductor. If a net charge Q flows across any cross-section in time t, the current is I = Q/t, so Q = It.

The SI unit of current is the ampere (A). One ampere is the flow of one coulomb of charge per second, so 1 A = 1 C/s. Small currents are given in milliampere (1 mA = 10⁻³ A) or microampere (1 µA = 10⁻⁶ A).

Which way does current flow?

Electrons were not known when electricity was first studied, so current was taken as a flow of positive charge. By convention, the direction of current is still taken as opposite to the direction of flow of electrons.

A continuous and closed path of current is an electric circuit; if it is broken anywhere, the current stops. Current is measured with an ammeter, which is always connected in series.

A rectangular loop with a cell at the bottom left, a plug key K along the bottom, a bulb at the top and an ammeter A on the right. Arrows labelled I show current leaving the positive terminal, passing through the bulb and ammeter, and returning through the key.

See Fig. 11.1 in your NCERT textbook

Worked example 1. A bulb filament draws 0.5 A for 10 minutes. Find the charge that flows through the circuit.

Given: I = 0.5 A; t = 10 min = 600 s. Formula: Q = It. Substitute: Q = 0.5 A × 600 s. Answer: Q = 300 C flows in 600 s.

What is potential difference, and how is it measured?

Water in a horizontal tube flows when there is a pressure difference between its ends. Similarly, electrons in a wire move when there is a difference of electric pressure, the potential difference, along it. A cell or battery produces this by chemical action.

The potential difference (V) between two points in a current-carrying circuit is the work done to move a unit charge from one point to the other: V = W/Q, so the work done is W = QV.

The SI unit of potential difference is the volt (V). One volt is the potential difference between two points when 1 joule of work is done to move 1 coulomb of charge between them, so 1 V = 1 J/C. A 6 V battery gives 6 J of energy to each coulomb passing through it.

Which instrument measures potential difference?

Potential difference is measured with a voltmeter, which is always connected in parallel across the two points. Moving 2 C across 12 V, for example, needs W = 12 V × 2 C = 24 J of work.

What does Ohm's law state?

The German physicist Georg Simon Ohm (1787 to 1854) found the relationship between the current in a metallic wire and the potential difference across its ends.

Definition: Ohm's law states that the potential difference V across the ends of a given metallic wire is directly proportional to the current I flowing through it, provided its temperature remains the same.

So V ∝ I and V/I is a constant, the resistance R of the wire at that temperature: V = IR. Rearranged, R = V/I and I = V/R, with V in volts, I in amperes and R in ohms.

What is resistance?

Resistance is the property of a conductor to resist the flow of charges through it: electrons are held back by the attraction of the atoms among which they move.

The SI unit of resistance is the ohm (Ω). A conductor has a resistance of 1 Ω if a potential difference of 1 V across it drives a current of 1 A, so 1 ohm = 1 volt/1 ampere.

How does resistance control current?

From I = V/R, current is inversely proportional to resistance: doubling the resistance halves the current. A rheostat is a variable resistance used to change the current without changing the voltage source.

Worked example 2. What current do (a) a bulb filament of 1200 Ω and (b) a heater coil of 100 Ω draw from a 220 V source?

Given: V = 220 V; R = 1200 Ω and 100 Ω. Formula: I = V/R. Substitute: (a) 220 V/1200 Ω; (b) 220 V/100 Ω. Answer: (a) 0.18 A; (b) 2.2 A.

Worked example 3. A heater draws 4 A when the potential difference across it is 60 V. What current will it draw at 120 V?

Given: V₁ = 60 V; I₁ = 4 A; V₂ = 120 V. Formula: R = V₁/I₁, then I₂ = V₂/R. Substitute: R = 60 V/4 A = 15 Ω; I₂ = 120 V/15 Ω. Answer: I₂ = 8 A.

How is Ohm's law verified experimentally?

The wire used is nichrome, an alloy of nickel, chromium, manganese and iron whose resistivity changes very little with temperature.

Apparatus: a nichrome wire XY about 0.5 m long, an ammeter, a voltmeter, a plug key, connecting wires and four 1.5 V cells.

