Ohm’s Law | ICSE Class 10 Physics Notes

This note covers electric charge and current, potential difference, Ohm’s law, experimental verification, voltage-current graphs, resistance and resistivity, ohmic and non-ohmic behaviour, superconductors, series and parallel combinations, simple resistor networks, electromotive force and internal resistance.

What are electric charge, current and potential difference?

Electric charge is the electrical property carried by particles such as electrons. Electrons carry negative charge. A conductor allows electric charge to flow through it; a metallic wire is a familiar example. An electric circuit is a continuous, closed conducting path for current.

Electric current is the rate of flow of charge through a cross-section, meaning a cut across a conductor. Let Q represent the charge passing through this section, t the time taken, and I the current.

I = Q/t

The SI, or International System of Units, unit of charge is the coulomb, written C. The SI unit of current is the ampere, written A. Time is measured in seconds, written s. Thus, 1 A = 1 C/s.

In metallic wires, electrons carry the moving charge. Conventional current has the direction in which positive charge would flow, opposite to electron flow. Through the external circuit, conventional current travels from the positive terminal of the cell towards its negative terminal.

What does potential difference measure?

A cell converts chemical energy into electrical energy to maintain a potential difference. A battery is a source made using one or more cells. The two electrical connections of a cell are its terminals.

Definition: Potential difference between two points is the work done per unit charge in moving charge between those points. It is also called voltage. Work is energy transferred; let W denote this work and V denote the potential difference.

V = W/Q

The SI unit of potential difference is the volt, also written V. The SI unit of work is the joule, written J. Hence 1 V = 1 J/C. In a formula V represents voltage; following a numerical value, V denotes volts.

Current measures charge passing per second, whereas potential difference measures energy transferred per coulomb. A cell can maintain a potential difference across its terminals even when the external circuit is open, meaning the conducting path is broken and no current flows.

Worked example 1. A charge of 2 C moves between points at a potential difference of 12 V. Find the work done.

Formula: W = VQ. Substitute: W = 12 × 2. Answer: W = 24 J. Each coulomb transfers 12 J, so two coulombs transfer 24 J.

What does Ohm’s law state, and when does it apply?

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

Directly proportional means that the ratio of the two quantities remains constant. The symbol ∝ means “is proportional to”. Thus V ∝ I, or V/I is constant, for the given wire at a fixed temperature.

This constant is the wire’s resistance, denoted by R. Resistance is the property of a conductor that opposes the flow of charge. A resistor is a conducting component with appreciable resistance.

V = IR

R = V/I

I = V/R

The SI unit of resistance is the ohm, written Ω. A conductor has resistance 1 Ω when a potential difference of 1 V produces a current of 1 A through it. Thus, 1 ohm = 1 V/A.

How should proportionality be used?

For a fixed resistance, doubling the potential difference doubles the current. Halving the potential difference halves the current. When comparing different resistances at the same potential difference, current is inversely proportional to resistance: doubling resistance halves current.

These are different comparisons. In the first, resistance remains constant while voltage changes. In the second, voltage remains constant while resistance changes. State what is held constant before describing the relationship.

Note: The temperature condition is part of Ohm’s law. If a wire becomes hotter while current passes, its resistance can change. A calculation that assumes constant resistance must retain that assumption.

Worked example 2. A heater draws 4 A when the potential difference across it is 60 V. Find its resistance and its current at 120 V, assuming its resistance remains constant.

Formula: R = V/I; I = V/R. Substitute: R = 60/4 = 15 Ω; I = 120/15. Answer: R = 15 Ω and the new current is 8 A. The voltage doubles and, under the stated condition, the current doubles.

How is Ohm’s law verified experimentally?

An ammeter measures current. It is connected in series, meaning in the same conducting path as the component whose current is measured. A voltmeter measures potential difference. It is connected in parallel, meaning across the same two points as the component.

