Current Electricity | ISC Class 12 Physics Notes

Current Electricity covers charge flow, drift velocity, mobility, Ohm’s law, resistance, resistivity, temperature effects, electrical energy and power, cells and their combinations, Kirchhoff’s laws, Wheatstone bridge, metre bridge and potentiometer measurements.

What is electric current and how does charge flow in a conductor?

Definition: Electric current is the net electric charge crossing a cross-section, a surface cutting across a conductor, per unit time.

Let I denote current, Q the net charge transferred and t the time interval. For steady current, which remains constant with time, I = Q/t. In the International System of Units (SI), current is measured in amperes, symbol A; 1 A = 1 C s⁻¹.

The coulomb, symbol C, is the SI unit of charge, and the second, symbol s, measures time. Thus 1 C = 1 A s. For varying current, instantaneous current is I = dQ/dt, where dQ/dt means the rate of change of transferred charge with time.

Which particles carry the current?

In a metal, some electrons are practically free to move while positive ions, atoms or groups with net charge, remain at fixed sites. In electrolytic solutions, positive and negative ions can carry current.

Without an applied electric field, meaning force per unit positive test charge, electrons still have random thermal motion, but their average velocity is zero. Equal average motion in opposite directions produces no net current.

Conventional current follows the direction of positive charge flow. It is opposite to electron drift in a metal.

A steady current needs a maintained electric field and a closed conducting path. A cell or battery supplies energy to maintain this flow; charge redistribution without a continuing supply produces only a temporary current.

How do drift velocity and mobility determine electric current?

Electric field E is force per unit positive test charge. Its SI unit is volt per metre, V m⁻¹. Potential difference V is energy transferred per unit charge between two points; its SI unit is volt, with 1 V = 1 J C⁻¹, where J denotes joule, the energy unit.

Drift velocity is the average velocity acquired by charge carriers under an applied field. We use v⃗ᵈ for electron drift velocity and vᵈ for its positive magnitude, or drift speed, in m s⁻¹.

Let e be the positive magnitude of electron charge, so an electron has charge −e; m is electron mass in kilograms. The acceleration vector a⃗, the rate of change of velocity, is a⃗ = −eE⃗/m. An arrow denotes a vector, with magnitude and direction.

Relaxation time τ, measured in seconds, is the average interval between successive collisions. Collisions occur at random times. Averaging the field-induced motion gives v⃗ᵈ = −eτE⃗/m and vᵈ = eEτ/m. Electrons drift opposite to the field, towards higher potential.

Derivation: Relation between current and drift speed

Let n be the number of free electrons per unit volume, in m⁻³, and A the uniform cross-sectional area, in m², normal to the drift direction. Here A means area; as a unit, A means ampere.

1. During a short time interval Δt, electrons within distance vᵈΔt of the cross-section can cross it through drift.
2. The volume is AvᵈΔt, containing nAvᵈΔt electrons.
3. The magnitude of charge crossing is ΔQ = neAvᵈΔt, where ΔQ is the transferred charge magnitude.
4. Divide this charge by Δt to obtain the magnitude of current.

I = neAvᵈ

Mobility μ is drift speed per unit electric field: μ = vᵈ/E = eτ/m. It is positive and has SI unit m² V⁻¹ s⁻¹. Large electron number density allows substantial current even when drift speed is small.

Solid segments show a repeated-collision path from A to B. A dotted path ends at B′ when the field arrow points left. The slight drift is opposite to the electric field.

See Fig. 3.3 in your NCERT textbook

Worked example 1. A copper wire carries 1.5 A through area 1.0 × 10⁻⁷ m². Use n = 8.5 × 10²⁸ m⁻³ and e = 1.6 × 10⁻¹⁹ C to find electron drift speed.

Formula: vᵈ = I/(neA). Substitute: vᵈ = 1.5/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷). Answer: approximately 1.1 × 10⁻³ m s⁻¹, or 1.1 mm s⁻¹, opposite to conventional current.

What does Ohm’s law state and how is it explained microscopically?

Ohm’s law states that current through a conductor is directly proportional to the potential difference across it, provided its physical state remains unchanged. In particular, temperature must remain constant. Resistance R is the ratio V/I; its SI unit is ohm, Ω, with 1 Ω = 1 V A⁻¹.

