JEE Main chapters · Class 12 · Physics

Current Electricity: JEE Main Physics formulas and notes

Current Electricity is one of the most scoring chapters of Class 12 and usually gives 2 questions in JEE Mains: a network of resistors, a cell or grouping question, Kirchhoff’s laws, a meter bridge or potentiometer, the conversion of a galvanometer, and heating or an RC circuit. Most questions are short calculations that depend on drawing the circuit properly and on a few standard results. About 8 hours spread over 6 days is enough.

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Quick revision

I = nAvd e; J = I/A = σE; vd = μE = eEτ/m; σ = ne²τ/m = neμ
R = ρl/A; stretched n times: R′ = n²R; R = R₀(1 + αΔT); metals α > 0, semiconductors α < 0
Series: ΣR, V divides as R. Parallel: Σ1/R, I divides as 1/R
Cube: 7R/12, 3R/4, 5R/6; infinite ladder x² − ax − ab = 0; ring x(1 − x)R
V = E − Ir; series nE, nr; parallel E, r/m; mixed best when R = nr/m
P(load) = E²R/(R + r)²; maximum E²/4r at R = r, efficiency 50 %
KCL: sum of currents at a node = 0; KVL: sum of potential changes in a loop = 0
Bridge P/Q = R/S; meter bridge X = R l/(100 − l) (X on the left); r = R(l₁ − l₂)/l₂
Shunt S = Ig G/(I − Ig); series R = V/Ig − G; P = VI = I²R = V²/R; H = I²Rt; τ = RC

Formulas with solved examples

Current, drift and conduction

Current, charge and current density

I = (dq)/(dt)
charge through a section q = ∫I dt = area under the I–t graph
I = nAvd e (n free electrons per m³, area A, drift speed vd)
current density J = (I)/(A) = nevd (vector, along the direction of E for the conventional current)
I = ∫J·dA
conventional current is opposite to the electron flow
current is a scalar and it is the same at every section of a series wire (conservation of charge)

How to apply. Find q from I(t) by integrating, or read I from q(t) by differentiating. For a wire use I = nAvd e; for a non-uniform wire the current is the same everywhere but J and vd change with A.

Example. I = 3t² A: charge in the first 2 s is q = ∫3t² dt = t³ = 8 C. A current of 5 A in a wire of cross-section 2 mm² gives J = 5 ÷ (2 × 10⁻⁶) = 2.5 × 10⁶ A/m².

Current due to rotating charges

A ring or particle carrying charge q and making f turns per second: I = qf = (qω)/(2π) = (q)/(T)
for an electron in a Bohr orbit I = e f
the loop has magnetic moment M = IA (see Magnetism)
for a charged ring with surface or line charge use the total charge Q and the frequency of rotation

How to apply. Find the number of times per second the charge passes a fixed point of its path: that number is f. Then I = Qf.

Example. A ring with 10 μC spinning at 50 revolutions per second: I = 10 × 10⁻⁶ × 50 = 5 × 10⁻⁴ A = 0.5 mA. An electron making 6.6 × 10¹⁵ revolutions per second: I = 1.6 × 10⁻¹⁹ × 6.6 × 10¹⁵ = 1.06 mA.

Drift velocity and mobility

vd = aτ = (eE)/(m) τ, where a = eE/m and τ = mean time between collisions (relaxation time)
mobility μ = (vd)/(E) = (eτ)/(m) (m²/V s)
I = nAeμE
vd ∝ E ∝ V/l, vd ∝ I/(nA)
vd ≈ 10⁻⁴ m/s while thermal speed ≈ 10⁵ m/s and the signal (field) travels at about c
when temperature rises τ falls, so the mobility and vd (at the same E) fall

How to apply. Use I = nAvd e to get vd from the current. For a fixed current, vd ∝ 1/A and vd ∝ 1/n. If E is given, vd = μE. Drift speed is tiny; the bulb lights at once because the field is set up at nearly the speed of light.

Example. Copper wire, I = 2 A, A = 1 mm², n = 8.5 × 10²⁸: vd = 2 ÷ (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 10⁻⁶) = 1.47 × 10⁻⁴ m/s. With E = ρJ = 1.7 × 10⁻⁸ × 2 × 10⁶ = 0.034 V/m, μ = 1.47 × 10⁻⁴ ÷ 0.034 = 4.3 × 10⁻³ m²/V s. In a 2 m wire the electron takes 2 ÷ 1.47 × 10⁻⁴ ≈ 1.4 × 10⁴ s (3.8 h) to cross it.

Conductivity and relaxation time

J = σE (microscopic Ohm’s law)
σ = (1)/(ρ) = (ne²τ)/(m) = neμ
ρ = (m)/(ne²τ)
in a semiconductor σ = e(neμe + nhμh)
for a metal ρ rises with temperature because τ falls (n is almost constant)
for a semiconductor ρ falls because n rises quickly

How to apply. Write ρ = m/(ne²τ). Then compare two situations by looking at which of n and τ has changed. In a metal only τ changes with temperature.

Example. Copper, n = 8.5 × 10²⁸, τ = 2.5 × 10⁻¹⁴ s: σ = 8.5 × 10²⁸ × (1.6 × 10⁻¹⁹)² × 2.5 × 10⁻¹⁴ ÷ 9.1 × 10⁻³¹ = 5.98 × 10⁷ S/m, so ρ = 1.67 × 10⁻⁸ Ω m. If τ is halved, ρ doubles.