A plug key is a switch that opens or closes the circuit. Nichrome is a metal alloy, meaning a mixture containing metals, used as the test wire. The experiment compares the voltage across this wire with the current through it.

The wire between X and Y is labelled R. The ammeter and key K lie in the main circuit with the cells. The voltmeter is connected across X and Y. Arrows indicate the conventional current direction.

See Fig. 11.2 in your NCERT textbook

What is the procedure?

1. Connect a nichrome wire, an ammeter, a plug key and a cell in series. Connect the voltmeter across the wire. Keep the same wire throughout the investigation.
2. Close the key and record the current through the wire and the potential difference across it. These two readings form one corresponding pair.
3. Repeat with two cells, then three cells and then four cells, using cells rated 1.5 V each. Record the actual meter readings for each arrangement.
4. Calculate V/I for every measured pair. Compare these ratios while keeping the wire’s temperature unchanged.
5. Plot potential difference against current. Approximately equal ratios and a straight line through the origin support Ohm’s law for the wire under these conditions.

The origin is the graph point where both plotted quantities are zero. The experimental ratios are approximately the same; they should not be described as perfectly identical measured values. The ideal relationship is V/I = R.

Use the measured potential difference across the wire rather than replacing it with the sum of the cell ratings. The voltmeter measures the quantity that belongs in the ratio. Open the key after taking readings, and avoid heating that changes the test wire’s resistance.

How can resistance regulate current?

A rheostat is a variable resistor used to change circuit resistance and regulate current without changing the voltage source. Its function differs from that of a meter: it adjusts resistance, whereas an ammeter or voltmeter measures an electrical quantity.

How does a voltage-current graph give resistance?

On a graph of V against I, put current on the horizontal axis and potential difference on the vertical axis. An axis is a reference line carrying the scale for a plotted quantity. Label each axis with its quantity and unit.

Current in amperes is marked horizontally and potential difference in volts vertically. The plotted line rises straight from the origin, showing the linear relationship between voltage and current.

See Fig. 11.3 in your NCERT textbook

The slope is the vertical change divided by the corresponding horizontal change. Choose two separated points on the straight line. Let V₁ and V₂ denote their voltages, and I₁ and I₂ their currents. The subscripts identify the first and second points.

R = (V₂ − V₁)/(I₂ − I₁)

Here the minus sign means subtraction. Since the slope has units of volts per ampere, it gives resistance in ohms. With the same axis scales, a steeper straight V-against-I line represents a greater resistance.

Why do the axis labels matter?

If the axes are reversed and current is plotted vertically against voltage horizontally, the slope becomes current change divided by voltage change. It is then 1/R, the reciprocal of resistance. A reciprocal is one divided by a quantity.

Read the labels before using a slope. The phrase “voltage-current graph” by itself does not specify which quantity is vertical. The resistance-from-slope rule applies directly when voltage is on the vertical axis and current on the horizontal axis.

Table: Current and potential difference readings for graph practice.

These readings support an approximately straight relationship. Retain the recorded values when plotting; do not replace them with numbers chosen to make the ratios identical. A straight line representing the overall trend gives a resistance of approximately 3.3 Ω.

What determines resistance, and how does it differ from resistivity?

For a uniform metallic conductor at a fixed temperature, resistance depends on its length, cross-sectional area and material. Uniform means that the wire’s cross-section and material are the same along its length.

Let l denote length and A denote cross-sectional area, the area exposed by a cut across the wire. In this formula A represents area, whereas A written after a current value means ampere. Keeping these uses distinct prevents unit errors.

Resistance is directly proportional to length when material, area and temperature remain the same. It is inversely proportional to cross-sectional area when material, length and temperature remain the same. Combining these relationships introduces the material property called resistivity.

R = ρl/A

Resistivity, also called specific resistance, is denoted by the Greek letter ρ, pronounced rho. It characterises the material at the specified temperature. The SI unit of resistivity is the ohm metre, written Ω m. Length is in metres, written m, and area in square metres, written m².