V = IR expresses Ohm’s law when R is constant over the range considered. Merely calculating V/I does not prove ohmic behaviour. An ohmic conductor obeys this proportionality; a non-ohmic device does not.

How can the law be checked experimentally?

An ammeter measures current and is connected in series with the test wire, so the same current passes through both. A voltmeter measures potential difference and is connected in parallel across the wire’s ends.

Use a rheostat, an adjustable resistance, to change current. Record paired current and voltage readings, opening the key between observations to limit heating. A key is a switch that opens or closes the circuit.

For an ohmic conductor, the graph is a straight line through the origin. With V vertically and I horizontally, slope is ΔV/ΔI = R. With I vertically and V horizontally, slope is ΔI/ΔV = 1/R. The symbol Δ denotes a change.

Derivation: Conductivity and the microscopic form of Ohm’s law

Current density j is current per unit area normal to flow, j = I/A, for uniform flow. Its SI unit is A m⁻². Conductivity σ measures current density produced per unit field in an ohmic material.

1. Substitute vᵈ = eEτ/m into I = neAvᵈ to obtain I = ne²AτE/m.
2. Divide by A to obtain j = ne²τE/m. Conventional current density points along the field.
3. Identify the proportionality constant σ = ne²τ/m, assuming n and τ are independent of applied field.
4. For a uniform wire of length l, measured in metres, E = V/l. Rearrangement gives V = [ml/(ne²τA)]I.

σ = ne²τ/m; j⃗ = σE⃗

Ohm’s law is not a fundamental law of nature. Voltage may vary non-linearly with current; reversing voltage may change the current magnitude; or a current may correspond to more than one voltage. A diode, a direction-dependent conducting device, provides a non-ohmic example.

Voltage is vertical and current horizontal. A dashed straight line represents Ohm’s law; the solid curve bends away from that line at larger current.

See Fig. 3.5 in your NCERT textbook

How do resistance, resistivity, conductance and conductivity differ?

Resistivity ρ is a material property relating resistance to dimensions: R = ρl/A. For a uniform conductor, ρ = RA/l. The SI unit of resistivity is ohm metre, Ω m. Resistivity depends on temperature and pressure, but not on conductor dimensions.

Conductance G is the reciprocal of resistance: G = 1/R. Its SI unit is siemens, S, with 1 S = 1 Ω⁻¹. Conductivity is the reciprocal of resistivity: σ = 1/ρ; its SI unit is S m⁻¹.

At constant material and temperature, resistance is proportional to length and inversely proportional to cross-sectional area. The drift model gives ρ = m/(ne²τ).

Metals have low resistivities; insulators such as rubber have much higher values. Semiconductors lie between these classes. A semiconductor has conductivity sensitive to temperature or suitable impurities.

Worked example 2. A wire of length 15 m and uniform area 6.0 × 10⁻⁷ m² has resistance 5.0 Ω when measured with a negligibly small current. Find its resistivity at the experimental temperature.

Formula: ρ = RA/l. Substitute: ρ = (5.0 × 6.0 × 10⁻⁷)/15. Answer: ρ = 2.0 × 10⁻⁷ Ω m, equivalently 0.00000020 V m A⁻¹.

How does temperature affect resistance and resistivity?

For a metallic conductor, over a limited temperature range that is not too large, resistivity is approximately ρₜ = ρ₀[1 + α(T − T₀)]. Here T is temperature, T₀ the reference temperature, ρₜ and ρ₀ the respective resistivities, and α the temperature coefficient of resistivity.

Temperature coefficient α is the fractional increase in resistivity per unit temperature increase in the range where the relation is linear. Its unit is K⁻¹, or °C⁻¹ for Celsius temperature differences. K denotes kelvin, the SI temperature unit; °C denotes degree Celsius.

For dimensions treated as unchanged, Rₜ = R₀[1 + α(T − T₀)], where Rₜ and R₀ are resistances at the corresponding temperatures.

Why do metals and semiconductors behave differently?