Four cells in series at the top, with dashed taps so that one to four cells can be used. Current passes to X, through the nichrome wire XY (a zigzag marked R), through the ammeter A and the plug key K, and back to the cells. The voltmeter V is connected across X and Y, with its + terminal at X.

See Fig. 11.2 in your NCERT textbook

What are the steps?

1. Connect the ammeter in series with the wire XY and the voltmeter in parallel across XY.
2. Using one cell, plug the key and record the current I and the potential difference V.
3. Take out the key after each reading.
4. Repeat with two, three and then four cells, recording I and V each time.
5. Calculate V/I for each pair of readings.
6. Plot a graph of V (y-axis) against I (x-axis).

What is observed and concluded?

V/I is approximately the same each time, and the V-I graph is a straight line through the origin. So V ∝ I at constant temperature, verifying Ohm's law; the constant ratio is the resistance.

Instead of changing the number of cells, a rheostat in series can be used to change the current.

Note: Ohm's law holds only while the wire's temperature stays the same. Removing the key between readings limits the time current flows, and so limits heating.

How is resistance found from the slope of a V-I graph?

For an ohmic conductor, a graph of V (y-axis) against I (x-axis) is a straight line through the origin. Its slope, the change in V divided by the change in I, equals V/I, so R = ΔV/ΔI.

A steeper V-I line means a larger resistance. If I is plotted against V instead, the slope is 1/R, and a steeper line means a smaller resistance.

The y-axis is potential difference (V), marked 0 to 2.0 V; the x-axis is current (A), marked 0 to 0.6 A. The plotted points, small crosses, lie on a straight line from the origin: V increases linearly as I increases.

See Fig. 11.3 in your NCERT textbook

How is the slope read?

1. Draw the best straight line through the points and the origin.
2. Choose two points on the line that are far apart.
3. Read the change in V (ΔV) and the change in I (ΔI) between them.
4. Calculate R = ΔV/ΔI in ohms.

Worked example 4. For a resistor, I = 0.5, 1.0, 2.0, 3.0 and 4.0 A gave V = 1.6, 3.4, 6.7, 10.2 and 13.2 V. Find its resistance.

Given: five readings close to a straight line. Formula: R = ΔV/ΔI. Substitute: ΔV = 13.2 V − 1.6 V = 11.6 V; ΔI = 4.0 A − 0.5 A = 3.5 A; R = 11.6 V/3.5 A. Answer: R ≈ 3.3 Ω; each V/I ratio lies between 3.2 and 3.4 V per A.

What are ohmic and non-ohmic conductors?

Ohm's law holds for a large class of materials but is not a fundamental law of nature. An ohmic conductor obeys it, giving a straight-line V-I graph through the origin, as nichrome wire does at constant temperature. A non-ohmic conductor does not; such devices are widely used in electronics.

In what ways can a conductor fail to obey Ohm's law?

• V stops being proportional to I: the graph curves away from a straight line.
• The relation depends on the sign of V: reversing V does not give the same size of current in the opposite direction, as in a diode.
• The relation is not unique: more than one V gives the same I, as in gallium arsenide (GaAs).

In Figure 3.5, the V against I curve for a good conductor follows the dashed Ohm's law line, then bends above it. In Figure 3.6, diode current rises steeply, to about 1.5 mA, for positive voltages near 0.2 V, but is only microamperes for negative voltages. In Figure 3.7, GaAs current rises, levels off, falls in a negative resistance region, then rises again.

See Figs. 3.5, 3.6 and 3.7 in your NCERT textbook

Heating also matters. A nichrome toaster element measures 75.3 Ω at 27.0 °C, but on 230 V its current settles at 2.68 A, so its hot resistance is 230 V/2.68 A ≈ 85.8 Ω. As R rises with heating, its V-I graph over this range is not straight.

Table: Ohmic and non-ohmic conductors compared.

What factors affect the resistance of a conductor?

A board carries four wires connected one at a time: (1) nichrome of length l, (2) nichrome of the same thickness and length 2l, (3) thicker nichrome of length l, and (4) copper of the same length and cross-section as wire (1). A cell, an ammeter A and a plug key K complete the circuit.

