PhysicsISC Class 12 · Chapter 3

Current Electricity ISC Class 12: Notes, Derivations and Numericals

Q: Is the colour code of carbon resistors in the ISC 2027 syllabus?

No. The revised 2027 Physics syllabus removed carbon resistors and their colour code, and the derivation of the series and parallel formulas for resistors. You still use those formulas in numericals.

Q: Do I need to prove the Wheatstone bridge condition using Kirchhoff's laws?

No. The syllabus asks you to take I_g = 0 at balance and derive P/Q = R/S directly; it notes that Kirchhoff's laws are not necessary for this.

Q: What is new in this chapter for 2027?

Ohm's law in vector form, \vec{J} = \sigma\vec{E}, is now named in the syllabus. Expect questions linking current density, field, conductivity and drift velocity.

Q: Why is a potentiometer better than a voltmeter for measuring emf?

At balance the potentiometer draws no current from the cell, so it measures the true emf. A voltmeter draws a small current, so it reads the terminal p.d., which is less than the emf by Ir.

Q: How do I increase the sensitivity of a potentiometer?

Reduce the potential gradient: use a longer wire or increase the series resistance in the primary circuit, keeping the p.d. across the wire above the emf to be measured.

Q: How many marks does this chapter carry?

Electrostatics and Current Electricity together carry 14 of the 70 theory marks. Kirchhoff, potentiometer and cell questions appear almost every year.

Updated 14 October 2026 · Fact-checked

Current Electricity is Chapter 3 of ISC Class 12 Physics. It covers drift velocity and mobility, Ohm's law in vector form $J = \sigma E$, resistivity and its temperature dependence, power and billing, emf and internal resistance, grouping of cells, Kirchhoff's laws, the Wheatstone and metre bridges, and the potentiometer for comparing emfs and finding internal resistance.

What Current Electricity is about

This chapter explains what happens inside a conductor when current flows, then gives you the tools to analyse circuits.

In a metal, free electrons move randomly; an applied field gives them a small average drift velocity $v_{d} = \frac{e E τ}{m}$ opposite to E, where τ is the relaxation time. From this you derive $I = n A e v_{d}$ , the conductivity $σ = \frac{n e ^{2} τ}{m}$ , resistivity $ρ = \frac{m}{n e ^{2} τ}$ and Ohm's law in vector form, $J = σ E$ (newly listed for 2027). Resistance $R = ρl / A$ rises with temperature for metals (τ falls) and falls for semiconductors (n rises). You also need electrical energy and power ( $P = V I = I^{2} R = V^{2} / R$ ), the kilowatt-hour and electricity bills.

A cell has an emf ε (work per unit charge done by the source) and an internal resistance r, so its terminal p.d. is $V = ε - I r$ . You derive the results for identical cells in series, parallel and mixed grouping, two unequal cells in parallel and n unequal cells in series.

Kirchhoff's laws (junction rule: charge conservation; loop rule: energy conservation) solve networks. The Wheatstone bridge balance condition $P / Q = R / S$ is derived with $I_{g} = 0$ (Kirchhoff's laws not needed), and the metre bridge uses it to measure an unknown resistance. The potentiometer works on $V \propto l$ for a uniform wire carrying a steady current: it compares emfs and measures internal resistance without drawing current from the cell under test.

The 2027 syllabus has dropped the carbon-resistor colour code and the derivation of series and parallel resistor formulas. This chapter shares 14 theory marks with electrostatics, and its numericals (Kirchhoff, bridges, potentiometer, cells) are reliable Section B and C items.