In a metal, n is not temperature-dependent to any appreciable extent. With increasing temperature, collisions become more frequent and τ decreases, so resistivity increases. For metals, α is positive in the linear description.

For semiconductors, the increase in available charge carriers with temperature more than compensates for the reduction in relaxation time. Their resistivity therefore decreases as temperature rises.

Nichrome, an alloy of nickel, iron and chromium, exhibits a very weak temperature dependence of resistivity. Manganin and constantan have similar properties. Their resistance changes very little with temperature, making them useful in wire-wound standard resistors.

The copper graph curves upwards in the low-temperature range shown. Nichrome has a gently rising line. The typical semiconductor curve falls as temperature increases.

See Figs. 3.8, 3.9 and 3.10 in your NCERT textbook

Worked example 3. A nichrome toaster element has resistance 75.3 Ω at 27.0 °C with negligible heating. On a 230 V supply, its steady current is 2.68 A. Its average temperature coefficient over the range is 1.70 × 10⁻⁴ °C⁻¹. Find its steady temperature.

Formula: Rₜ = V/I; T = T₀ + (Rₜ − R₀)/(αR₀). Substitute: Rₜ = 230/2.68 = 85.8 Ω; T − 27.0 = (85.8 − 75.3)/(75.3 × 1.70 × 10⁻⁴) = 820 °C. Answer: T = 847 °C, using the rounded intermediate resistance.

The toaster’s initial current is slightly higher than its steady current. Heating increases resistance, causing a slight decrease in current. The temperature becomes steady when heating by the current equals heat loss to the surroundings.

Worked example 4. A platinum resistance thermometer reads 5 Ω at the ice point, 0 °C, and 5.23 Ω at the steam point, 100 °C. In a hot bath its resistance is 5.795 Ω. Assume the linear resistance-temperature relation.

Formula: T = [(Rₜ − R₀)/(R₁₀₀ − R₀)] × 100 °C, where R₁₀₀ is resistance at 100 °C. Substitute: T = [(5.795 − 5)/(5.23 − 5)] × 100 °C. Answer: the bath temperature is 345.65 °C.

How are electrical energy, power and electricity consumption calculated?

Electrical power P is electrical energy transferred per unit time. Let W represent transferred energy, measured in joules. For current I through potential difference V during time t, the transferred charge is It, giving W = VIt and P = W/t = VI.

The SI unit of power is watt, symbol W; 1 W = 1 J s⁻¹. Distinguish W as an energy symbol in a formula from W written after a numerical power value as the watt unit.

In a resistor, charge carriers gain energy from the field between collisions and transfer energy to the material during collisions. The material heats up. For a steady current through an ohmic resistance, substituting Ohm’s law gives P = I²R = V²/R.

The energy dissipated in time t is W = I²Rt = V²t/R. At fixed current, greater resistance gives greater power dissipation. At fixed voltage, greater resistance gives smaller power dissipation. Specify what is held constant before comparing two resistors.

How is energy expressed in commercial units?

A kilowatt-hour, kWh, is the energy transferred by a power of one kilowatt maintained for one hour. A kilowatt is 1000 W and an hour is 3600 s, so 1 kWh = 3.6 × 10⁶ J. One commercial unit of electricity means one kWh.

To calculate consumption, multiply each appliance’s power in kilowatts by its operating time in hours, then add the energies. With a stated uniform charge per unit, the energy charge equals the consumed kWh multiplied by that rate.

Note: A kilowatt is a unit of power; a kilowatt-hour is a unit of energy.

How do emf, internal resistance and terminal potential difference differ?

Electromotive force ε, abbreviated emf, is the work supplied by a source per unit charge moved internally from lower to higher potential. Thus ε = dW/dQ, where dW is a small amount of source work and dQ the charge moved. Emf is measured in volts, not force units.

A cell maintains current by supplying energy. A source of emf can draw on chemical, electrical, mechanical, thermal or radiant energy. Internal resistance r is resistance within the cell, measured in ohms. Terminal potential difference V is the voltage between its terminals.

Derivation: Current supplied by a cell

Consider a cell delivering steady current I through an external resistance R. Assume connecting-wire resistance is negligible, or included in R.