Resistance and resistivity

Ohm’s law, resistance and resistivity

V = IR for an ohmic conductor (V–I graph is a straight line through the origin)
R = (ρl)/(A) = (l)/(σA)
ρ depends on the material and temperature only, R also on l and A
R = (ml)/(ne²τA)
V = IR is also written as E = ρJ
conductance G = 1/R (siemens)
1 Ω = 1 V/A

How to apply. First find R from the shape (R = ρl/A), then the current or voltage from V = IR. For equal lengths and equal potential difference, I ∝ A/ρ.

Example. A copper wire (ρ = 1.7 × 10⁻⁸ Ω m) 10 m long with A = 1 mm²: R = 1.7 × 10⁻⁸ × 10 ÷ 10⁻⁶ = 0.17 Ω. At 3 A the potential drop is 0.51 V.

Stretching, cutting and joining a wire

The volume lV = Al is constant when a wire is stretched or drawn
stretched to n times the length: A becomes A/n, R′ = n²R
R ∝ l² ∝ 1/A² ∝ 1/r⁴ at constant volume
small change: (ΔR)/(R) ≈ 2(Δl)/(l)
a wire cut into n equal pieces: each has R/n, and joined in parallel gives R/n²
same material and mass: R ∝ l²

How to apply. At constant volume write R = ρl²/V. For a wire stretched by a given percent use ΔR/R = 2Δl/l (valid for small changes). For equal mass compare lengths squared.

Example. A 10 Ω wire is drawn to 3 times its length: R′ = 9 × 10 = 90 Ω. A 16 Ω wire cut into 4 equal parts: each is 4 Ω, and the four in parallel give 4 ÷ 4 = 1 Ω = 16 ÷ 4². If l increases by 1 % at constant volume, R increases by 2 %.

Temperature dependence of resistance

R = R₀(1 + αΔT), ρ = ρ₀(1 + αΔT)
(R₂)/(R₁) = (1 + αt₂)/(1 + αt₁) when R₀ is the value at 0 °C
conductors: α > 0 (R rises); semiconductors, insulators, carbon, electrolytes: α < 0 (R falls)
nichrome, manganin, constantan have very small α and large ρ (standard resistors, heaters)
series combination with no change of R: R₁α₁ + R₂α₂ = 0 (one with +α, one with −α)
a small-α alloy is used for resistance boxes

How to apply. Use the linear formula with the reference temperature at which α is given. For two temperatures use the ratio form. If the question gives R at two temperatures, find α first.

Example. R = 10 Ω at 20 °C and α = 0.004 K⁻¹: R(70 °C) = 10 (1 + 0.004 × 50) = 12 Ω. The resistance doubles when αΔT = 1, that is ΔT = 250 K.

Colour code of carbon resistors

bands in order: 1st digit, 2nd digit, multiplier (×10ⁿ), tolerance
colours = digits: black 0, brown 1, red 2, orange 3, yellow 4, green 5, blue 6, violet 7, grey 8, white 9
tolerance: gold ±5 %, silver ±10 %, no band ±20 %
R = (10a + b) × 10ⁿ Ω ± tolerance
a gold multiplier is ×0.1, silver ×0.01

How to apply. Read from the end that has the bands closer together. Write the two digits, then the power of ten, then the tolerance. The value lies between R(1 − tol) and R(1 + tol).

Example. Yellow, violet, red, gold: 47 × 10² = 4700 Ω ± 5 %, so between 4465 Ω and 4935 Ω. Brown, black, orange, silver: 10 × 10³ = 10 kΩ ± 10 %.

Resistance of non-uniform conductors

divide into thin slices and add: R = ∫(ρ dx)/(A(x))
truncated cone with end radii a and b, length l: R = (ρl)/(πab)
thin cylindrical shell between radii a and b, length l, current flowing radially: R = (ρ)/(2πl) ln(b)/(a)
concentric spherical shells: R = (ρ)/(4π)((1)/(a) − (1)/(b))

How to apply. Choose a thin slice of thickness dx (or dr) at a distance x (or r). Its resistance is ρ dx/A, with A as a function of position. Add (integrate) the slices because the current passes through them one after the other.

Example. Cone with ρ = 2 × 10⁻⁶ Ω m, l = 0.1 m, a = 1 mm, b = 2 mm: R = 2 × 10⁻⁶ × 0.1 ÷ (π × 10⁻³ × 2 × 10⁻³) = 0.032 Ω. Radial flow, ρ = 100 Ω m, a = 1 cm, b = 2 cm, l = 1 m: R = 100 × ln2 ÷ (2π) = 11.0 Ω.

Non-ohmic conductors: static and dynamic resistance

static resistance R = (V)/(I) (the chord from the origin)
dynamic (differential) resistance r = (dV)/(dI) (the slope of the tangent of the V–I graph)
ohmic: both are equal and constant
diode, filament, thermistor: non-linear V–I graph
lamp filament: R rises with I because the temperature rises
terminal graph of a cell: V = E − Ir, a line of slope −r

How to apply. Read the V and I at the point asked. For a formula I(V), use R = V/I for the static value and r = 1/(dI/dV) for the dynamic one.