Weightage: Current Electricity is part of the unit Electrostatics; Current Electricity, which carries 14 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Drift velocity, current and mobility: Free electrons in a metal move randomly at high speed, colliding with ions; with no field their average velocity is zero. A field E gives each electron an acceleration $a = e E / m$ between collisions. Averaged over the relaxation time τ (mean time between collisions), they acquire a drift velocity $v_{d} = \frac{e E τ}{m}$ , opposite to E. In time t, all electrons within a length $v_{d} t$ pass a cross-section, so $I = n A e v_{d}$ (n = free electrons per m³). Mobility $μ = v_{d} / E = e τ / m$ (unit m² V⁻¹ s⁻¹). Since $E = V / l$ , doubling V doubles $v_{d}$ ; doubling l at fixed V halves it. Drift speeds are tiny (about $1 0^{- 4} m s^{- 1}$ ); a lamp lights at once because the field is set up along the whole wire almost instantly.; Worked example
A copper wire of cross-section $1 m m^{2}$ carries 1.6 A. Copper has $8.5 \times 1 0^{28}$ free electrons per m³. Find the drift speed.
Try it first, then show the solutionSolution
$v_{d} = \frac{I}{n A e}$
$= \frac{1.6}{8.5 \times 1 0 ^{28} \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19}} = \frac{1.6}{1.36 \times 1 0 ^{4}}$
$= 1.18 \times 1 0^{- 4} m s^{- 1}$
Answer: About $1.2 \times 1 0^{- 4} m s^{- 1}$ (0.12 mm per second)
Ohm's law, resistivity and J = σE: Current density $J = I / A$ (A m⁻²). Using $I = n A e v_{d}$ and $v_{d} = e E τ / m$ : $J = \frac{n e ^{2} τ}{m} E$ , i.e. $J = σ E$ with conductivity $σ = \frac{n e ^{2} τ}{m}$ and resistivity $ρ = \frac{1}{σ} = \frac{m}{n e ^{2} τ}$ . This is Ohm's law in vector form. Putting $J = I / A$ and $E = V / l$ gives $V = I \frac{ρl}{A} = I R$ , so $R = ρl / A$ . ρ depends on the material and temperature, not on the shape. Ohmic conductors give a straight-line V-I graph through the origin (slope gives R); non-ohmic devices (diode, filament lamp, electrolyte) do not. On a J-E graph, the steeper line has the larger σ and so the smaller ρ.; Worked example
For copper, $n = 8.5 \times 1 0^{28} m^{- 3}$ and $τ = 2.5 \times 1 0^{- 14} s$ . Find its conductivity, resistivity and electron mobility. ( $e = 1.6 \times 1 0^{- 19} C$ , $m = 9.1 \times 1 0^{- 31} k g$ )
Try it first, then show the solutionSolution
$σ = \frac{n e ^{2} τ}{m} = \frac{8.5 \times 1 0 ^{28} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2} \times 2.5 \times 1 0 ^{- 14}}{9.1 \times 1 0 ^{- 31}} = 5.98 \times 1 0^{7} S m^{- 1}$
$ρ = 1/ σ = 1.67 \times 1 0^{- 8} Ω m$
$μ = \frac{e τ}{m} = \frac{1.6 \times 1 0 ^{- 19} \times 2.5 \times 1 0 ^{- 14}}{9.1 \times 1 0 ^{- 31}} = 4.4 \times 1 0^{- 3} m^{2} V^{- 1} s^{- 1}$
Answer: $σ \approx 6.0 \times 1 0^{7} S m^{- 1}$ , $ρ \approx 1.67 \times 1 0^{- 8} Ω m$ , $μ \approx 4.4 \times 1 0^{- 3} m^{2} V^{- 1} s^{- 1}$
Temperature dependence of resistance: In a metal, n hardly changes with temperature but ions vibrate more, so τ falls and ρ rises roughly linearly: $R_{t} = R_{0} (1 + α t)$ , where α is the temperature coefficient of resistance (positive for metals). Alloys such as nichrome and manganin have very small α, so they are used for standard resistors and heating coils. In a semiconductor, heating frees many more charge carriers (n rises steeply), which outweighs the fall in τ, so ρ falls (negative α). The ρ-T graph for a metal is a gently rising curve; for a semiconductor, a falling one.; Worked example
A platinum wire has a resistance of 10 Ω at 20 °C and 12 Ω at 70 °C. Find its temperature coefficient of resistance (using the 20 °C value as reference).
Try it first, then show the solutionSolution
$R_{2} = R_{1} [1 + α (t_{2} - t_{1})] \Rightarrow α = \frac{R _{2} - R _{1}}{R _{1} ( t _{2} - t _{1} )}$
$α = \frac{12 - 10}{10 \times 50} = 4 \times 1 0^{- 3}^{\circ} C^{- 1}$
Answer: $4 \times 1 0^{- 3}$ per °C
Electrical energy, power and billing: Energy supplied in time t: $W = V I t$ . With Ohm's law, $W = I^{2} R t = \frac{V ^{2}}{R} t$ . Power $P = V I = I^{2} R = \frac{V ^{2}}{R}$ (watt). Which form to use depends on what is common: in series (same I), $P \propto R$ ; in parallel (same V), $P \propto 1/ R$ . A bulb rated (V, P) has $R = V^{2} / P$ , so the lower-wattage bulb has the higher resistance; in series it glows brighter. Commercial unit: 1 kWh = 1 'unit' $= 3.6 \times 1 0^{6} J$ . Bill = number of units × rate.; Worked example
A 1.5 kW heater is used for 2 hours a day for 30 days. Electricity costs ₹7 per unit. Find the energy used and the cost.
Try it first, then show the solutionSolution
Energy $= 1.5 k W \times 2 h \times 30 = 90 k W h$
Cost $= 90 \times 7 = ₹630$
Answer: 90 units; ₹630
Emf, internal resistance and terminal potential difference: The emf of a source is the work done per unit charge in moving charge from its negative to its positive terminal inside it: $ε = d W / d q$ . The energy may come from chemical, mechanical, thermal or radiant sources. With external resistance R and internal resistance r, the work done per unit time equals the heat produced outside and inside: $ε I d t = I^{2} R d t + I^{2} r d t$ , so $I = \frac{ε}{R + r}$ and the terminal p.d. is $V = I R = ε - I r$ . Ir is the 'lost volts' inside the cell. On open circuit (I = 0) V = ε; on a V-I graph the intercept on the V-axis is ε and the slope magnitude is r. While a cell is being charged, $V = ε + I r$ .; Worked example
A battery of emf 12 V and internal resistance 0.5 Ω is connected to a 5.5 Ω resistor. Find the current and the terminal potential difference.
Try it first, then show the solutionSolution
$I = \frac{ε}{R + r} = \frac{12}{5.5 + 0.5} = 2 A$
$V = ε - I r = 12 - 2 \times 0.5 = 11 V$ (check: $I R = 2 \times 5.5 = 11 V$ )
Answer: 2 A; 11 V
Grouping of cells: n identical cells in series (each ε, r): emf nε, internal resistance nr, $I = \frac{n ε}{R + n r}$ ; useful when R ≫ r. If one cell is reversed, the emf becomes $(n - 2) ε$ but the resistance is still nr. m identical cells in parallel: emf ε, internal resistance r/m, $I = \frac{ε}{R + r / m}$ ; useful when R ≪ r. Mixed grouping (m rows of n in series): $I = \frac{n ε}{R + n r / m}$ ; current is maximum when $R = n r / m$ . n unequal cells in series: $ε = ε_{1} + ε_{2} + \dots$ (with signs), $r = r_{1} + r_{2} + \dots$ . Two unequal cells in parallel: $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}$ and $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ .; Worked example
Four cells, each of emf 1.5 V and internal resistance 0.5 Ω, drive a 4 Ω resistor. Find the current when they are (a) all in series and (b) all in parallel.
Try it first, then show the solutionSolution
(a) Series: emf 6 V, internal resistance 2 Ω: $I = \frac{6}{4 + 2} = 1 A$
(b) Parallel: emf 1.5 V, internal resistance $0.5/4 = 0.125 Ω$ : $I = \frac{1.5}{4.125} = 0.364 A$
Answer: (a) 1 A (b) 0.36 A; series is better here because R is much larger than r
Kirchhoff's laws: Junction rule (first law): the algebraic sum of currents meeting at a junction is zero; current in = current out. It expresses conservation of charge. Loop rule (second law): the algebraic sum of potential changes round any closed loop is zero, $\sum ε = \sum I R$ . It expresses conservation of energy. Sign convention: going through a resistor in the direction of current, $Δ V = - I R$ (going 'downhill'); against the current, $+ I R$ . Going through a cell from − to +, $Δ V = + ε$ ; from + to −, $- ε$ . Assume a direction for each current; a negative answer simply means the current flows the other way.; Worked example
Two cells, $ε_{1} = 12 V$ and $ε_{2} = 6 V$ (negligible internal resistance), are in two parallel branches with resistors of 2 Ω each; a third branch between the same two junctions has 4 Ω. Both cells have their positive terminals towards the top junction. Find the current in each branch.
Try it first, then show the solutionSolution
Let the top junction be at potential V (bottom at 0). Branch currents towards the top junction: $I_{1} = \frac{12 - V}{2}$ , $I_{2} = \frac{6 - V}{2}$ ; down through 4 Ω: $I_{3} = \frac{V}{4}$ .
Junction rule: $I_{1} + I_{2} = I_{3} \Rightarrow 2 (12 - V) + 2 (6 - V) = V \Rightarrow 36 = 5 V \Rightarrow V = 7.2 V$ .
$I_{1} = 2.4 A$ , $I_{2} = - 0.6 A$ (0.6 A flows into the 6 V cell's positive terminal: it is being charged), $I_{3} = 1.8 A$ . Check: $2.4 - 0.6 = 1.8$ .
Answer: 2.4 A from the 12 V cell, 0.6 A into the 6 V cell (charging it), 1.8 A through the 4 Ω resistor
Wheatstone bridge and metre bridge: Four resistors P, Q, R, S form a loop ABCD, with a cell across AC and a galvanometer across BD. At balance $I_{g} = 0$ , so B and D are at the same potential. Then P and Q carry the same current $I_{1}$ , R and S the same current $I_{2}$ ; $I_{1} P = I_{2} R$ and $I_{1} Q = I_{2} S$ ; dividing, $\frac{P}{Q} = \frac{R}{S}$ (no Kirchhoff's laws needed). Balance does not change if the cell and galvanometer are interchanged. The metre bridge is a Wheatstone bridge in which a uniform 1 m wire provides the ratio arms: with a known R in the left gap and unknown S in the right, balance at length l from the left gives $\frac{R}{S} = \frac{l}{100 - l}$ , so $S = R \frac{100 - l}{l}$ . Balance near the middle of the wire gives the most accurate result.; Worked example
In a metre bridge a 4 Ω resistor is in the left gap and an unknown resistor in the right gap. Balance is at 40 cm from the left end. Find the unknown resistance.
Try it first, then show the solutionSolution
$\frac{R}{S} = \frac{l}{100 - l} \Rightarrow S = R \frac{100 - l}{l}$
$S = 4 \times \frac{60}{40} = 6 Ω$
Answer: 6 Ω
Potentiometer: principle, emf comparison and internal resistance: A long uniform wire AB carries a steady current from a driver cell. The fall of potential along it is proportional to length: $V = K l$ , where K is the potential gradient (V m⁻¹). A cell of emf ε connected (+ to A) through a galvanometer to a jockey is balanced at length l when no current flows through it: $ε = K l$ . Because the test cell supplies no current at balance, the potentiometer measures its true emf (a voltmeter draws current and reads less). Comparing emfs: $\frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}}$ . Internal resistance: balance length $l_{1}$ with the cell on open circuit and $l_{2}$ with a resistance R across it: $r = R \frac{l _{1} - l _{2}}{l _{2}}$ . A smaller potential gradient (longer wire, larger series resistance) means greater sensitivity. Balance is impossible if the driver's p.d. across the wire is less than the emf being measured.; Worked example
A cell of emf 1.5 V balances at 300 cm on a potentiometer wire. Another cell balances at 240 cm. Find its emf.
Try it first, then show the solutionSolution
$\frac{ε _{2}}{ε _{1}} = \frac{l _{2}}{l _{1}}$
$ε_{2} = 1.5 \times \frac{240}{300} = 1.2 V$
Answer: 1.2 V