How should the effects be compared?

A longer wire and a shorter wire made of the same material can have different resistances but the same resistivity at the same temperature. Resistance describes the particular conductor; resistivity describes its material.

The resistivity of an alloy is generally higher than that of its constituent metals. Copper and aluminium are generally used for electrical transmission lines. Both uses reflect the importance of material choice when selecting a conductor for a particular purpose.

Worked example 3. A wire has resistance 26 Ω, length 1 m and diameter 0.3 mm at 20°C. Find its resistivity. Here mm means millimetre and °C means degrees Celsius. Let d denote diameter; π is the circle constant, approximately 3.14, and a superscript ² means squared.

Formula: A = πd²/4; ρ = RA/l. Substitute: d = 0.0003 m; ρ = 26 × π × (0.0003)²/(4 × 1). Answer: ρ = 1.84 × 10⁻⁶ Ω m, to the stated precision. Here 10⁻⁶ means one millionth.

How do ohmic, non-ohmic and superconducting materials differ?

An ohmic conductor obeys Ohm’s law over the conditions being considered. Its voltage-current ratio remains constant, giving a straight line through the origin. A non-ohmic device does not maintain this proportional relationship.

Calculating V/I at one operating point does not by itself establish ohmic behaviour. An operating point is a particular corresponding voltage and current. The test is whether the ratio stays constant as the applied voltage changes under the required conditions.

What can cause departures from proportionality?

Some devices have a curved voltage-current relationship. In others, reversing the applied voltage changes the behaviour. A diode is an electronic device whose current-voltage relationship depends on the direction of the applied voltage, and it provides an example of non-ohmic behaviour.

Ohm’s law is obeyed by many substances, but it is not a fundamental law of nature applying to every conducting device. Even materials that obey it over some range can depart from it outside that range.

Temperature also matters. The resistivity of metals increases with temperature. Nichrome, manganin and constantan are alloys whose resistivity has a very weak temperature dependence; “very weak” does not mean zero. Their resistance changes very little with temperature.

Semiconductors are materials whose resistivity lies between that of conductors and insulators. An insulator offers very high resistance compared with a conductor of the same size. Unlike metals, semiconductors have resistivity that decreases as temperature increases.

What is a superconductor?

A superconductor is a material with zero electrical resistivity in its superconducting state. Certain metals and alloys show this behaviour at very low temperatures. Some ceramic materials and mixed oxides also show superconductivity.

This is different from merely having a small resistivity. A good ordinary conductor still has resistance; a superconducting material has zero resistivity in that state. Do not describe all cold metals as superconductors or treat superconductivity as another word for good conduction.

How are resistors in series combined?

Resistors connected end to end along one path form a series combination. The same current passes through every resistor. The total potential difference across the combination equals the sum of the potential differences across its individual resistors.

An equivalent resistance is a single resistance that can replace a combination while preserving the current drawn at the same applied voltage. Let R₁, R₂ and R₃ denote three resistances, and V₁, V₂ and V₃ their respective potential differences.

Three resistors labelled R₁, R₂ and R₃ lie consecutively between X and Y. The voltmeter spans X and Y, while the ammeter, battery and key are in the main loop.

See Fig. 11.6 in your NCERT textbook

Derivation: Equivalent resistance in series

Let Rₛ denote the equivalent series resistance, with the subscript s indicating series. Let V be the total potential difference and I the common current. Each resistor obeys Ohm’s law under the conditions used.

1. The potential differences add: V = V₁ + V₂ + V₃.
2. Apply Ohm’s law to each resistor: V₁ = IR₁, V₂ = IR₂ and V₃ = IR₃.
3. For the equivalent resistor, V = IRₛ. Substitute to obtain IRₛ = IR₁ + IR₂ + IR₃.
4. Divide by the common non-zero current I to obtain the equivalent resistance.