Example. A device obeys I = 0.02 V² (A, V in volts). At V = 5 V: I = 0.5 A, static R = 5 ÷ 0.5 = 10 Ω; dI/dV = 0.04 V = 0.2 A/V, so dynamic r = 5 Ω.

Combinations of resistors

Resistors in series

Rₑq = R₁ + R₂ + R₃ + …
same current I in each
V divides in the ratio of resistances: V₁ = V(R₁)/(R₁ + R₂)
power in each P ∝ R
Rₑq is greater than the largest one
a switch or ideal ammeter in series adds no resistance

How to apply. Redraw the circuit, add the resistors that carry the same current, then find I = V/Rₑq and the drop across each.

Example. 2 Ω, 3 Ω and 5 Ω in series across 20 V: Rₑq = 10 Ω, I = 2 A, the drops are 4 V, 6 V and 10 V (they add to 20 V).

Resistors in parallel

(1)/(Rₑq) = (1)/(R₁) + (1)/(R₂) + …
two resistors: Rₑq = (R₁R₂)/(R₁ + R₂)
n equal resistors: R/n
same potential difference V across each
current divides inversely as resistance: I₁ = I(R₂)/(R₁ + R₂)
Rₑq is less than the smallest one
power in each P ∝ 1/R

How to apply. Find the points at the same potential. Add the conductances (1/R). For two branches use the current-divider formula, taking the other resistance in the numerator.

Example. 6 Ω and 3 Ω in parallel: Rₑq = 18 ÷ 9 = 2 Ω. With 9 A in the main line: I₆ = 9 × 3 ÷ 9 = 3 A and I₃ = 9 × 6 ÷ 9 = 6 A (they add to 9 A).

Balanced networks and equipotential points

if two points have the same potential, no current flows between them: remove the resistor joining them (or short them, the result is the same)
balanced Wheatstone bridge: (P)/(Q) = (R)/(S), the middle resistor carries no current
use the symmetry of the figure: points that are mirror images across the line joining the battery terminals have equal potential
a wire of zero resistance joining two points makes them one point (short circuit)

How to apply. Mark the potentials. By symmetry or by the ratio test find the points at equal potential and remove (or merge) accordingly. Then reduce by series and parallel steps.

Example. Bridge with arms 2 Ω, 4 Ω (top) and 3 Ω, 6 Ω (bottom) and a 10 Ω resistor between the middle points: 2/4 = 3/6, so it is balanced and the 10 Ω carries no current. Rₑq = (2 + 4) ∥ (3 + 6) = 6 ∥ 9 = 3.6 Ω.

Cube of resistors

twelve edges, each of resistance R
across an edge (adjacent corners): (7R)/(12)
across a face diagonal: (3R)/(4)
across a body diagonal (opposite corners): (5R)/(6)
for a body diagonal the three neighbours of each corner are at the same potential and may be joined

How to apply. Join the points of equal potential layer by layer: for the body diagonal there are 3 edges in parallel, then 6 in parallel, then 3 in parallel: R/3 + R/6 + R/3 = 5R/6.

Example. Each edge is 12 Ω: edge 7 Ω, face diagonal 9 Ω, body diagonal 10 Ω.

Infinite ladder

if the network repeats without end, removing one section does not change Rₑq
series arm a, shunt arm b: Rₑq = a + (b Rₑq)/(b + Rₑq) so Rₑq² − aRₑq − ab = 0
Rₑq = (a + √(a² + 4ab))/(2)
for a = b = R: Rₑq = (1 + √5)/(2) R ≈ 1.618R
it is larger than a

How to apply. Call the unknown total resistance x. The part to the right of the first section also has the value x. Write x = a + (b ∥ x), solve the quadratic and reject the negative root.

Example. a = 2 Ω and b = 4 Ω: x² − 2x − 8 = 0, so x = 4 Ω (check: 2 + 4 ∥ 4 = 2 + 2 = 4 Ω). For a = b = 2 Ω: x = 2 × 1.618 = 3.24 Ω.

Star–delta and uniform ring

delta to star: arm = (product of the two delta sides meeting there)/(sum of the three sides)
three equal delta sides R give star arms R/3
uniform ring of resistance R, points dividing it into arcs of x and (1 − x) of the ring: Rₑq = x(1 − x)R
diametrically opposite points: R/4
the same idea for a wire bent into a loop

How to apply. Replace the triangle by a star when no pair is in series or parallel. For a ring write the two arcs as two parallel branches.

Example. Three 9 Ω resistors in a triangle: each star arm is 3 Ω, so between two corners Rₑq = 6 Ω (check: 9 ∥ 18 = 6 Ω). A ring of 16 Ω: between points 90° apart the arcs are 4 Ω and 12 Ω, Rₑq = 4 × 12 ÷ 16 = 3 Ω; between opposite points 16/4 = 4 Ω.

Cells and EMF

EMF, internal resistance and terminal voltage

E = work done per unit charge by the cell (the open-circuit voltage)
I = (E)/(R + r)
discharging: V = E − Ir
charging: V = E + Ir and I = (V − E)/(R + r)
open circuit: V = E
short circuit: I = (E)/(r)
V–I graph: intercept E on the V axis, slope −r
V = (ER)/(R + r)

How to apply. Add the internal resistance in series with the external resistance to get I. The terminal voltage is the voltage across the external circuit. During charging the current enters the positive terminal and V > E.