Physics in real life: Current Electricity

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Electrical energy and billing

Suppose your family runs a 1.5 kW room heater for 2 hours every evening through a 30-day winter month, and the tariff is ₹7 per unit.

How many units does the heater use, and what does it add to the bill? How much would it save to run it for 1.5 hours a day instead?

Try it first, then show the solutionSolution

Units $= 1.5 \times 2 \times 30 = 90 k W h$ ; cost $= 90 \times 7 = ₹630$
At 1.5 h a day: $1.5 \times 1.5 \times 30 = 67.5 k W h$ , cost ₹472.50; saving ₹157.50

Answer: 90 units costing ₹630; cutting to 1.5 h a day saves ₹157.50

Memory hook: Kilowatts × hours = units; units × rate = rupees.

Internal resistance and terminal p.d.

Imagine the headlights of a car dimming for a moment when the driver turns the key to start the engine.

Suppose the car battery has emf 12 V and internal resistance 0.02 Ω, and the starter motor draws 150 A. What is the terminal voltage while cranking?

Try it first, then show the solutionSolution

Lost volts $= I r = 150 \times 0.02 = 3 V$
Terminal p.d. $V = ε - I r = 12 - 3 = 9 V$

Answer: About 9 V, so the headlights (sharing the same terminals) dim until the starter stops.

Memory hook: Big current, big Ir, small terminal voltage.

Temperature dependence of resistance

Suppose you notice that filament bulbs usually fail at the moment they are switched on, not while glowing.

A tungsten filament has 20 Ω at 20 °C, and α = $4.5 \times 1 0^{- 3}^{\circ} C^{- 1}$ . Estimate its resistance at 2520 °C and compare the switch-on and running currents on 230 V.

Try it first, then show the solutionSolution

$R = 20 [1 + 4.5 \times 1 0^{- 3} \times 2500] = 20 \times 12.25 = 245 Ω$
Switch-on (cold): $I = 230/20 = 11.5 A$ ; running (hot): $I = 230/245 = 0.94 A$

Answer: About 245 Ω hot; the cold filament takes over 12 times the running current for an instant, which stresses it.

Memory hook: Hot metal resists more, so the cold switch-on surge is the danger.