1. In a time interval t the source supplies energy εIt to the circuit.
2. The external resistor dissipates I²Rt, while the internal resistance dissipates I²rt.
3. Conservation of energy gives εIt = I²Rt + I²rt, hence ε = I(R + r).
4. The external voltage is V = IR. Substituting the current relation gives V = ε − Ir.

I = ε/(R + r); V = ε − Ir

The product Ir is the internal voltage drop, also called back emf in this circuit description. In an open circuit, where no current flows, terminal voltage equals emf. During discharge with non-zero internal resistance, terminal voltage is lower than emf.

During charging, an external source drives current into the cell’s positive terminal. If I is the positive magnitude of charging current, V = ε + Ir.

Worked example 5. A battery of emf 10 V and internal resistance 3 Ω delivers 0.5 A to a resistor. Neglect connecting-wire resistance. Find the external resistance and terminal voltage.

Formula: R = ε/I − r; V = ε − Ir. Substitute: R = 10/0.5 − 3; V = 10 − (0.5 × 3). Answer: R = 17 Ω and V = 8.5 V. Checking IR gives 0.5 × 17 = 8.5 V.

Worked example 6. An 8.0 V storage battery with internal resistance 0.5 Ω is charged from a 120 V direct-current supply through a 15.5 Ω series resistor. Direct current has a fixed direction. Neglect other resistance. Find charging current and battery terminal voltage.

Formula: I = (Vₛ − ε)/(R + r); V = ε + Ir, where Vₛ is supply voltage. Substitute: I = (120 − 8.0)/(15.5 + 0.5) = 7.0 A; V = 8.0 + (7.0 × 0.5). Answer: V = 11.5 V. The series resistor limits charging current.

How are cells combined in series, parallel and mixed groups?

An equivalent cell represents a combination; denote its emf by εₑ and internal resistance by rₑ. For external resistance R, the supplied current is I = εₑ/(R + rₑ).

How is the series formula obtained?

For two aiding cells, whose emfs act in the same direction, let ε₁, ε₂ be their emfs and r₁, r₂ their internal resistances. Series connection joins the negative terminal of one to the positive terminal of the next, with the same current through both.

Adding their terminal voltages gives V = (ε₁ − Ir₁) + (ε₂ − Ir₂). Comparison with V = εₑ − Irₑ yields εₑ = ε₁ + ε₂; rₑ = r₁ + r₂.

For any number of unequal series cells, add their emfs algebraically and add all internal resistances. A cell opposing the chosen source direction contributes a negative emf. Its resistance still adds positively.

How is the parallel formula obtained?

In a parallel connection, like terminals join together and both cells have common terminal voltage V. Define I₁ and I₂ as the currents supplied by the first and second cells. Then I = I₁ + I₂.

Since I₁ = (ε₁ − V)/r₁ and I₂ = (ε₂ − V)/r₂, adding and rearranging gives V = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) − I[r₁r₂/(r₁ + r₂)]. This applies to non-zero internal resistances.

εₑ = (ε₁r₂ + ε₂r₁)/(r₁ + r₂); rₑ = r₁r₂/(r₁ + r₂)

Unequal parallel-cell emfs therefore cannot simply be added. If a calculated branch current is negative, its actual direction is opposite to the initially assumed outward direction.

How are identical cells grouped?

Let each identical cell have emf ε and internal resistance r. Let N be their total number. In a mixed arrangement, let Nₛ be the number in each series row and Nₚ the number of identical rows in parallel, so N = NₛNₚ.

For the mixed result, first replace each series row by emf Nₛε and resistance Nₛr. The identical rows share current equally. Their common emf stays Nₛε, while their parallel internal resistance becomes Nₛr/Nₚ. Substitute these into the equivalent-cell current formula.

How do Kirchhoff’s laws solve electrical circuits?

A junction is a connection where circuit branches meet. A branch is a conducting path between junctions. A closed loop is a route through a circuit that returns to its starting point.

Kirchhoff’s junction law states that the sum of currents entering a junction equals the sum leaving it. It expresses conservation of charge: steady current cannot produce continuing charge accumulation at a junction.