Example. E = 12 V, r = 1 Ω and R = 5 Ω: I = 12 ÷ 6 = 2 A, V = 12 − 2 × 1 = 10 V. A 12 V battery (r = 1 Ω) charged from 15 V through 2 Ω: I = 3 ÷ 3 = 1 A, terminal voltage 12 + 1 = 13 V.

Cells in series

n identical cells: Eₑq = nE, rₑq = nr
I = (nE)/(R + nr)
good when R ≫ r
cells with opposite polarity subtract: Eₑq = ΣE (taken with signs), rₑq = Σr always adds
a cell with reversed terminals is being charged by the others

How to apply. Take the sum of EMFs with signs (going round the loop) and the sum of all internal resistances.

Example. 3 cells, each 1.5 V and 0.5 Ω, with R = 2.5 Ω: I = 4.5 ÷ (2.5 + 1.5) = 1.125 A. A 12 V (1 Ω) cell opposed to a 6 V (2 Ω) cell with R = 3 Ω: I = (12 − 6) ÷ (3 + 1 + 2) = 1 A.

Cells in parallel

m identical cells: Eₑq = E, rₑq = (r)/(m)
I = (E)/(R + r/m)
good when R ≪ r
unequal cells: Eₑq = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂), rₑq = (r₁r₂)/(r₁ + r₂)
the EMF does not increase; only the current capacity and the lowering of r

How to apply. For identical cells the EMF is that of one cell. For unequal cells use the node-voltage form above. Check the current in each cell.

Example. Cells 4 V (1 Ω) and 2 V (1 Ω) in parallel: Eₑq = (4 + 2) ÷ 2 = 3 V, rₑq = 0.5 Ω. Four identical 1.5 V, 2 Ω cells in parallel across 0.5 Ω: I = 1.5 ÷ (0.5 + 0.5) = 1.5 A (one cell alone gives 0.6 A).

Mixed grouping of cells

N = mn cells: n in series in each row, m rows in parallel
I = (nmE)/(mR + nr)
the current is greatest when the total internal resistance equals the external one: R = (nr)/(m)
then Imax = (nE)/(2R) = (E√N)/(2√(rR))

How to apply. Put N = mn into I and choose n and m so that nr/m is as near as possible to R. If R ≫ r use all in series; if R ≪ r use all in parallel.

Example. 24 cells, E = 2 V, r = 1 Ω each, R = 6 Ω: need n/m = R/r = 6 with nm = 24, so n = 12 and m = 2. I = 24 × 2 ÷ (2 × 6 + 12 × 1) = 48 ÷ 24 = 2 A.

Power in a circuit and maximum power transfer

P(load) = I²R = (E²R)/((R + r)²)
maximum when R = r: Pₘₐₓ = (E²)/(4r)
efficiency η = (R)/(R + r) = (V)/(E) = 50 % at maximum power
total power of the cell = EI, lost in the cell I²r
P(load) is the same for two values R₁ and R₂ if R₁R₂ = r²
a graph of P against R rises, peaks at R = r and falls

How to apply. Write P(R). For the maximum set R = r. If a particular R is given, put it in the formula. Remember that maximum power is not maximum efficiency.

Example. E = 12 V, r = 3 Ω: at R = 3 Ω P = 144 ÷ 12 = 12 W (maximum). At R = 12 Ω, P = 144 × 12 ÷ 15² = 7.68 W. R = 1 Ω and R = 9 Ω give the same power (1 × 9 = 3²): 144 × 1 ÷ 16 = 9 W and 144 × 9 ÷ 144 = 9 W.

Kirchhoff and bridges

Kirchhoff’s laws

junction (current) law: ΣIᵢₙ = ΣIₒᵤₜ (charge is conserved)
loop (voltage) law: ΣΔV = 0 round any closed loop (energy is conserved)
signs: going through a resistor along the current ΔV = −IR, against it +IR
through a cell from − to +: +E, from + to −: −E
a negative answer means the current is opposite to the assumed direction

How to apply. Assume a current in each branch, apply the junction law at one node, then the loop law to as many independent loops as there are unknowns. Solve, and check with the junction law. For two nodes the node-voltage method is quicker.

Example. Branch 1: 8 V cell and 1 Ω; branch 2: 4 V cell and 2 Ω; load 2 Ω, all between the same two nodes. With the lower node at 0 V and the upper at V: (8 − V)/1 + (4 − V)/2 = V/2, so V = 5 V. I₁ = 3 A, I₂ = −0.5 A (the 4 V cell is being charged), load current 2.5 A (3 − 0.5 = 2.5 A).

Wheatstone bridge

balanced when no current flows through the galvanometer: (P)/(Q) = (R)/(S)
so for an unknown S: S = R(Q)/(P)
at balance the bridge is the same as two parallel branches (P + Q) and (R + S)
the bridge is most sensitive when all four arms are of the same order
when it is not balanced, use Kirchhoff’s laws (or the star–delta change)

How to apply. Check P/Q = R/S first. If balanced, remove the galvanometer arm. If not, assign currents and use the loop law, or use the Wheatstone network as two loops.

Example. P = 10 Ω, Q = 20 Ω, R = 15 Ω: S = R Q/P = 15 × 20 ÷ 10 = 30 Ω. Rₑq of the balanced bridge = 30 ∥ 45 = 18 Ω.