Topics in this chapter

Current Electricity formula sheet

Current: $I = \frac{Q}{t} = n A e v_{d}$; n = free electrons per m³.
Drift velocity and mobility: $v_{d} = \frac{e E τ}{m}, μ = \frac{v _{d}}{E} = \frac{e τ}{m}$; Unit of μ: m² V⁻¹ s⁻¹.
Ohm's law in vector form: $J = σ E, J = \frac{I}{A}$; Equivalent to V = IR for a uniform wire.
Conductivity and resistivity: $σ = \frac{n e ^{2} τ}{m}, ρ = \frac{m}{n e ^{2} τ}$; σ in S m⁻¹, ρ in Ω m.
Resistance: $R = \frac{V}{I} = \frac{ρl}{A}$; Stretching a wire n times (volume fixed) multiplies R by n².
Temperature dependence: $R_{t} = R_{0} (1 + α t)$; α > 0 for metals, α < 0 for semiconductors.
Energy and power: $W = V I t = I^{2} R t = \frac{V ^{2}}{R} t, P = V I = I^{2} R = \frac{V ^{2}}{R}$; 1 kWh = $3.6 \times 1 0^{6} J$ .
Cell in a circuit: $I = \frac{ε}{R + r}, V = ε - I r$; Charging: $V = ε + I r$ .
Identical cells: $I_{ser i es} = \frac{n ε}{R + n r}; I_{p a r a l l e l} = \frac{ε}{R + r / m}; I_{mi x e d} = \frac{n ε}{R + n r / m}$; Mixed: m rows of n cells; maximum current when R = nr/m.
Two unequal cells in parallel: $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}, r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$; Positive terminals joined together.
Kirchhoff's laws: $\sum I = 0 (j u n c t i o n); \sum ε = \sum I R (l oo p)$; Charge and energy conservation.
Wheatstone bridge balance: $\frac{P}{Q} = \frac{R}{S}$; Condition $I_{g} = 0$ .
Metre bridge: $S = R \frac{100 - l}{l}$; l in cm from the end next to R.
Potentiometer: $V = K l; \frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}}; r = R \frac{l _{1} - l _{2}}{l _{2}}$; K = potential gradient = (current × wire resistance) / wire length.

Current Electricity: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Wheatstone bridge and metre bridge

Derivation of the balance condition, and 'draw a labelled diagram of a metre bridge' in 2- or 3-mark questions.

Draw

Wheatstone: a diamond ABCD with P (A to B), Q (B to C), R (A to D), S (D to C); cell and key across A and C; galvanometer across B and D.
Mark currents $I_{1}$ through P and Q, $I_{2}$ through R and S, and $I_{g} = 0$ .
Metre bridge: a 100 cm wire on a scale between two thick copper strips; gaps for R (left) and S (right); galvanometer from the middle strip to a jockey on the wire; cell and key across the wire's ends.

Label

P, Q, R, S
G
cell and key K
jockey J
balance length l and (100 − l)

Watch out: Put the galvanometer between the junction of the two gaps and the jockey; draw the cell across the full wire.

Potentiometer circuits

'Draw a labelled circuit to compare the emfs of two cells' or 'to find the internal resistance of a cell'.

Draw

Primary circuit: driver cell, key and rheostat (or series resistance) in series with the wire AB, positive terminal to A.
Comparing emfs: cells $ε_{1}$ and $ε_{2}$ with positive terminals to A, through a two-way key to a galvanometer and jockey.
Internal resistance: the test cell (+ to A) with a resistance box R and key in parallel with it, then galvanometer and jockey.

Label

driver cell
rheostat Rh
wire AB
jockey J
G
$ε_{1}$ , $ε_{2}$ or test cell with R

Watch out: All positive terminals go to the same end A; a reversed test cell gives no balance point.

Graphs: V-I for ohmic and non-ohmic, ρ-T, and terminal p.d. against current

Graph interpretation items (finding ε and r from a V-I graph; which material from a ρ-T curve).

Draw

Ohmic: straight line through the origin; non-ohmic (filament lamp): curve bending towards the I-axis.
ρ against T: metal rising; semiconductor falling steeply.
Terminal p.d. V against I for a cell: straight line falling from ε on the V-axis, slope −r.

Label

axes with units
intercept = ε
slope = −r

Watch out: When reading ε and r from a graph, quote the two points you used.

How the board asks Current Electricity

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
One-mark concepts (Section A)	Effect of V or l on drift speed; emf of identical cells in parallel; which bulb glows brightest in series; balance point when cell and galvanometer are swapped.	Write the governing relation ( $v_{d} \propto V / l$ , $P = I^{2} R$ , $P / Q = R / S$ ) before choosing.
Derivations (3 marks)	$I = n A e v_{d}$ ; $σ = n e^{2} τ / m$ and $J = σ E$ ; $V = ε - I r$ ; cells in series, parallel or mixed; Wheatstone balance condition.	State the physical idea (relaxation time, energy balance, $I_{g} = 0$ ), then the algebra, then the final result.
Kirchhoff's-law networks	Two- or three-loop circuits with two cells; find a branch current or an unknown emf.	Mark currents, apply the junction rule first to cut unknowns, then write loop equations with a consistent sign convention; solve and interpret negative signs.
Bridges and potentiometer numericals	Unknown resistance from balance length; shunting the unknown; emf ratio; internal resistance; potential gradient from the driver circuit.	For K, find the current in the wire using the full primary-circuit resistance (wire + series resistor + driver's r).
Graph-based items	V-I graph of a cell (find ε and r), J-E graph of two wires (which has higher ρ or $v_{d}$ ), ρ-T curves.	Intercepts and slopes carry the physics: identify which quantity each gives before calculating.
Power and heating	Bulbs in series and parallel, heating coil cut or doubled, monthly bill.	Find each R from the rating; use $I^{2} R$ in series and $V^{2} / R$ in parallel.

Tricks to remember Current Electricity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Never Accept Every Velocity
$I = n A e v_{d}$ : 'n A e v' read as 'Never Accept Every Velocity'.
Use it when: Recalling the current-drift relation in order.
Series cells add, parallel cells don't
Identical cells: series adds both ε and r; parallel keeps ε, divides r.
Use it when: One-mark items on battery emf.
Junction = charge, loop = energy
'J for Junction, C for Charge; L for Loop, E for Energy' (the first law conserves charge, the second energy).
Use it when: Stating what each Kirchhoff law expresses.
Downhill with the current
Walking with the current through a resistor is going downhill: −IR. Walking from − to + inside a cell climbs: +ε.
Use it when: Getting loop signs right.
Low watt, high R, bright in series
$R = V^{2} / P$ : the 40 W bulb has the largest R, so with the same current it glows brightest in series.
Use it when: Bulb brightness questions.
Potentiometer: l is proportional to V
Every potentiometer result is a ratio of lengths: emfs as $l_{1} : l_{2}$ ; internal resistance from $(l_{1} - l_{2}) / l_{2}$ .
Use it when: Fast setup of potentiometer numericals.