For example, if I₃ enters a junction while I₁ and I₂ leave, I₃ = I₁ + I₂, where each subscript identifies a branch current. Currents are added algebraically with chosen signs, not as spatial vectors.

Kirchhoff’s loop law states that the algebraic sum of potential changes around a closed circuit loop is zero. It expresses conservation of energy: returning to the same point gives no net potential change.

What is a reliable solution method?

1. Mark cell polarities and assign a direction to each unknown branch current.
2. Apply the junction law to express related currents and reduce the unknowns.
3. Choose closed loops giving independent equations and record signed voltage changes, including internal resistance where present.
4. Solve the simultaneous equations and interpret any negative current as flow opposite to its assumed direction.

For the single-cell discharge circuit, traversal in the current direction gives ε − IR − Ir = 0. Rearranging reproduces I = ε/(R + r).

How does a balanced Wheatstone bridge measure resistance?

A Wheatstone bridge has four resistances arranged around a closed quadrilateral, a cell across one diagonal and a galvanometer across the other. A galvanometer is a device for detecting small currents. A null deflection is a zero indication.

Label the vertices A, B, C and D, with the cell across A and C. Define R₁ along AD, R₂ along AB, R₃ along DC and R₄ along BC.

A diamond has A at the left, B at the top, C at the right and D below. The galvanometer G joins B and D. The cell connects A and C; R₄ is labelled unknown and R₃ standard.

See Fig. 3.18 in your NCERT textbook

How is the balance condition derived?

Let Iɢ denote galvanometer current. Start with the balanced condition Iɢ = 0, so B and D have equal potential. Let Iₐ be the current along ADC and Iᵦ the current along ABC.

1. No current passes between B and D, so the same Iₐ passes through R₁ and R₃, and the same Iᵦ through R₂ and R₄.
2. Equal potentials at B and D imply equal drops from A: IₐR₁ = IᵦR₂.
3. The drops from B and D to C are also equal: IₐR₃ = IᵦR₄.
4. Dividing the two relations eliminates both branch currents and gives the bridge ratio.

R₁/R₂ = R₃/R₄; R₄ = R₂R₃/R₁

Adjust the standard resistance until the galvanometer shows null deflection, then calculate the unknown resistance. Balance does not mean zero current through the four arms. The ratio condition applies at null; an unbalanced bridge requires circuit equations including galvanometer current.

How does a metre bridge determine an unknown resistance?

A metre bridge is a practical form of Wheatstone bridge. It uses a one-metre wire of uniform cross-sectional area and material, mounted alongside a scale. The wire’s two parts supply a resistance ratio proportional to their lengths.

Connect unknown resistance X in the left gap and known resistance R in the right gap. X denotes the resistance to be measured. A galvanometer joins the gap junction to a jockey, a movable contact touching the bridge wire. Connect the source across the wire’s ends.

Let l₁ be the length from the left end to the balance point and l₂ the remaining length. At the null point, the galvanometer current is zero. The bridge condition then gives X/R = l₁/l₂, hence X = Rl₁/l₂.

If l₁ is expressed in centimetres, l₂ = 100 − l₁ centimetres. Use the same unit for both lengths. The equality of resistance and length ratios requires the wire to be uniform and its temperature sufficiently steady.

How is a useful balance obtained?

Choose the known resistance so the balance lies near the middle of the wire, for example within 30 cm to 70 cm. Touch the jockey briefly. Opposite galvanometer deflections at the two ends indicate that a null lies between them.

Interchange the known and unknown resistances and repeat, accounting for the exchanged positions in the ratio. This reduces the effect of unaccounted terminal resistances. Do not keep the original left-gap formula unchanged after exchanging the gaps.

To calculate resistivity of the tested wire, use ρ = XAₓ/Lₓ, where Aₓ is that wire’s cross-sectional area and Lₓ its measured length.

How does a potentiometer measure voltage, compare emfs and find internal resistance?

A potentiometer balances an unknown voltage against the potential drop along a uniform wire carrying steady current. Its principle is that potential difference between two points on this wire is proportional to their separation along the wire.

Define potential gradient k as potential drop per unit wire length, in V m⁻¹. For balance length l, the distance from the starting end to the null point, V = kl. Uniform area, uniform material and constant current keep k constant.