Meter bridge

uniform wire of length 100 cm: resistance proportional to length
balance at length l from the left end: (resistance in the left gap)/(resistance in the right gap) = (l)/(100 − l)
unknown X in the left gap: X = R(l)/(100 − l)
unknown X in the right gap: X = R(100 − l)/(l)
specific resistance ρ = (Xπd²)/(4L)
best accuracy when l is near 50 cm
interchanging X and R and averaging removes end errors
the heating of the wire must be avoided (press the key briefly)

How to apply. Be sure which gap holds the unknown. The segment next to a resistor is in proportion to that resistor: (left resistance) / (right resistance) = l/(100 − l). Use balance near 50 cm.

Example. R = 10 Ω in the right gap, X in the left gap, balance at l = 40 cm: X = 10 × 40 ÷ 60 = 6.67 Ω. With X in the right gap instead: X = 10 × 60 ÷ 40 = 15 Ω. Wire of length 0.5 m and diameter 1 mm with X = 15 Ω: ρ = 15 × π × (10⁻³)² ÷ (4 × 0.5) = 2.36 × 10⁻⁵ Ω m.

Potentiometer and meters

Potentiometer: principle and potential gradient

a long uniform wire of length L with a steady current: potential difference V ∝ length
potential gradient k = (V)/(L) = (I ρ)/(A) = (E R(wire))/((R(wire) + Rₕ + r) L)
a balanced cell of EMF E: E = kl (no current drawn from the cell at balance, so it measures the true EMF)
sensitivity rises when k is smaller (a series resistance Rₕ, longer wire)
the driver cell EMF must be greater than the EMF measured
the potentiometer is an ideal voltmeter

How to apply. Find the current in the wire from the driving circuit (E of the driver divided by the total resistance), then k = I × (resistance per unit length). The balancing length gives the unknown: E = kl.

Example. Wire of 10 m and 20 Ω, driver cell 3 V (no internal resistance) and rheostat 10 Ω in series: I = 3 ÷ 30 = 0.1 A, V(wire) = 2 V, k = 0.2 V/m. A cell balancing at 4.5 m has E = 0.2 × 4.5 = 0.9 V.

Potentiometer: comparing EMFs and internal resistance

comparing two cells: (E₁)/(E₂) = (l₁)/(l₂)
sum and difference method: (E₁ + E₂)/(E₁ − E₂) = (l₁)/(l₂) (aiding and opposing)
internal resistance: l₁ is the balance with the cell on open circuit (E = kl₁), l₂ with a known R across it (V = kl₂): r = R(l₁ − l₂)/(l₂)
the cell must give no current when l₁ is found
the reading with R gives the terminal voltage

How to apply. Open circuit gives E ∝ l₁; closing the key across R gives V = E R/(R + r) ∝ l₂. The ratio l₁/l₂ = (R + r)/R gives r.

Example. Balancing lengths 60 cm and 40 cm: E₁/E₂ = 3/2, so 1.5 V and 1.0 V. l₁ = 5 m, l₂ = 4 m with R = 8 Ω across the cell: r = 8 × (5 − 4) ÷ 4 = 2 Ω.

Galvanometer to ammeter: the shunt

a small resistance S (the shunt) is connected in parallel with the galvanometer (resistance G, full-scale current Ig)
S = (Ig G)/(I − Ig) = (G)/(n − 1) for a range I = nIg
ammeter resistance Rₐ = (GS)/(G + S) ≈ S (very small)
connected in series with the circuit
an ideal ammeter has zero resistance

How to apply. The galvanometer takes Ig and the shunt takes I − Ig at the same potential difference: Ig G = (I − Ig) S. Solve for S.

Example. G = 100 Ω, Ig = 1 mA, range 1 A: S = 0.001 × 100 ÷ 0.999 = 0.1001 Ω ≈ 0.1 Ω, and Rₐ ≈ 0.1 Ω. For a range 10 times larger than Ig: S = G/9.

Galvanometer to voltmeter: the series resistance

a large resistance Rₛ in series with the galvanometer
Rₛ = (V)/(Ig) − G
voltmeter resistance Rᵥ = G + Rₛ = (V)/(Ig) (very large)
connected in parallel across the element
an ideal voltmeter has infinite resistance
a range extended n times (V = nVg): Rₛ = (n − 1)G

How to apply. Full-scale: V = Ig (G + Rₛ). Solve for Rₛ. The voltmeter resistance is the full V divided by Ig.

Example. G = 50 Ω, Ig = 2 mA, range 10 V: Rₛ = 10 ÷ 0.002 − 50 = 4950 Ω and Rᵥ = 5000 Ω.

Errors due to the meters

voltmeter across R: reading is across R ∥ Rᵥ, so R(measured) = (R Rᵥ)/(R + Rᵥ) < R
ammeter in series: R(measured) = R + Rₐ > R
for a small R use the ammeter outside the voltmeter, for a large R put the voltmeter outside
the voltmeter reads less than the true value because it draws current; the ammeter reads less because it adds resistance

How to apply. Treat the meter as a resistor of its own resistance and recalculate the circuit.

Example. R = 1000 Ω with a 9000 Ω voltmeter across it: the circuit sees 1000 ∥ 9000 = 900 Ω, a 10 % error. A 0.5 Ω ammeter in series with a 10 Ω resistor reads 10.5 Ω.