See Fig. 11.5 in your NCERT textbook

What do the observations show?

Doubling the length halves the ammeter reading; a thicker wire of the same material gives a larger reading; a different material of the same size gives a different reading.

So resistance depends on length, area of cross-section and the nature of the material. For a uniform metallic conductor, R ∝ l and R ∝ 1/A.

Derivation: why resistance depends on length and area

Take a slab of an ohmic material of length l and cross-section A, with resistance R = V/I.

1. Two identical slabs end to end carry the same current I, with V across each, so 2V across both: resistance 2V/I = 2R. So R ∝ l.
2. Cutting one slab lengthwise gives two halves of area A/2, each carrying I/2 with V across it.
3. Each half has resistance V/(I/2) = 2R, so halving the area doubles the resistance: R ∝ 1/A.
4. Combining, R ∝ l/A, so R = ρl/A, where ρ depends on the material.

R = ρl/A: resistance increases with length and decreases as the area of cross-section increases, for a given material at a given temperature.

How does temperature affect resistance?

Resistance and resistivity both vary with temperature. In metals, resistivity rises with temperature because electrons collide more often. Alloys such as nichrome, manganin and constantan change very little, so they are used in standard resistors. In semiconductors, resistivity falls as temperature rises.

Table: Factors affecting the resistance of a wire.

Worked example 5. A wire of length l and area A has a resistance of 4 Ω. Find the resistance of a wire of the same material with length l/2 and area 2A.

Given: R₁ = ρl/A = 4 Ω. Formula: R₂ = ρ(l/2)/(2A). Substitute: R₂ = (1/4) × 4 Ω. Answer: R₂ = 1 Ω, a quarter of the original.

What is specific resistance (resistivity)?

In R = ρl/A, the constant ρ (rho) is the electrical resistivity, or specific resistance, of the material. It depends on the material and its temperature, not on the dimensions of the conductor.

Rearranging gives ρ = RA/l. With l = 1 m and A = 1 m², ρ = R, so resistivity is the resistance of a conductor of unit length and unit area of cross-section. The SI unit of resistivity is the ohm metre (Ω m).

Metals and alloys have resistivities of 10⁻⁸ Ω m to 10⁻⁶ Ω m; insulators such as rubber and glass, roughly 10¹⁰ to 10¹⁷ Ω m.

Table: Electrical resistivity of some substances at 20 °C.

Why are different materials chosen for different jobs?

• Alloys such as nichrome, manganin and constantan generally have a higher resistivity than their constituent metals and do not oxidise readily at high temperatures, so they are used in electric irons, toasters and heaters.
• Tungsten is used almost exclusively for bulb filaments.
• Copper and aluminium are generally used for electrical transmission lines.

Worked example 6. A metal wire 1 m long and 0.3 mm in diameter has a resistance of 26 Ω at 20 °C. Find its resistivity and identify the metal.

Given: R = 26 Ω; l = 1 m; d = 3 × 10⁻⁴ m. Formula: ρ = RA/l, with A = πd²/4. Substitute: ρ = 26 Ω × 3.14 × (3 × 10⁻⁴ m)² / (4 × 1 m). Answer: ρ = 1.84 × 10⁻⁶ Ω m, matching manganese.

How are materials classified by resistivity, and what are superconductors?

Materials are classified as conductors, semiconductors and insulators, in increasing order of resistivity. Insulators such as ceramic, rubber and plastics have resistivities 10¹⁸ times greater than metals or more. Semiconductors such as silicon lie in between.

What is a superconductor?

A superconductor has zero resistivity, or infinite conductivity. Earlier, only metals and alloys at very low temperatures, 0 to 15 K, were known to be superconductors. Now some ceramic materials and mixed oxides show superconductivity at up to 150 K.

A material is superconducting only below its critical temperature. With zero resistance, current flows without energy being dissipated as heat.

A superconductor is also a perfect diamagnet: magnetic field lines are completely expelled from it, the Meissner effect. It repels a magnet, and superconducting magnets can be used to run magnetically levitated superfast trains.