How is the null method arranged?

A driver cell supplies the current through the potentiometer wire. Connect the positive terminal of the cell being measured to the wire’s higher-potential end, and its negative terminal through a galvanometer to the jockey.

At balance, the galvanometer shows zero current: the wire’s potential drop equals the measured voltage. With the test cell otherwise open, no current is drawn from it at balance, so the measured value is its emf.

For use as a voltmeter, first establish k and find the length balancing the required potential difference. Then V = kl gives that voltage. The whole wire’s potential drop must exceed the voltage to be balanced to obtain a balance within its length.

How are the emfs of two cells compared?

Let ε₁ and ε₂ denote the two emfs, and l₁ and l₂ their respective balance lengths from the same starting end. With the wire current unchanged, ε₁ = kl₁ and ε₂ = kl₂. Dividing gives ε₁/ε₂ = l₁/l₂.

Switch the test cells in turn without moving the rheostat contact between the paired measurements. Otherwise k can change, and it cannot be cancelled from the ratio. The driver’s emf must exceed both test-cell emfs, with sufficient voltage across the wire itself.

How is a cell’s internal resistance obtained?

1. Keep the test cell’s external load disconnected and measure balance length l₀ for its emf ε.
2. Connect a known load resistance R across that cell and measure balance length l for its terminal voltage V.
3. Keep potentiometer-wire current unchanged, so ε/V = l₀/l. Here l₀ and l refer to open and loaded conditions respectively.
4. Since ε/V = (R + r)/R, solve for the cell’s internal resistance r.

r = R(l₀/l − 1) = R(l₀ − l)/l

Sensitivity here means the ability to distinguish small voltage differences through detectable changes of balance length. Lowering k increases sensitivity. A longer wire at the same total wire voltage lowers k; reducing wire current also lowers k.

Glossary

• Electric current — Net electric charge passing through a specified cross-section per unit time.
• Drift velocity — Average velocity acquired by charge carriers under an applied electric field.
• Relaxation time — Average time interval between successive collisions of conduction electrons in the drift model.
• Mobility — Positive ratio of charge-carrier drift speed to the magnitude of applied electric field.
• Current density — Current per unit area normal to flow, directed along conventional current.
• Resistance — Ratio of potential difference across a conductor to current through it.
• Resistivity — Material property equal to conductor resistance multiplied by cross-sectional area and divided by length.
• Conductance — Reciprocal of the resistance of a conductor, measured in siemens.
• Conductivity — Reciprocal of resistivity, relating current density to electric field in an ohmic material.
• Electromotive force — Energy supplied by a source per unit charge moved internally to higher potential.
• Internal resistance — Resistance within a cell that contributes to voltage drop when current flows.
• Kilowatt-hour — Energy transferred by one kilowatt of power operating for one hour.
• Wheatstone bridge — Four-resistance network used to determine unknown resistance by obtaining zero galvanometer current.
• Potential gradient — Potential drop per unit length along a potentiometer wire carrying steady current.
• Balance length — Length of potentiometer wire whose potential drop equals the voltage being measured.

Common errors and misconceptions

• Misconception: Zero current means electrons are stationary. Correct: Random thermal motion continues; without a preferred average direction, it produces no net current.
• Misconception: Electron drift and conventional current point in the same direction. Correct: In a metal, negative electrons drift opposite to conventional current and the electric field.
• Misconception: Writing V = IR proves that a device obeys Ohm’s law. Correct: Ohmic behaviour requires R to remain constant over the range considered, with physical conditions unchanged.
• Misconception: A longer wire must have greater resistivity. Correct: Length affects resistance; resistivity is a material property evaluated at specified temperature and pressure.
• Misconception: Metal resistivity is exactly linear in temperature everywhere. Correct: The linear expression is an approximation over a limited temperature range.
• Misconception: Cell terminal voltage equals emf in every circuit. Correct: It equals ε − Ir during discharge and ε + Ir during charging, using the corresponding positive current magnitude.
• Misconception: All bridge currents vanish at balance. Correct: Galvanometer current vanishes; currents can still flow through the four resistance arms.
• Misconception: Any two potentiometer balance lengths give the emf ratio. Correct: Both lengths must be measured from the same starting end with the same potential gradient.