Rₛ = R₁ + R₂ + R₃

For positive resistances, the series total exceeds any individual resistance. The current is common, but the voltages need not be equal. Since the voltage across each resistor is current multiplied by its resistance, the larger resistance has the larger potential difference.

Worked example 4. A lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected in series across 6 V. Find their equivalent resistance, current and individual potential differences. Use the stated resistances as constant values.

Formula: Rₛ = R₁ + R₂; I = V/Rₛ; V₁ = IR₁; V₂ = IR₂. Substitute: Rₛ = 20 + 4 = 24 Ω; I = 6/24 = 0.25 A. Answer: the resistance is 24 Ω, the current is 0.25 A, and the potential differences are 5 V across the lamp and 1 V across the conductor.

The two voltage drops add to the applied 6 V. Here a voltage drop means the potential difference across a resistor in the direction of conventional current. It does not mean that current is consumed along the path.

How are resistors in parallel combined?

A parallel combination connects resistors between the same two points, giving separate paths called branches. Every branch has the same potential difference. The total current entering the combination equals the sum of the branch currents.

Let I₁, I₂ and I₃ denote the currents through resistors R₁, R₂ and R₃ respectively. Let Rₚ denote the equivalent parallel resistance, with subscript p indicating parallel. The common potential difference across the branches is V.

Three branches containing R₁, R₂ and R₃ join the common points X and Y. A voltmeter is connected across X and Y. The battery, key and ammeter are in the external path supplying the combination.

See Fig. 11.7 in your NCERT textbook

Derivation: Equivalent resistance in parallel

1. Add the branch currents to obtain the total current: I = I₁ + I₂ + I₃.
2. Use the common potential difference: I₁ = V/R₁, I₂ = V/R₂ and I₃ = V/R₃.
3. Write the current through the equivalent resistance as I = V/Rₚ.
4. Substitute into the current sum and divide by the common non-zero voltage V.

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

For two resistors this becomes Rₚ = R₁R₂/(R₁ + R₂). The formula follows by adding the two reciprocal terms and then taking the reciprocal of the result.

For positive resistances, the parallel equivalent is smaller than the smallest individual resistance. Each extra branch adds another route for current. At the same applied voltage, a smaller equivalent resistance draws a larger total current.

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

Formula: I₁ = V/R₁; I₂ = V/R₂; I₃ = V/R₃; I = I₁ + I₂ + I₃; Rₚ = V/I. Substitute: I₁ = 12/5 = 2.4 A; I₂ = 12/10 = 1.2 A; I₃ = 12/30 = 0.4 A. Answer: the branch currents are 2.4 A, 1.2 A and 0.4 A; the total is 4 A and Rₚ = 12/4 = 3 Ω.

The branch currents differ because the resistances differ. Their voltages are equal because all three resistors connect across the same pair of points. Identify the connections before deciding which quantity is common.

How are simple series-parallel networks solved?

A resistor network is an arrangement containing several connected resistors. A simple series-parallel network can be reduced by replacing an identifiable series or parallel group with its equivalent resistance, then examining the simpler circuit that remains.

Which relationships should be checked first?

Resistors drawn near each other are not necessarily in series. Follow the connecting wires. If both ends of two resistors join the same two points, those resistors are in parallel even if the drawing places them far apart.

An upper group contains R₁ and R₂ in parallel. A lower group contains R₃, R₄ and R₅ in parallel. The two groups lie in series in the circuit containing the battery, ammeter and key.

See Fig. 11.12 in your NCERT textbook

Worked example 6. A 10 Ω resistor and a 40 Ω resistor form one parallel group. A second parallel group contains 30 Ω, 20 Ω and 60 Ω resistors. The two groups are connected in series across 12 V. Find total resistance and current.

Let R′ and R″ denote the equivalent resistances of the first and second groups; the prime marks distinguish them. Formula: 1/R′ = 1/10 + 1/40; 1/R″ = 1/30 + 1/20 + 1/60; R = R′ + R″; I = V/R.