Current Electricity: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Drift speed when the wire is longer · 1 mark · MCQBoard level

The length of a metal wire is doubled while the potential difference across it is kept the same. The drift speed of the electrons:
(a) doubles
(b) halves
(c) stays the same
(d) becomes one-fourth

Try it first, then show the solutionSolution

$v_{d} = \frac{e E τ}{m}$ and $E = V / l$ . Doubling l at the same V halves E, so $v_{d}$ halves.

Answer: (b) halves

Example 2: Bulbs in series · 1 mark · Assertion-reasonBoard level

Assertion: When a 40 W and a 100 W bulb (both rated for 230 V) are connected in series to 230 V, the 40 W bulb glows brighter.
Reason: In series, both bulbs carry the same current, and the 40 W bulb has the greater resistance.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

$R = V^{2} / P$ , so the 40 W bulb has the larger R. Same current, so $P = I^{2} R$ is larger for it; the reason explains the assertion.

Answer: (a)

Example 3: Unequal cells in parallel · 2 marks · Short answerHarder board level

A cell of emf 2 V and internal resistance 1 Ω is connected in parallel (positive to positive) with a cell of emf 1 V and internal resistance 1 Ω. The combination drives a 1 Ω resistor. Find the current in the resistor. [2]

Try it first, then show the solutionSolution

$ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}} = \frac{2 \times 1 + 1 \times 1}{2} = 1.5 V$ ; $r_{e q} = \frac{1 \times 1}{2} = 0.5 Ω$
$I = \frac{1.5}{1 + 0.5} = 1 A$

Answer: 1 A

Example 4: Derivation: conductivity and J = σE · 3 marks · Short answerHarder board level

Starting from the drift velocity of free electrons, show that the current density in a conductor is $J = σ E$ , and obtain an expression for σ in terms of the relaxation time. [3]

Try it first, then show the solutionSolution

Acceleration of an electron: $a = - e E / m$ . Averaged over the relaxation time τ, drift velocity $v_{d} = - \frac{e E τ}{m}$ .
In time t, the charge crossing area A is $n e A v_{d} t$ , so $I = n e A v_{d}$ and $J = \frac{I}{A} = n e v_{d}$ .
Substituting $v_{d} = \frac{e E τ}{m}$ : $J = \frac{n e ^{2} τ}{m} E$ ; in vector form $J = σ E$ (J is along E since the electrons' charge is negative) with $σ = \frac{n e ^{2} τ}{m}$ .

Answer: $J = σ E$ , $σ = \frac{n e ^{2} τ}{m}$ (so $ρ = \frac{m}{n e ^{2} τ}$ )

Example 5: Two-loop Kirchhoff problem · 3 marks · Short answerHarder board level

In a circuit, a 12 V cell with a 2 Ω resistor and a 6 V cell with a 2 Ω resistor are connected as two parallel branches between junctions X (top) and Y (bottom), positive terminals towards X. A 4 Ω resistor forms a third branch between X and Y. The cells have negligible internal resistance. Using Kirchhoff's laws, find the current in the 4 Ω resistor and in the 6 V cell. [3]

Try it first, then show the solutionSolution

Let $I_{1}$ flow up through the 12 V branch, $I_{2}$ up through the 6 V branch, and $I_{3} = I_{1} + I_{2}$ down through 4 Ω (junction rule at X).
Loop with 12 V and 4 Ω: $12 = 2 I_{1} + 4 (I_{1} + I_{2}) \Rightarrow 6 I_{1} + 4 I_{2} = 12$ . Loop with 6 V and 4 Ω: $6 = 2 I_{2} + 4 (I_{1} + I_{2}) \Rightarrow 4 I_{1} + 6 I_{2} = 6$ .
Solving: $I_{1} = 2.4 A$ , $I_{2} = - 0.6 A$ , so $I_{3} = 1.8 A$ . The negative sign shows 0.6 A flows down through the 6 V cell, i.e. into its positive terminal (it is being charged).

Answer: 1.8 A in the 4 Ω resistor; 0.6 A through the 6 V cell from its positive to negative terminal

Example 6: Potentiometer: principle and internal resistance · 5 marks · Long answerHarder board level

(a) State the principle of a potentiometer. [1]
(b) Draw a labelled circuit and derive an expression to find the internal resistance of a cell using a potentiometer. [2]
(c) With the cell on open circuit the balance length is 280 cm; when a 6 Ω resistor is connected across the cell it is 240 cm. Find the internal resistance. Why is the potentiometer preferred to a voltmeter here? [2]

Try it first, then show the solutionSolution

(a) When a steady current flows through a wire of uniform cross-section, the fall of potential across any length is proportional to that length: $V = K l$ .
(b) Circuit: driver cell, key and rheostat across AB; test cell (+ to A) with a resistance box R and key $K_{2}$ across it, then galvanometer and jockey. Open $K_{2}$ : $ε = K l_{1}$ . Close $K_{2}$ : the terminal p.d. $V = K l_{2}$ .
$\frac{ε}{V} = \frac{l _{1}}{l _{2}}$ and $ε = I (R + r)$ , $V = I R$ , so $\frac{R + r}{R} = \frac{l _{1}}{l _{2}} \Rightarrow r = R \frac{l _{1} - l _{2}}{l _{2}}$ .
(c) $r = 6 \times \frac{280 - 240}{240} = 1 Ω$ .
At balance no current is drawn from the test cell, so the potentiometer measures its true emf; a voltmeter draws current and reads the lower terminal p.d.