Exam-style questions with model answers

Q1. Distinguish the emf of a cell from its terminal potential difference while it supplies current through an external resistor. State when the two are equal. [2 marks]

1. Emf ε is the energy supplied by the cell per unit charge. Terminal potential difference V is the voltage between its terminals and across the external resistor.
2. During discharge, V = ε − Ir, where I is supplied current and r is internal resistance. In an open circuit I is zero, so V equals ε.

Q2. A copper wire carries 1.5 A and has cross-sectional area 1.0 × 10⁻⁷ m². Its free-electron number density is 8.5 × 10²⁸ m⁻³. Taking electron charge magnitude as 1.6 × 10⁻¹⁹ C, calculate drift speed and state its direction relative to current. [3 marks]

1. Use I = neAvᵈ, where I is current, n electron number density, e electron charge magnitude, A cross-sectional area and vᵈ drift speed. Rearranging gives vᵈ = I/(neA).
2. Substitution gives vᵈ = 1.5/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷) ≈ 1.1 × 10⁻³ m s⁻¹.
3. Electron drift is opposite to conventional current because electrons carry negative charge.

Q3. A uniform wire of length 15 m and cross-sectional area 6.0 × 10⁻⁷ m² has resistance 5.0 Ω, measured using negligible current. Calculate its resistivity and explain why the measuring current is kept small. [3 marks]

1. For a uniform conductor, R = ρl/A, where R is resistance, ρ resistivity, l length and A area. Rearrangement gives ρ = RA/l.
2. Substituting the supplied dimensions and resistance gives ρ = (5.0 × 6.0 × 10⁻⁷)/15 = 2.0 × 10⁻⁷ Ω m.
3. A negligibly small measuring current limits heating, so the measured resistance corresponds to the experimental temperature rather than a substantially heated wire.

Q4. Using the free-electron drift model, derive the conductivity and resistivity of a metal. Let n be electron number density, e electron charge magnitude, m electron mass, τ relaxation time and E electric-field magnitude. Assume n and τ are independent of E. [5 marks]

1. The electric force on an electron is opposite to the field, with magnitude eE. Its acceleration magnitude between collisions is therefore eE/m.
2. Averaging over collisions gives drift-speed magnitude vᵈ = eEτ/m. Random thermal velocities average to zero, leaving the field-induced drift contribution.
3. For uniform cross-sectional area A normal to flow, current magnitude is I = neAvᵈ. Current density magnitude j = I/A is consequently nevᵈ.
4. Substitute the drift-speed expression to obtain j = ne²τE/m. Comparison with j = σE defines conductivity σ = ne²τ/m.
5. Resistivity ρ is the reciprocal of conductivity, so ρ = m/(ne²τ). With n and τ independent of E, current density is proportional to the applied field.

Q5. A battery of emf 10 V and internal resistance 3 Ω supplies 0.5 A to an external resistor. Neglect connecting-wire resistance. Calculate the external resistance, terminal voltage and voltage drop inside the battery. [4 marks]

1. For discharge, I = ε/(R + r), where I is current, ε emf, R external resistance and r internal resistance. Thus R + r = ε/I = 10/0.5 = 20 Ω.
2. The external resistance is R = 20 − 3 = 17 Ω.
3. The internal voltage drop is Ir = 0.5 × 3 = 1.5 V.
4. The terminal voltage is V = ε − Ir = 10 − 1.5 = 8.5 V, also equal to IR.

Q6. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is charged from a 120 V direct-current supply through a series resistor of 15.5 Ω. Neglect other resistance. Find the charging current and battery terminal voltage, and explain the resistor’s purpose. [4 marks]

1. The supply opposes the battery’s emf during charging, leaving a driving voltage of 120 − 8.0 = 112 V across the total resistance.
2. The total resistance is 15.5 + 0.5 = 16.0 Ω, so the charging current I is 112/16.0 = 7.0 A.
3. Battery terminal voltage is emf plus internal drop: V = 8.0 + 7.0 × 0.5 = 11.5 V.
4. The series resistor limits the charging current by increasing resistance in the charging circuit.