Substitute: 1/R′ = 5/40, so R′ = 8 Ω; 1/R″ = 6/60, so R″ = 10 Ω. Answer: R = 8 + 10 = 18 Ω and I = 12/18 ≈ 0.67 A. The symbol ≈ means approximately equal to.

Retain exact intermediate values where possible, then round the final answer. The first group’s equivalent, 8 Ω, is below 10 Ω; the second group’s equivalent, 10 Ω, is below 20 Ω. Their series total is greater than either group’s equivalent.

What are electromotive force and internal resistance?

Electromotive force, abbreviated emf and denoted here by E, is the potential difference between the terminals of a cell when no current flows through it. Despite its name, emf is a potential difference, not a mechanical force, and is measured in volts.

A cell contains electrodes, the conducting parts connected to its terminals, and an electrolyte, a substance through which charge is carried by moving ions. Ions are charged atoms or groups of atoms. Current inside the cell encounters resistance.

Internal resistance, denoted by r, is the resistance within the cell. The resistance of the connected circuit outside the cell is its external resistance, denoted here by R. Both resistances are measured in ohms.

How does the terminal voltage change when current flows?

The terminal voltage V is the actual potential difference across the cell’s terminals. For a cell supplying current I to an external resistor, part of the emf corresponds to the potential difference Ir across its internal resistance.

V = E − Ir

Thus the terminal voltage is lower than the emf when the cell supplies current and its internal resistance is non-zero. With an open external circuit, I = 0 and the terminal voltage equals the emf. This relation describes a cell supplying current.

Since the external resistor has V = IR, substitution gives E = IR + Ir. Therefore I = E/(R + r). The denominator includes both external and internal resistance. Using E/R instead would omit the internal resistance.

The quantity Ir is sometimes called the lost volts, meaning the voltage associated with internal resistance. It is not a loss of electric charge. Internal resistance may be neglected in a calculation when Ir is much smaller than E.

Worked example 7. A battery of emf 10 V and internal resistance 3 Ω supplies a current of 0.5 A to an external resistor. Find the terminal voltage and external resistance.

Formula: V = E − Ir; R = V/I. Substitute: V = 10 − (0.5 × 3) = 8.5 V; R = 8.5/0.5. Answer: the terminal voltage is 8.5 V and the external resistance is 17 Ω. The internal voltage drop is 1.5 V.

Keep the symbols distinct: uppercase R refers to the external resistance and lowercase r to the internal resistance. The full circuit contains both, while the terminal voltage belongs across the external part alone.

Glossary

• Electric current — The rate at which electric charge passes through a cross-section of a conductor.
• Potential difference — Work done per unit charge in moving charge between two points in a circuit.
• Resistance — The property of a conductor that opposes charge flow, measured in ohms.
• Ohmic conductor — A conductor whose voltage is proportional to current under the specified constant conditions.
• Non-ohmic device — A device for which voltage and current do not maintain the proportional relationship required by Ohm’s law.
• Resistivity — A material property relating a uniform conductor’s resistance to its length and cross-sectional area.
• Ammeter — An instrument connected in series to measure current through a circuit or component.
• Voltmeter — An instrument connected in parallel to measure potential difference between two points.
• Rheostat — A variable resistor used to regulate current by changing the resistance in a circuit.
• Equivalent resistance — A single resistance replacing a combination while preserving current at the same applied voltage.
• Series combination — Resistors connected end to end along one path, carrying the same current.
• Parallel combination — Resistors connected between the same two points, each having the same potential difference.
• Electromotive force — The potential difference across a cell’s terminals when no current flows through it.
• Internal resistance — Resistance within a cell that contributes a voltage drop when the cell supplies current.
• Superconductor — A material having zero electrical resistivity when it is in its superconducting state.