Table: Classes of material by resistivity.

What are the emf, terminal voltage and internal resistance of a cell?

An electrolytic cell keeps a steady current in a circuit. Its positive (P) and negative (N) electrodes dip in an electrolytic solution.

Part (a) shows a vessel of electrolyte with electrodes P and N, and points A and B in the electrolyte close to them; a resistor R outside carries current I from C to D. Part (b) shows the cell symbol, + for P and − for N, with R across it.

See Fig. 3.12 in your NCERT textbook

What is electromotive force (emf)?

The electromotive force (emf), ε, of a cell is the potential difference between its electrodes in open circuit, when no current is drawn. It is a potential difference, not a force; the name is historical. It is work done per unit charge, and the SI unit of emf is the volt.

What is internal resistance?

The current through the external resistor also flows through the electrolyte, from N to P. This electrolyte has a finite resistance r, the internal resistance. It varies from cell to cell, and is much higher for dry cells than for common electrolytic cells.

Internal resistance is smaller when the electrodes have a larger area in the electrolyte, larger when they are farther apart, and depends on the nature and temperature of the electrolyte.

Derivation: terminal voltage and current drawn from a cell

A cell of emf ε and internal resistance r drives a current I through an external resistor R.

1. Part of the emf drives the current through r, so the potential difference across the terminals, the terminal voltage, is V = ε − Ir.
2. The same V is across the external resistor, so V = IR.
3. Equating, IR = ε − Ir.
4. Collecting terms, I(R + r) = ε.

I = ε/(R + r): the current depends on the total resistance R + r, and the maximum current, when R = 0, is ε/r.

The terminal voltage, V = ε − Ir, equals the emf only when no current flows. Internal resistance may be neglected when ε is much greater than Ir.

Table: Emf and terminal voltage compared.

Worked example 7. A battery of emf 10 V and internal resistance 3 Ω drives 0.5 A through a resistor. Find the resistance and the terminal voltage.

Given: ε = 10 V; r = 3 Ω; I = 0.5 A. Formula: I = ε/(R + r), so R + r = ε/I; then V = ε − Ir. Substitute: R + 3 Ω = 10 V/0.5 A = 20 Ω; V = 10 V − (0.5 A × 3 Ω). Answer: R = 17 Ω and V = 8.5 V; check IR = 0.5 A × 17 Ω = 8.5 V.

How are resistors combined in series?

Resistors joined end to end, so that the same current passes through each in turn, are in series.

R₁, R₂ and R₃ are joined end to end between X and Y, with a battery, a plug key K and an ammeter A in the loop. Figure 11.6 has one voltmeter across X and Y; Figure 11.8 has voltmeters V₁, V₂ and V₃ across R₁ (X to P), R₂ (Q to S) and R₃ (T to Y).

See Figs. 11.6 and 11.8 in your NCERT textbook

The ammeter reads the same wherever it is placed, so the current is the same through each resistor. The potential difference across the combination is the sum of those across the resistors: V = V₁ + V₂ + V₃.

Derivation: equivalent resistance in series

Let Rₛ carry the same current I with the same V across it as the combination.

1. For the whole circuit, V = IRₛ.
2. For each resistor, V₁ = IR₁, V₂ = IR₂ and V₃ = IR₃.
3. Since V = V₁ + V₂ + V₃, IRₛ = IR₁ + IR₂ + IR₃.
4. Divide every term by I.

Rₛ = R₁ + R₂ + R₃: the equivalent resistance in series is the sum of the individual resistances, and is greater than any one of them.

What are the drawbacks of series circuits?

A bulb and a heater cannot sensibly be joined in series, as they need very different currents. If one component fails, the circuit breaks and none works, as with series fairy lights.

Worked example 8. A 20 Ω lamp and a 4 Ω conductor are in series with a 6 V battery. Find the total resistance, the current and the potential difference across each.

Given: R₁ = 20 Ω; R₂ = 4 Ω; V = 6 V. Formula: Rₛ = R₁ + R₂, I = V/Rₛ and V₁ = IR₁. Substitute: Rₛ = 24 Ω; I = 6 V/24 Ω; V₁ = 0.25 A × 20 Ω; V₂ = 0.25 A × 4 Ω. Answer: 24 Ω; 0.25 A; 5 V and 1 V, which add up to 6 V.