Answer: (c) r = 1 Ω

Example 7: Shunting the unknown in a metre bridge · 3 marks · Short answerHOTS / competency

A metre bridge with 4 Ω in the left gap balances at 40 cm. When a resistor X is connected in parallel with the unknown resistor in the right gap, the balance point moves to 50 cm. Find the unknown resistance and X. [3]

Try it first, then show the solutionSolution

First balance: $S = 4 \times \frac{60}{40} = 6 Ω$ .
Second balance at 50 cm: the right gap now equals the left: $S^{'} = 4 Ω$ .
$\frac{6 X}{6 + X} = 4 \Rightarrow 6 X = 24 + 4 X \Rightarrow X = 12 Ω$ .

Answer: Unknown 6 Ω; X = 12 Ω

Example 8: One cell connected the wrong way round · 2 marks · Short answerHOTS / competency

Six cells, each of emf 1.5 V and internal resistance 0.5 Ω, are meant to be in series, but one is connected the wrong way round. The battery drives a 3 Ω resistor. Find the current. [2]

Try it first, then show the solutionSolution

The reversed cell cancels one correct cell: net emf $= (6 - 2) \times 1.5 = 6 V$ . Internal resistance is still $6 \times 0.5 = 3 Ω$ .
$I = \frac{6}{3 + 3} = 1 A$

Answer: 1 A (instead of 1.5 A if all were the right way round)

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Setting up a potentiometer

Suppose a student sets up a potentiometer with a 4 m uniform wire of resistance 8 Ω. The driver cell has emf 4 V and internal resistance 1 Ω, and a 1 Ω resistor is in series with the wire.

(i) · 1 mark
Find the current in the potentiometer wire.
Answer
Answer: 0.4 A
Total resistance = 8 + 1 + 1 = 10 Ω, so $I = 4/10 = 0.4 A$ .
(ii) · 1 mark
Find the potential gradient along the wire.
Answer
Answer: $0.8 V m^{- 1}$
p.d. across wire $= 0.4 \times 8 = 3.2 V$ ; $K = 3.2/4 = 0.8 V m^{- 1}$ .
(iii) · 1 mark
Where will a cell of emf 1.2 V balance?
Answer
Answer: 1.5 m from A
$l = ε / K = 1.2/0.8 = 1.5 m$ .
(iv) · 2 marks
The student wants greater sensitivity. Should the series resistor be increased or decreased? What is the risk of increasing it too far?
Answer
Answer: Increased; if the p.d. across the wire falls below the emf being measured, no balance point is found.
A larger series resistance lowers the current, so K falls and the balance length for the same emf grows, giving a more precise reading. But the p.d. across the whole wire must stay above 1.2 V: $\frac{4 \times 8}{9 + R _{s}} \geq 1.2$ , so $R_{s}$ must not exceed about 17.7 Ω.

Finding emf and internal resistance from readings

Suppose a student connects a cell to a variable resistor, with an ammeter in series and a voltmeter across the cell's terminals. She records 1.45 V at 0.10 A and 1.25 V at 0.50 A.

(i) · 1 mark
Find the internal resistance of the cell.
Answer
Answer: 0.5 Ω
$V = ε - I r$ : $r = \frac{1.45 - 1.25}{0.50 - 0.10} = 0.5 Ω$
(ii) · 1 mark
Find the emf.
Answer
Answer: 1.5 V
$ε = V + I r = 1.45 + 0.10 \times 0.5 = 1.5 V$ (the V-axis intercept of the V-I graph)
(iii) · 2 marks
What is the external resistance in the second reading, and what current would flow if the terminals were short-circuited?
Answer
Answer: 2.5 Ω; 3 A
$R = V / I = 1.25/0.5 = 2.5 Ω$ . Short circuit: $I = ε / r = 1.5/0.5 = 3 A$ .
(iv) · 1 mark
For what external resistance is the power delivered to it greatest, and what is that power?
Answer
Answer: R = r = 0.5 Ω; 1.125 W
$P = \frac{ε ^{2} R}{( R + r ) ^{2}}$ is greatest when R = r; then $P = \frac{ε ^{2}}{4 r} = \frac{2.25}{2} = 1.125 W$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A cell of emf 1.5 V balances at 300 cm on a potentiometer wire. Another cell balances at 240 cm. Find the emf of the second cell. (2 marks)

Step	Marks
Writes $\frac{ε _{1}}{ε _{2}} = \frac{l _{1}}{l _{2}}$	1
$ε_{2} = 1.5 \times \frac{240}{300} = 1.2 V$ , with unit	1

A cell gives 0.5 A through a 4 Ω resistor and 0.25 A through a 9 Ω resistor. Find its emf and internal resistance. (3 marks)

Step	Marks
$ε = 0.5 (4 + r)$ and $ε = 0.25 (9 + r)$	1
$2 + 0.5 r = 2.25 + 0.25 r \Rightarrow r = 1 Ω$	1
$ε = 0.5 \times 5 = 2.5 V$	1

State Kirchhoff's two laws and the conservation principle behind each. Use them to find the three branch currents for: 12 V cell with 2 Ω, 6 V cell with 2 Ω and a 4 Ω resistor, all in parallel branches between two junctions (cells' positive terminals to the same junction, negligible internal resistance). (5 marks)

Step	Marks
Junction rule (sum of currents at a junction is zero) — conservation of charge	1
Loop rule (sum of emfs = sum of IR round a loop) — conservation of energy	1
Correct loop equations: $6 I_{1} + 4 I_{2} = 12$ and $4 I_{1} + 6 I_{2} = 6$ (with $I_{3} = I_{1} + I_{2}$ )	1
Solves: $I_{1} = 2.4 A$ , $I_{2} = - 0.6 A$	1
$I_{3} = 1.8 A$ ; interprets the negative sign (6 V cell is being charged)	1

Where students lose marks in Current Electricity

The slips examiners see most often, and what to write instead.