Common errors and misconceptions

• Misconception: Ohm’s law applies to every device without conditions. Correct: It requires proportionality between voltage and current under specified conditions, including unchanged temperature for a metallic wire.
• Misconception: A voltmeter goes in series and an ammeter across the resistor. Correct: Connect the ammeter in series and the voltmeter in parallel across the relevant points.
• Misconception: The slope of any electrical graph gives resistance. Correct: Voltage plotted vertically against current gives resistance as the slope. Reversing those axes gives the reciprocal of resistance.
• Misconception: Resistance and resistivity are interchangeable. Correct: Resistance depends on a conductor’s dimensions and material; resistivity characterises the material at the stated temperature.
• Misconception: Series resistors must have equal voltage drops. Correct: They carry the same current. Different resistances can have different voltage drops, which add to the total.
• Misconception: Parallel resistances are added directly. Correct: Add their reciprocals and then take the reciprocal. Parallel branches share voltage, while their currents add.
• Misconception: Emf is a force and always equals terminal voltage. Correct: Emf is measured in volts. A supplying cell has terminal voltage V = E − Ir.
• Misconception: Any material with low resistance is a superconductor. Correct: A superconductor has zero resistivity in its superconducting state, which differs from merely low resistivity.

Exam-style questions with model answers

Q1. State Ohm’s law for a metallic wire and give the condition required for its validity. [2 marks]

1. The potential difference across a given metallic wire is directly proportional to the current through it.
2. The wire’s temperature must remain the same, so its resistance remains constant during the comparison.

Q2. A heater draws 4 A at 60 V. Its resistance remains constant when the voltage is raised to 120 V. Calculate its resistance and new current, and explain the change. [3 marks]

1. Using resistance R = V/I, where V is potential difference and I is current, R = 60/4 = 15 Ω.
2. The resistance remains 15 Ω, so at the new voltage the current is I = V/R = 120/15 = 8 A.
3. The voltage has doubled while resistance stays constant. Ohm’s law therefore predicts that current doubles from 4 A to 8 A.

Q3. Describe an experiment using a metallic wire, cells, a plug key, an ammeter and a voltmeter to verify Ohm’s law. Include the connections, readings, temperature condition and graph. [5 marks]

1. Connect the wire, cells, plug key and ammeter in series. The ammeter measures the current through the test wire in this conducting path.
2. Connect the voltmeter in parallel across the ends of the wire. Its reading gives the potential difference across the component being investigated.
3. Close the key, record the voltage and current together, and repeat with different numbers of cells to obtain several corresponding pairs of readings.
4. Keep the wire’s temperature unchanged and open the key between readings. Calculate voltage divided by current for each pair; the values should be approximately the same.
5. Plot voltage vertically against current horizontally. A straight line through the origin supports Ohm’s law, and its slope gives the wire’s resistance.

Q4. A lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected in series across 6 V. Treat both resistances as constant. Find the equivalent resistance, current, and potential difference across each component. [4 marks]

1. The equivalent series resistance is the sum of the two resistances: Rₛ = 20 + 4 = 24 Ω.
2. The same current passes through both components. Using the full applied voltage, I = V/Rₛ = 6/24 = 0.25 A.
3. The potential difference across the lamp is its resistance multiplied by the common current: 20 × 0.25 = 5 V.
4. The potential difference across the conductor is 4 × 0.25 = 1 V. Together the drops give 5 + 1 = 6 V.

Q5. Resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel across 12 V. Explain the branch voltages, calculate each branch current, and find the total current and equivalent resistance. [6 marks]

1. All three resistors join the same two points. Each therefore has a potential difference of 12 V, irrespective of its resistance.
2. Using current equal to voltage divided by resistance, the current through the 5 Ω branch is 12/5 = 2.4 A.
3. The 10 Ω branch has the same applied voltage but a larger resistance. Its current is 12/10 = 1.2 A.
4. The current through the 30 Ω branch is 12/30 = 0.4 A. It is the smallest branch current because this resistance is greatest.
5. The total current is the sum of the branch currents: 2.4 + 1.2 + 0.4 = 4 A.
6. The equivalent resistance is applied voltage divided by total current: 12/4 = 3 Ω, which is below the smallest branch resistance.