Moving-coil galvanometer: sensitivity

deflection θ = (NBA)/(k) I (k = torsion constant, radial field)
current sensitivity (θ)/(I) = (NBA)/(k)
voltage sensitivity (θ)/(V) = (NBA)/(kR)
doubling N doubles the current sensitivity, but R doubles too, so the voltage sensitivity is unchanged
a radial field and soft iron core make θ ∝ I
increasing sensitivity: large N, B, A, small k

How to apply. Use θ ∝ I. For the effect of changing the coil, list how N, A, R change and apply the formula.

Example. N = 50, B = 0.05 T, A = 2 × 10⁻⁴ m², k = 10⁻⁶ N m/rad: θ/I = 50 × 0.05 × 2 × 10⁻⁴ ÷ 10⁻⁶ = 500 rad/A = 0.5 rad/mA. With R = 50 Ω: θ/V = 10 rad/V.

Power and heating

Electric power, energy and bulbs

P = VI = I²R = (V²)/(R)
energy W = Pt; 1 kWh = 3.6 × 10⁶ J
rated bulb (V, P): R = (V²)/(P)
at a different voltage V′: P′ = P(V′²)/(V²) (R assumed constant)
in series P ∝ R: the bulb of lower rating (higher R) glows brighter
in parallel P ∝ 1/R: the bulb of higher rating glows brighter
power is greatest when the resistance is smallest at a fixed V

How to apply. Find each R from its rating, then the current in the actual circuit and the power with the form that suits the circuit: I²R for series, V²/R for parallel.

Example. 100 W and 60 W bulbs (220 V): R₁ = 484 Ω, R₂ = 806.7 Ω. In series across 220 V: I = 220 ÷ 1290.7 = 0.1704 A, P₁ = I²R₁ = 14.1 W and P₂ = 23.4 W, so the 60 W bulb is brighter. A 100 W bulb on 110 V gives 25 W.

Joule heating

H = I²Rt = VIt = (V²t)/(R) (joule)
H (cal) = (I²Rt)/(4.2)
heating a body: Pt × η = ms ΔT
the heat in resistors in series is ∝ R, in parallel ∝ 1/R
the heat produced in a wire of fixed length and V falls if R rises
power loss in transmission lines = I²R, so the voltage is raised and I lowered

How to apply. Equate the heat to the energy needed, ms ΔT (add latent heat for a change of state). Allow for the efficiency η.

Example. A 1 kW heater warms 2 kg of water by 40 K: t = 2 × 4200 × 40 ÷ 1000 = 336 s. If only 80 % of the heat reaches the water, t = 420 s. 2 Ω and 3 Ω in parallel give heat in the ratio 3 : 2.

Energy bill and units of electricity

energy in kWh = (P(watt) × t(hour))/(1000)
cost = units × rate
P = VI, so a 2 kW heater on 220 V draws 9.1 A
the electricity meter measures energy (kWh), not power
1 unit = 1 kWh = 3.6 × 10⁶ J

How to apply. Add up P × t for every appliance, divide by 1000 and multiply by the rate.

Example. 5 bulbs of 40 W for 6 h a day for 30 days: 200 W × 6 h = 1.2 kWh a day, 36 kWh in the month; at ₹6 per unit the bill is ₹216.

RC circuits and household wiring

Charging a capacitor through a resistor

time constant τ = RC
charging: q = CE(1 − e^−t/τ), Vᴄ = E(1 − e^−t/τ), I = (E)/(R) e^−t/τ
at t = 0 the capacitor acts as a short (I = E/R); at t → ∞ as an open circuit (I = 0)
after t = τ the charge is 63 % of the final value, after 2τ 86 %
energy supplied by the cell CE², half stored (½CE²) and half lost as heat in R, independent of R (C and U = ½CV²: see Electrostatics)

How to apply. Find the equivalent resistance seen by the capacitor, then τ = RC. Put t/τ in the exponential. Use the initial and final values (0 and ∞) to check the answer.

Example. R = 1 MΩ, C = 2 μF: τ = 2 s. At t = 2 s the capacitor holds 63.2 % of its final charge. With E = 10 V the cell supplies CE² = 2 × 10⁻⁶ × 100 = 200 μJ, of which 100 μJ is stored and 100 μJ becomes heat.

Discharging and the steady state

discharging: q = q₀e^−t/τ, I = I₀e^−t/τ
half of the charge is left after t = τ ln 2 = 0.693τ
in the steady state no current flows through a capacitor branch: remove it, find the currents, then Vᴄ = the voltage across the branch it is connected to
q = CVᴄ
then redraw for t = 0 (capacitor = short if uncharged)

How to apply. For the steady state, replace each capacitor by an open circuit. The voltage across the capacitor is that across the resistor (or part) in parallel with it.

Example. τ = 2 s: the charge falls to half in 2 × 0.693 = 1.39 s. A 12 V cell, R₁ = 2 Ω in series with R₂ = 4 Ω, and a 3 μF capacitor across R₂: steady current 2 A, Vᴄ = 8 V, Q = 3 × 8 = 24 μC.

Household circuits and the fuse

appliances are in parallel across the 220 V supply, each at full voltage and working independently
I = (P)/(V) for each, and the total current is ΣP/V
fuse (live wire) of low melting point; the rating is a little above the normal current
fuse wire: the current at which it melts, I ∝ r^3/2 (heating I²ρl/πr² equals the loss from the surface 2πrl)
earth wire to the metal body protects from shock; MCB cuts the circuit in an overload or a short circuit
the switch goes in the live wire

How to apply. Add the powers of everything that is on, divide by the voltage to get the current and choose the next standard rating. For two fuse wires of the same material and length use I ∝ r^3/2.