Mistake: Wrong loop equations in Kirchhoff problems
CISCE examiners' comments, 2024 and 2025: candidates could not frame correct loop equations, failed to apply the junction rule, used the wrong sign convention, made errors solving the simultaneous equations, or just applied Ohm's law to each loop.
Wrong working
Loop 1: $12 = 2 I_{1} + 4 I_{1}$ (current through the shared 4 Ω branch taken as $I_{1}$ only).
Right working
Shared branch carries $I_{1} + I_{2}$ : $12 = 2 I_{1} + 4 (I_{1} + I_{2})$ .
Do this: Apply the junction rule first, label every branch current on the figure, then go round each loop in one direction.
Usually costs: 2-3 marks in a 3-mark question.
Mistake: Leaving out the wire's own resistance in the potentiometer's primary circuit
CISCE examiners' comments, 2024: when finding the current in the potentiometer wire, candidates omitted the resistance of the wire itself, and so could not find the p.d. across it or the unknown emf.
Wrong working
$I = \frac{6}{20 + 2}$ (series resistor and driver only).
Right working
$I = \frac{ε}{R _{w i r e} + R _{s} + r}$ ; then $V_{w i r e} = I R_{w i r e}$ and $K = V_{w i r e} / L$ .
Do this: List every resistance in the primary circuit before dividing.
Usually costs: 1-2 marks.
Mistake: Answering 3e for identical cells in parallel
CISCE examiners' comments, 2024: for three identical cells in parallel, candidates applied resistor rules and answered 3e.
Wrong working
Parallel: emf = 3e.
Right working
Parallel: emf = e (only the internal resistance changes, to r/3).
Do this: Emfs add only in series.
Usually costs: 1 mark.
Mistake: Choosing the highest-wattage bulb as brightest in series
CISCE examiners' comments, 2024: with 40 W, 60 W and 100 W bulbs in series, candidates used the parallel idea and chose the 100 W bulb.
Wrong working
100 W is the most powerful, so it glows brightest.
Right working
Same I in series; $P = I^{2} R$ and $R = V^{2} / P_{r a t e d}$ is largest for 40 W: the 40 W bulb glows brightest.
Do this: Decide first whether I or V is common.
Usually costs: 1 mark.
Mistake: Misnaming meters and misreading a V-I graph
CISCE examiners' comments, 2025: candidates called the ammeter a galvanometer or potentiometer, did not realise V = ε when I = 0, confused emf with terminal voltage, and omitted units.
Wrong working
emf = slope of the V-I graph.
Right working
emf = intercept on the V-axis (I = 0); internal resistance = magnitude of the slope.
Do this: Write $V = ε - I r$ beside the graph and match it to y = c + mx.
Usually costs: 1-2 marks.
Mistake: Reading a J-E graph backwards
CISCE examiners' comments, 2025: candidates could not relate J and E on a graph for two wires and picked the wrong wire as having higher resistivity, or could not link drift velocity with current density.
Wrong working
Steeper J-E line means higher resistivity.
Right working
$J = σ E$ : the steeper line has larger σ, so smaller ρ.
Do this: Identify what the slope is ( $σ = 1/ ρ$ ) before comparing.
Usually costs: 1-2 marks.

Practice questions on Current Electricity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of electron mobility is:
(a) $m s^{- 1}$
(b) $m^{2} V^{- 1} s^{- 1}$
(c) $V m^{- 1}$
(d) $m^{2} s^{- 1}$
Hint
$μ = v_{d} / E$ .
Answer
(b) $m^{2} V^{- 1} s^{- 1}$
Q2 · 1 mark · MCQBoard level
As the temperature of a pure semiconductor rises, its resistivity:
(a) increases
(b) decreases
(c) stays the same
(d) first increases then decreases
Hint
Which changes more: n or τ?
Answer
(b) decreases, because many more charge carriers are freed
Q3 · 1 mark · MCQBoard level
Three identical cells of emf ε and internal resistance r are joined in parallel. The emf and internal resistance of the combination are:
(a) 3ε, 3r
(b) ε, r/3
(c) 3ε, r/3
(d) ε, 3r
Hint
Parallel: same p.d.
Answer
(b) ε, r/3
Q4 · 1 mark · Very short answerBoard level
Write Ohm's law in vector form and name the quantities.
Hint
Current density and field.
Answer
$J = σ E$ : J is current density, σ conductivity, E electric field.
Q5 · 2 marks · Short answerBoard level
Find the resistance of a 230 V, 60 W bulb and a 230 V, 100 W bulb. Which glows brighter when they are connected in series to 230 V?
Hint
$R = V^{2} / P$ .
Answer
$R_{60} = 23 0^{2} /60 = 882 Ω$ , $R_{100} = 23 0^{2} /100 = 529 Ω$ . In series the current is common, so the 60 W bulb (larger R) dissipates more power and glows brighter.
Q6 · 2 marks · Short answerHarder board level
A copper wire of cross-section $1 m m^{2}$ carries 1.6 A. With $n = 8.5 \times 1 0^{28} m^{- 3}$ , find the drift speed. How long would an electron take to drift 1 m?
Hint
$v_{d} = I / n A e$ .
Answer
$v_{d} = 1.18 \times 1 0^{- 4} m s^{- 1}$ ; time $= 1/1.18 \times 1 0^{- 4} \approx 8.5 \times 1 0^{3} s$ (about 2.4 hours).
Q7 · 2 marks · Short answerHarder board level
A wire has resistance 10 Ω at 20 °C and 12 Ω at 70 °C. Find α, and the temperature at which its resistance is 11 Ω.
Hint
Use 20 °C as reference.
Answer
$α = \frac{2}{10 \times 50} = 4 \times 1 0^{- 3}^{\circ} C^{- 1}$ . $11 = 10 [1 + 0.004 (t - 20)] \Rightarrow t - 20 = 25$ , so t = 45 °C.
Q8 · 2 marks · Short answerHOTS / competency
Why can no balance point be found on a potentiometer if the emf of the driver cell is less than the emf of the cell under test? Suggest one change that fixes it.
Hint
Compare the largest p.d. available on the wire with ε.
Answer
The p.d. across the whole wire is less than the driver's emf, hence less than the test emf; no length of wire gives an equal p.d., so the galvanometer always deflects one way. Fix: use a driver cell of larger emf (or reduce the series resistance so the wire's p.d. exceeds the test emf).
Q9 · 3 marks · Short answerHarder board level
A cell gives a current of 0.5 A through a 4 Ω resistor and 0.25 A through a 9 Ω resistor. Find its emf and internal resistance.
Hint
Write $ε = I (R + r)$ twice.
Answer
$0.5 (4 + r) = 0.25 (9 + r) \Rightarrow r = 1 Ω$ ; $ε = 2.5 V$ .
Q10 · 3 marks · Short answerHarder board level
A heating coil is rated 1000 W, 220 V. It is cut into two equal halves and the halves are connected in parallel across 220 V. Find the new power.
Hint
Find R first; each half has R/2.
Answer
$R = 22 0^{2} /1000 = 48.4 Ω$ ; each half 24.2 Ω; in parallel 12.1 Ω; $P = 22 0^{2} /12.1 = 4000 W$ (four times).
Q11 · 3 marks · Short answerHarder board level
Twelve identical cells, each of emf 1.5 V and internal resistance 0.5 Ω, are arranged in 3 parallel rows of 4 cells in series. They drive a 2 Ω resistor. Find the current. Is this arrangement giving the maximum possible current for this resistor?
Hint
$I = \frac{n ε}{R + n r / m}$ .
Answer
emf $= 4 \times 1.5 = 6 V$ ; internal resistance $= \frac{4 \times 0.5}{3} = 0.67 Ω$ ; $I = \frac{6}{2.67} = 2.25 A$ . Maximum current needs $R = n r / m$ , i.e. $0.67 \neq = 2$ , so not quite: 6 rows of 2 cells (internal resistance 0.17 Ω, emf 3 V) give 1.38 A, 2 rows of 6 (emf 9 V, internal resistance 1.5 Ω) give 2.57 A, which is larger.
Q12 · 5 marks · Long answerHarder board level
(a) State Kirchhoff's two laws. [2]
(b) Derive the balance condition of a Wheatstone bridge. [2]
(c) In a Wheatstone bridge P = 10 Ω, Q = 20 Ω and R = 15 Ω. Find S for balance. [1]
Hint
For (b), take $I_{g} = 0$ .
Answer
(a) Junction: algebraic sum of currents at a junction is zero. Loop: algebraic sum of emfs in a closed loop equals the sum of IR products. (b) At balance B and D are at the same potential, so $I_{1} P = I_{2} R$ and $I_{1} Q = I_{2} S$ ; dividing gives $P / Q = R / S$ . (c) $S = \frac{QR}{P} = \frac{20 \times 15}{10} = 30 Ω$ .
Q13 · 2 marks · Short answerHarder board level
Three cells of emf 1.5 V, 2 V and 2.5 V with internal resistances 0.2 Ω, 0.3 Ω and 0.5 Ω are connected in series (all the same way) to a 5 Ω resistor. Find the current and the terminal p.d. of the battery.
Hint
Add emfs and internal resistances.
Answer
$ε = 6 V$ , $r = 1 Ω$ , $I = \frac{6}{6} = 1 A$ ; terminal p.d. $= 6 - 1 \times 1 = 5 V$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define drift velocity.Show answer
The average velocity with which free electrons drift opposite to an applied field: $v_{d} = e E τ / m$ .
2.What is relaxation time?Show answer
The average time between successive collisions of a free electron with the ions of the lattice.
3.Expression for resistivity in terms of τ?Show answer
$ρ = \frac{m}{n e ^{2} τ}$ .
4.Define emf of a cell.Show answer
The work done per unit charge by the source in moving charge from its negative to its positive terminal inside it.
5.Relation between terminal p.d., emf and internal resistance?Show answer
$V = ε - I r$ (discharging).
6.State the principle of a potentiometer.Show answer
For a steady current through a uniform wire, the potential drop across any length is proportional to that length.
7.Balance condition of a Wheatstone bridge?Show answer
$P / Q = R / S$ when the galvanometer current is zero.
8.Why is a metre bridge most accurate near the middle?Show answer
End errors and the percentage error in measuring l are smallest when l and (100 − l) are similar in size.
9.1 kWh in joules?Show answer
$3.6 \times 1 0^{6} J$ .