Q6. Derive the equivalent resistance of three resistors R₁, R₂ and R₃ connected in parallel across a common potential difference V. Assume that each obeys Ohm’s law. [4 marks]

1. Let I₁, I₂ and I₃ be their branch currents and I the total current. The currents add: I = I₁ + I₂ + I₃.
2. Since each resistor has voltage V across it, Ohm’s law gives I₁ = V/R₁, I₂ = V/R₂ and I₃ = V/R₃.
3. Let Rₚ be the equivalent resistance. Its current is I = V/Rₚ, so V/Rₚ = V/R₁ + V/R₂ + V/R₃.
4. Divide by the common non-zero voltage. The result is 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃.

Q7. A battery of emf 10 V and internal resistance 3 Ω supplies a current of 0.5 A to an external resistor. Calculate the internal voltage drop, terminal voltage and external resistance. [3 marks]

1. The internal voltage drop is Ir, where I is current and r is internal resistance: 0.5 × 3 = 1.5 V.
2. The battery supplies current, so terminal voltage equals emf minus the internal drop: V = 10 − 1.5 = 8.5 V.
3. This terminal voltage is across the external resistor. Its resistance is R = V/I = 8.5/0.5 = 17 Ω.

Q8. For a uniform metallic wire at constant temperature, explain the effects on resistance of doubling its length at unchanged area, and increasing its area at unchanged length. Keep the material unchanged. Distinguish resistance from resistivity. [3 marks]

1. Resistance is proportional to length when material, area and temperature remain the same, so doubling the length doubles the resistance.
2. Resistance is inversely proportional to cross-sectional area when material, length and temperature remain the same, so increasing the area reduces resistance.
3. Resistance describes the particular wire and depends on its dimensions. Resistivity describes the material at that temperature and remains unchanged in these comparisons.

Key takeaways

• Current measures charge flow per second, while potential difference measures work done per coulomb between two points.
• Ohm’s law relates voltage and current proportionally for a given metallic wire provided its temperature remains the same.
• Connect an ammeter in series and a voltmeter across the component; compare corresponding readings during verification.
• The slope gives resistance when voltage is plotted vertically against current horizontally; check the axes before calculating.
• A uniform wire’s resistance increases with length and decreases with area, with material and temperature held constant.
• Series resistors share current and their resistances add; parallel resistors share voltage and their reciprocal resistances add.
• Non-ohmic behaviour departs from proportionality, while superconductivity means zero resistivity in the superconducting state.
• A cell supplying current has terminal voltage below its emf when internal resistance produces a non-zero internal voltage drop.

Test yourself

What does a current of 1 A mean?

One coulomb of charge passes through a cross-section of the conductor every second.

Why must the temperature condition accompany Ohm’s law?

A change in temperature can change resistance, so voltage and current need not remain proportional during heating.

What is the resistance of a wire carrying 4 A when 60 V is applied?

Resistance equals voltage divided by current: R = 60/4 = 15 Ω.

What does the slope represent when current is vertical and voltage is horizontal?

The slope is current change divided by voltage change, equal to the reciprocal of resistance for an ohmic conductor.

What remains common to all resistors in series?

The same current passes through each resistor along the single conducting path.

What remains common to resistors connected between the same two points?

They are in parallel and have the same potential difference across their ends.

When does the terminal voltage of a cell equal its emf?

In an open circuit, no current flows, so there is no internal voltage drop and terminal voltage equals emf.

How does a superconductor differ from an ordinary good conductor?

A superconductor has zero resistivity in its superconducting state; an ordinary good conductor has low but non-zero resistivity.