Example. Heater 1.5 kW, iron 500 W and four 60 W bulbs on 220 V: 2240 W, I = 10.2 A, so a 15 A fuse suits. A fuse wire of radius 0.2 mm that blows at 5 A: for radius 0.4 mm the current is 5 × 2^3/2 = 14.1 A.

Most asked question types

How do I solve “Drift velocity and current density” questions in JEE Main?

The current, the area and the number density of electrons are given. Find the drift speed, the current density, the mobility or the effect of a change of area. 1. I = nAvd e gives vd. 2. J = I/A and J = σE connect with the field. 3. For a fixed current, vd ∝ 1/A; μ = vd/E. Example: 2 A in 1 mm² copper: vd = 1.47 × 10⁻⁴ m/s.

How do I solve “Charge from a varying current; rotating charges” questions in JEE Main?

The current is given as a function of time, or a charged ring or electron is rotating. Find the charge through a section or the equivalent current. 1. q = ∫I dt, the area under the I–t graph. 2. Rotating charge: I = q f. 3. Use SI units and a complete cycle for f. Example: I = 3t² for 2 s gives 8 C; 10 μC at 50 rev/s gives 0.5 mA.

How do I solve “Stretched wire and change of resistance” questions in JEE Main?

A wire is stretched, cut or drawn into a thinner wire. Find the new resistance or the percentage change. 1. Volume is constant: A′l′ = Al. 2. R′ = n²R for n times the length. 3. A small change: ΔR/R = 2Δl/l. Example: A 10 Ω wire stretched to 3 times its length becomes 90 Ω.

How do I solve “Resistivity and its change with temperature” questions in JEE Main?

The resistance at two temperatures, α or a combination of materials is given. Find the temperature, α or the resistance at a new temperature. 1. R = R₀(1 + αΔT), or the ratio of (1 + αt). 2. Metals: R rises; semiconductors: R falls. 3. For a constant combination, R₁α₁ + R₂α₂ = 0. Example: 10 Ω at 20 °C with α = 0.004: 12 Ω at 70 °C.

How do I solve “Colour code and V–I graphs” questions in JEE Main?

The colours of a carbon resistor are given, or a V–I graph of a conductor or device is shown. Find the value, the tolerance, or the resistance at a point. 1. Digits, multiplier and tolerance in that order. 2. Static R = V/I; dynamic r = dV/dI. 3. A straight line through the origin is ohmic. Example: Yellow, violet, red, gold: 4700 Ω ± 5 %.

How do I solve “Series and parallel networks” questions in JEE Main?

A network of resistors is connected to a battery. Find the equivalent resistance, the current in a branch or the potential difference between two points. 1. Redraw by marking equal-potential nodes. 2. Combine series and parallel parts step by step. 3. Use the current divider and then V = IR. Example: 6 Ω ∥ 3 Ω = 2 Ω; 9 A divides 3 A and 6 A.

How do I solve “Balanced bridge and symmetry” questions in JEE Main?

A bridge, a cube, a square or a skeleton network is given. Find the equivalent resistance by finding the points at the same potential. 1. Test P/Q = R/S for a balanced bridge and remove the middle arm. 2. Join points that are equal by symmetry. 3. Cube: 7R/12, 3R/4, 5R/6. Example: Balanced bridge 2, 4, 3, 6 Ω: 3.6 Ω.

How do I solve “Infinite ladder and ring networks” questions in JEE Main?

A network repeats without end, or a uniform ring or wire loop is connected at two points. Find the equivalent resistance. 1. Let the whole be x and write x = a + (b ∥ x). 2. Solve the quadratic and take the positive root. 3. Ring: two arcs in parallel, x(1 − x)R. Example: a = 2 Ω, b = 4 Ω gives 4 Ω; ring 16 Ω at 90°: 3 Ω.

How do I solve “Terminal voltage and internal resistance” questions in JEE Main?

A cell with internal resistance drives a load, is on open circuit or short-circuited, or is being charged. Find the current, the terminal voltage or r. 1. I = E/(R + r), V = E − Ir. 2. Charging: V = E + Ir. 3. Read E and r from a V–I graph (intercept and slope). Example: E = 12 V, r = 1 Ω, R = 5 Ω: I = 2 A and V = 10 V.

How do I solve “Grouping of cells” questions in JEE Main?

Cells are joined in series, in parallel or in rows. Find the current in an external resistor, the best arrangement or the equivalent EMF. 1. Series: nE and nr. Parallel: E and r/m. 2. Mixed: I = nmE/(mR + nr). 3. Maximum current when R = nr/m. Example: 24 cells of 2 V and 1 Ω with R = 6 Ω: 12 in series, 2 rows, 2 A.

How do I solve “Maximum power transferred to the load” questions in JEE Main?

The power in a resistor connected across a cell is asked as a function of R. Find the maximum power, the R for it, or two resistors with the same power. 1. P = E²R/(R + r)². 2. Maximum at R = r: E²/4r. 3. Equal power for R₁R₂ = r². Example: E = 12 V, r = 3 Ω: Pmax = 12 W at R = 3 Ω.