Best order to study it

Derive drift velocity, $I = n A e v_{d}$ and $σ = n e^{2} τ / m$ ; learn $J = σ E$ and mobility.
Learn temperature dependence for metals and semiconductors, with graphs.
Practise power, bulb brightness and billing.
Derive $V = ε - I r$ and the cell-combination results; practise graph-based ε and r.
Master Kirchhoff's laws with three or four network problems.
Learn the Wheatstone and metre bridges, then the potentiometer (circuits, emf comparison, internal resistance, sensitivity).

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you derive $v_{d} = e E τ / m$ , $I = n A e v_{d}$ and $σ = n e^{2} τ / m$ ?
Can you say what happens to $v_{d}$ if V or l is doubled?
Can you write $R_{t} = R_{0} (1 + α t)$ and sketch ρ-T for a metal and a semiconductor?
Can you find bulb resistances and say which glows brighter in series and in parallel?
Can you derive $ε = I (R + r)$ and read ε and r from a V-I graph?
Can you combine identical and unequal cells in series and parallel?
Can you solve a two-loop Kirchhoff problem and interpret a negative current?
Can you derive $P / Q = R / S$ and use the metre-bridge formula?
Can you draw both potentiometer circuits and use $ε_{1} / ε_{2} = l_{1} / l_{2}$ and $r = R (l_{1} - l_{2}) / l_{2}$ ?

Current Electricity: frequently asked questions

Is the colour code of carbon resistors in the ISC 2027 syllabus?

No. The revised 2027 Physics syllabus removed carbon resistors and their colour code, and the derivation of the series and parallel formulas for resistors. You still use those formulas in numericals.

Do I need to prove the Wheatstone bridge condition using Kirchhoff's laws?

No. The syllabus asks you to take I_g = 0 at balance and derive P/Q = R/S directly; it notes that Kirchhoff's laws are not necessary for this.

What is new in this chapter for 2027?

Ohm's law in vector form, \vec{J} = \sigma\vec{E}, is now named in the syllabus. Expect questions linking current density, field, conductivity and drift velocity.

Why is a potentiometer better than a voltmeter for measuring emf?

At balance the potentiometer draws no current from the cell, so it measures the true emf. A voltmeter draws a small current, so it reads the terminal p.d., which is less than the emf by Ir.

How do I increase the sensitivity of a potentiometer?

Reduce the potential gradient: use a longer wire or increase the series resistance in the primary circuit, keeping the p.d. across the wire above the emf to be measured.

How many marks does this chapter carry?

Electrostatics and Current Electricity together carry 14 of the 70 theory marks. Kirchhoff, potentiometer and cell questions appear almost every year.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024