How are resistors combined in parallel?

Resistors connected between the same two points, each giving a separate path, are in parallel.

Resistors R₁, R₂ and R₃ branch from point X and rejoin at Y, with a voltmeter V across X and Y. A battery, a plug key K and an ammeter A in the main line complete the circuit, with current I entering at X and leaving at Y.

See Figs. 11.7 and 11.10 in your NCERT textbook

The potential difference V is the same across each resistor. The total current equals the sum of the branch currents: I = I₁ + I₂ + I₃.

Derivation: equivalent resistance in parallel

Let Rₚ draw the same total current I from the same V.

1. For the combination, I = V/Rₚ.
2. For each branch, I₁ = V/R₁, I₂ = V/R₂ and I₃ = V/R₃.
3. Since I = I₁ + I₂ + I₃, V/Rₚ = V/R₁ + V/R₂ + V/R₃.
4. Divide every term by V.

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃: the reciprocal of the equivalent resistance equals the sum of the reciprocals of the individual resistances.

For two resistors, Rₚ = R₁R₂/(R₁ + R₂). Adding a resistor in parallel lowers the total resistance, and Rₚ is less than the smallest resistance.

Why are household gadgets connected in parallel?

A parallel circuit divides the current and lowers the total resistance, which suits gadgets that need different currents. If one fails, the others stay connected.

Worked example 9. Resistors of 5 Ω, 10 Ω and 30 Ω are in parallel across 12 V. Find each current, the total current and the total resistance.

Given: R₁ = 5 Ω; R₂ = 10 Ω; R₃ = 30 Ω; V = 12 V across each. Formula: I₁ = V/R₁, I = I₁ + I₂ + I₃ and Rₚ = V/I. Substitute: 12 V/5 Ω; 12 V/10 Ω; 12 V/30 Ω; 1/Rₚ = 1/5 + 1/10 + 1/30 = 1/3. Answer: 2.4 A, 1.2 A and 0.4 A; total 4 A; Rₚ = 3 Ω, since 12 V/4 A = 3 Ω.

Table: Series and parallel combinations compared.

How are simple networks of resistors solved?

A network of series and parallel resistors is reduced step by step to one equivalent resistance; Ohm's law then gives the current.

1. Replace each group of resistors in parallel by its equivalent resistance.
2. Add resistances that are now in series.
3. Repeat until one equivalent resistance R remains.
4. Find the total current from I = V/R.

R₁ and R₂ are joined in parallel in the upper part of the circuit, and R₃, R₄ and R₅ in parallel in the lower part. The two groups are in series with a battery on the left, a plug key K at the bottom left and an ammeter A on the right.

See Fig. 11.12 in your NCERT textbook

Worked example 10. In this network, R₁ = 10 Ω, R₂ = 40 Ω, R₃ = 30 Ω, R₄ = 20 Ω and R₅ = 60 Ω, with a 12 V battery. Find the total resistance and current.

Given: five resistances; V = 12 V. Formula: 1/Rₚ = sum of reciprocals for each group; then R = R(upper) + R(lower) and I = V/R. Substitute: 1/10 + 1/40 = 5/40, giving 8 Ω; 1/30 + 1/20 + 1/60 = 6/60, giving 10 Ω; R = 18 Ω; I = 12 V/18 Ω. Answer: 18 Ω; 0.67 A.

How can a given resistance be made from identical resistors?

Three 6 Ω resistors give 9 Ω when two in parallel (3 Ω) are joined in series with the third. They give 4 Ω when two in series (12 Ω) are joined in parallel with the third: 12 × 6/(12 + 6) = 4 Ω.