Q7. A Wheatstone bridge has resistances R₁ on AD, R₂ on AB, R₃ on DC and R₄ on BC. A cell joins A and C, and a galvanometer joins B and D. At zero galvanometer current, derive its balance condition and obtain R₄ in terms of the other resistances. [5 marks]

1. Zero galvanometer current means points B and D are at equal potential. This is the balanced condition, not a condition of zero current throughout the circuit.
2. Let Iₐ be current along ADC and Iᵦ current along ABC. With no current through the galvanometer, each path has the same current through its two resistors.
3. The equal potential drops from A to D and A to B give IₐR₁ = IᵦR₂.
4. The equal potential drops from D to C and B to C give IₐR₃ = IᵦR₄. Dividing the two equations gives R₁/R₃ = R₂/R₄.
5. Rearranging gives the balance condition R₁/R₂ = R₃/R₄, and hence the unknown resistance is R₄ = R₂R₃/R₁.

Q8. A potentiometer balances a cell’s emf over length l₀ with its external load disconnected. After a known resistance R is connected across the cell, its terminal voltage balances over length l. Both lengths start at the same wire end and the potential gradient stays constant. Derive the cell’s internal resistance. [5 marks]

1. Let k denote the constant potential gradient, ε the cell’s emf, V its loaded terminal voltage and r its internal resistance. The unloaded balance gives ε = kl₀.
2. With resistance R connected across the cell, the new balance gives V = kl. Dividing cancels the common gradient: ε/V = l₀/l.
3. Let I denote the current supplied to the load. The terminal voltage is V = IR, while the complete cell circuit gives ε = I(R + r).
4. Divide these expressions to obtain ε/V = (R + r)/R = 1 + r/R. Equate this with the balance-length ratio l₀/l.
5. Solving yields r = R(l₀/l − 1) = R(l₀ − l)/l. The result requires the same gradient during both balance measurements.

Key takeaways

• Electric current measures net charge transfer; electron thermal motion alone produces no current because its average direction is zero.
• Electron drift speed connects microscopic motion to current through I = neAvᵈ, while mobility measures drift speed per unit field.
• Ohm’s law requires proportional voltage and current under unchanged physical conditions; resistance and resistivity describe different properties.
• Metal resistivity increases with temperature, but its linear temperature formula is an approximation over a limited range.
• Electrical power is VI and energy is VIt; commercial energy consumption is measured in kilowatt-hours.
• A discharging cell has terminal voltage ε − Ir; cell combinations require both equivalent emf and equivalent internal resistance.
• Kirchhoff’s laws express conservation of charge and energy, while bridge balance requires zero current through the galvanometer.
• Potentiometer measurements use a uniform potential gradient; unchanged wire current is essential when comparing two balance lengths.

Test yourself

Why does a small drift speed still permit a substantial current in a metal?

The number of conduction electrons per unit volume is enormous. Their combined charge transfer can give substantial current even when each electron’s drift speed is small.

For an ohmic wire, what is the slope when voltage is plotted vertically against current?

The slope is resistance, because the potential difference divided by current remains constant under the stated physical conditions.

Why does increasing temperature lower the resistivity of a semiconductor?

The increase in charge-carrier number density more than compensates for the reduction in relaxation time, so conductivity increases and resistivity decreases.

Why is a kilowatt-hour an energy unit?

It is power multiplied by time, representing the energy transferred by one kilowatt operating for one hour.

For N identical cells of emf ε and internal resistance r connected in parallel with like terminals joined, what are the equivalent values?

The equivalent emf remains ε and the equivalent internal resistance is r/N, because identical branches share current equally.

What does a negative current obtained from Kirchhoff’s equations mean?

The actual current flows opposite to the direction initially assigned to that branch. Its magnitude is the positive value of the calculated result.

In a balanced metre bridge, why can the two wire resistances be replaced by a length ratio?

Both parts belong to the same uniform wire. With the same material, area and temperature, each resistance is proportional to its length.

How can potentiometer sensitivity be increased while retaining a balance?

Reduce the potential gradient, for example by increasing wire length at fixed total wire voltage. The total drop must still exceed the voltage being balanced.