Glossary

• Electric current — the rate of flow of electric charge through a cross-section of a conductor, I = Q/t, measured in amperes with an ammeter in series.
• Potential difference — the work done to move a unit charge between two points of a current-carrying circuit, V = W/Q, measured in volts with a voltmeter in parallel.
• Resistance — the property of a conductor to resist the flow of charges through it; it equals V/I and is measured in ohms.
• Ohm's law — the potential difference across a metallic wire is directly proportional to the current through it, provided its temperature remains the same.
• Ohmic conductor — a conductor that obeys Ohm's law, giving a straight-line V-I graph through the origin, such as nichrome wire at constant temperature.
• Non-ohmic conductor — a material or device, such as a diode or gallium arsenide, for which potential difference is not proportional to current.
• Specific resistance — the resistivity ρ in R = ρl/A, a property of the material at a given temperature, measured in ohm metre.
• Superconductor — a material with zero resistivity, such as certain metals and alloys at very low temperatures, or some ceramics at up to 150 K.
• Electromotive force (emf) — the potential difference between the electrodes of a cell when no current is drawn; it is measured in volts and is not a force.
• Internal resistance — the finite resistance r of the electrolyte inside a cell, through which current flows from the negative to the positive electrode.
• Terminal voltage — the potential difference across a cell's terminals while it supplies current I, equal to ε − Ir and less than the emf.

Common errors and misconceptions

• Misconception: Electrons are used up as current flows round a circuit. Correct: Electrons are not consumed; the source supplies energy to move them through the gadgets, and that energy is what is paid for.
• Misconception: Conventional current flows in the same direction as electrons. Correct: It is taken as opposite to electron flow, from the positive to the negative terminal in the external circuit.
• Misconception: Any device for which V = IR can be written obeys Ohm's law. Correct: V = IR defines resistance for any conductor; Ohm's law requires R to stay constant, which a diode does not.
• Misconception: Resistance and resistivity mean the same thing. Correct: Resistance depends on length, area, material and temperature; resistivity is a property of the material at a given temperature, not of the conductor's dimensions.
• Misconception: A thicker wire has more resistance because it has more metal. Correct: A thicker wire has a larger area of cross-section, and R ∝ 1/A, so its resistance is lower.
• Misconception: The terminal voltage of a cell always equals its emf. Correct: They are equal only when no current is drawn; while current flows, V = ε − Ir, which is less than ε.
• Misconception: Adding a resistor in parallel increases the total resistance. Correct: It adds an extra path, so the total resistance decreases below the smallest resistance in the group.

Exam-style questions with model answers

Q1. State Ohm's law. Under what condition does it hold? [2 marks]

1. The potential difference V across the ends of a metallic conductor is directly proportional to the current I through it, so V = IR, where R is its resistance.
2. Condition: the law holds provided the temperature of the conductor remains the same.

Q2. A car battery has an emf of 12 V and an internal resistance of 0.4 Ω. Calculate the maximum current that can be drawn from it. [2 marks]

1. The current is greatest when the external resistance is zero, so I(max) = ε/r.
2. I(max) = 12 V/0.4 Ω = 30 A.

Q3. Distinguish between ohmic and non-ohmic conductors, with one example of each and the shape of their V-I graphs. [3 marks]

1. An ohmic conductor obeys Ohm's law: V/I is constant at constant temperature, and its V-I graph is a straight line through the origin. Example: nichrome wire at constant temperature.
2. A non-ohmic conductor does not obey Ohm's law: V is not proportional to I, or the relation changes when V is reversed.
3. Its V-I graph is curved, not a straight line through the origin. Example: a diode, which conducts readily one way but passes only microamperes when V is reversed.

Q4. Derive an expression for the equivalent resistance of three resistors R₁, R₂ and R₃ in series. [4 marks]

1. In series, the same current I flows through each resistor, and V = V₁ + V₂ + V₃.
2. By Ohm's law, V₁ = IR₁, V₂ = IR₂ and V₃ = IR₃.
3. For the equivalent resistor Rₛ carrying the same I with the same V, V = IRₛ.
4. So IRₛ = IR₁ + IR₂ + IR₃; dividing by I, Rₛ = R₁ + R₂ + R₃, greater than any single resistance.

Q5. A hot plate on a 220 V line has two coils A and B, each of 24 Ω, which may be used separately, in series or in parallel. Find the current in each case. [4 marks]

1. Separately: I = V/R = 220 V/24 Ω ≈ 9.17 A.
2. In series: Rₛ = 24 Ω + 24 Ω = 48 Ω, so I = 220 V/48 Ω ≈ 4.58 A.
3. In parallel: Rₚ = (24 × 24)/(24 + 24) = 12 Ω, so I = 220 V/12 Ω ≈ 18.33 A.

Q6. Describe an experiment to verify Ohm's law. Name the apparatus, describe the circuit, state the observations and explain how the graph is used. [5 marks]

1. Apparatus: a nichrome wire XY about 0.5 m long, an ammeter, a voltmeter, a plug key, connecting wires and four 1.5 V cells.
2. Circuit: the cells, wire XY, ammeter and plug key are joined in series; the voltmeter is connected in parallel across X and Y.
3. Using one cell, plug the key and record the current I and the potential difference V; take out the key after each reading.
4. Repeat with two, three and four cells, and calculate V/I for each pair of readings.
5. Observation: V/I is about the same each time, and a graph of V (y-axis) against I (x-axis) is a straight line through the origin, so V ∝ I at constant temperature, verifying Ohm's law.
6. The slope of the line, ΔV/ΔI, gives the resistance of the wire.

Q7. Derive the expression for the equivalent resistance of R₁, R₂ and R₃ in parallel. A lamp of 100 Ω, a toaster of 50 Ω and a water filter of 500 Ω are in parallel on a 220 V supply. What single resistance draws the same current, and what is that current? [6 marks]

1. In parallel, the potential difference V is the same across each resistor, and the current divides: I = I₁ + I₂ + I₃.
2. By Ohm's law, I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃, and for the equivalent resistor I = V/Rₚ.
3. So V/Rₚ = V/R₁ + V/R₂ + V/R₃; dividing by V gives 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃.
4. For the appliances: 1/Rₚ = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500, so Rₚ = 31.25 Ω.
5. Current: I = V/Rₚ = 220 V/31.25 Ω = 7.04 A.
6. So an electric iron of 31.25 Ω would draw the same 7.04 A as all three together.

Key takeaways

• Current is the rate of flow of charge, I = Q/t; one ampere is one coulomb per second, measured with an ammeter in series.
• Potential difference is work done per unit charge, V = W/Q; one volt is one joule per coulomb, measured by a voltmeter in parallel.
• Ohm's law: V is directly proportional to I for a metallic conductor, provided its temperature remains the same, so V = IR.
• The slope of a V-I graph, with V on the y-axis, gives the resistance; an ohmic conductor gives a straight line through the origin.
• Resistance depends on length, area of cross-section, material and temperature: R = ρl/A, where ρ is the specific resistance in ohm metre.
• Superconductors have zero resistivity: metals and alloys at very low temperatures, and some ceramics and mixed oxides at up to 150 K.
• Emf is a cell's potential difference in open circuit; while current flows, terminal voltage V = ε − Ir and I = ε/(R + r).
• In series, Rₛ = R₁ + R₂ + R₃ and current is the same; in parallel, 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ and potential difference is the same.

Test yourself

If the potential difference across a resistor is halved while its resistance stays the same, what happens to the current?

The current is also halved, because I = V/R and R is unchanged.

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

Alloys generally have a higher resistivity than their constituent metals and do not oxidise readily at high temperatures.

A 12 V battery drives 2.5 mA through an unknown resistor. What is its resistance?

R = V/I = 12 V/0.0025 A = 4800 Ω, that is, 4.8 kΩ.

A wire of resistance R is cut into five equal parts, which are joined in parallel. What is the new resistance?

Each part has resistance R/5; five in parallel give R/25, so the resistance falls to one twenty-fifth of R.

What are the highest and lowest resistances obtainable from coils of 4 Ω, 8 Ω, 12 Ω and 24 Ω?

Highest 48 Ω, with all in series; lowest 2 Ω, with all in parallel, since 1/4 + 1/8 + 1/12 + 1/24 = 1/2.

What is the Meissner effect?

The complete expulsion of magnetic field lines from a superconductor, which makes it a perfect diamagnet that repels a magnet.