How do I solve “Bulbs, power and brightness” questions in JEE Main?

Bulbs or heaters with ratings are used in series, in parallel or at another voltage. Find the power, the brightness order or the current. 1. R = V²/P from the rating. 2. Series: P = I²R, so bigger R is brighter. 3. Parallel: P = V²/R, so smaller R is brighter. Example: 100 W and 60 W bulbs in series on 220 V: 14.1 W and 23.4 W.

How do I solve “Joule heating and time to heat a liquid” questions in JEE Main?

A heater or a resistor warms water or a metal. Find the time taken, the final temperature, or the heat in a resistor. 1. P = V²/R or I²R. 2. ηPt = ms ΔT (add latent heat). 3. Series: H ∝ R; parallel: H ∝ 1/R. Example: 1 kW, 2 kg of water, 40 K: 336 s.

How do I solve “RC charging, discharging and steady state” questions in JEE Main?

A capacitor is charged or discharged through a resistor, or a circuit with a capacitor is in the steady state. Find the time, the charge or the current at an instant. 1. τ = RC; use e^−t/τ. 2. Steady state: capacitor = open circuit; t = 0: uncharged capacitor = short. 3. Q = C Vᴄ with Vᴄ found from the resistors. Example: R = 1 MΩ, C = 2 μF: 63 % charge at 2 s; steady circuit gives 24 μC.

How do I solve “Fuse and household wiring” questions in JEE Main?

Appliances are used on the mains, or fuse wires of different radius are compared. Find the total current, the fuse rating or the current that melts a fuse. 1. I = P/V for each appliance and add (parallel). 2. Fuse rating just above the load. 3. Fuse wire: I ∝ r^3/2. Example: 2240 W at 220 V: 10.2 A, use a 15 A fuse.

How do I solve “Kirchhoff’s laws in a two-loop circuit” questions in JEE Main?

Two or three cells and resistors are in several branches. Find a branch current, the potential difference between two points or find which cell is charging. 1. Assume currents and apply the junction rule. 2. Loop rule in each independent loop. 3. Or the node-voltage method for two nodes. Example: 8 V (1 Ω), 4 V (2 Ω), load 2 Ω: V = 5 V, currents 3 A, −0.5 A, 2.5 A.

How do I solve “Wheatstone bridge and meter bridge” questions in JEE Main?

The balance length of a meter bridge is given with a known resistance, or the four arms of a bridge are given. Find the unknown resistance or the effect of interchanging. 1. Balance: P/Q = R/S. 2. Check which gap holds X: ratio of resistances = ratio of lengths. 3. Interchange X and R to remove end errors. Example: R = 10 Ω, l = 40 cm, X on the right: 15 Ω.

How do I solve “Potentiometer: EMF and internal resistance” questions in JEE Main?

The balancing lengths are given for two cells, or for a cell with and without a resistor across it. Find the ratio of EMFs, the EMF, or the internal resistance. 1. E = kl with k = V/L. 2. E₁/E₂ = l₁/l₂. 3. r = R(l₁ − l₂)/l₂. Example: l₁ = 5 m, l₂ = 4 m, R = 8 Ω: r = 2 Ω.

How do I solve “Galvanometer to ammeter or voltmeter” questions in JEE Main?

A galvanometer of given resistance and full-scale current is changed to a meter of given range. Find the shunt or the series resistance. 1. Ammeter: S = Ig G/(I − Ig), in parallel. 2. Voltmeter: R = V/Ig − G, in series. 3. Resistance of the meter: ammeter ≈ S, voltmeter ≈ V/Ig. Example: G = 100 Ω, Ig = 1 mA: shunt 0.1 Ω for 1 A; G = 50 Ω, 2 mA: 4950 Ω for 10 V.

Common myths and the truth

Myth: Electrons in a wire move at the speed of light. Truth: The drift speed is about 10⁻⁴ m/s. The signal (the electric field) travels at nearly c, and that is why the lamp lights at once.
Myth: A battery supplies a fixed current. Truth: A battery supplies a nearly fixed EMF; the current depends on the total resistance of the circuit, I = E/(R + r).
Myth: The current gets used up as it passes through the resistors. Truth: The same current enters and leaves a resistor. The energy is transferred, and the potential drops.
Myth: Resistance depends only on the material of the wire. Truth: R = ρl/A depends on the length and area too; it is the resistivity ρ that is a property of the material and temperature.
Myth: When a wire is stretched to double the length its resistance doubles. Truth: The area halves as well, so R becomes four times.
Myth: In a series circuit the bulb of higher wattage is always brighter. Truth: In series the same I flows and P = I²R, so the bulb with the higher resistance (lower wattage) is brighter. In parallel the higher-wattage bulb is brighter.
Myth: Maximum power in the load means maximum efficiency. Truth: At R = r the power is maximum but the efficiency is only 50 %. Efficiency R/(R + r) rises with R.
Myth: A voltmeter reads the true EMF of a cell. Truth: A voltmeter draws some current, so it reads E − Ir. Only the potentiometer, which draws no current at balance, measures the EMF.
Myth: Resistance of every conductor rises with temperature. Truth: Metals have a positive α, but semiconductors, carbon and electrolytes have a negative α, and their resistance falls when they are heated.
Myth: Cells in parallel give a higher EMF. Truth: The EMF stays the same; the parallel group lowers the internal resistance and shares the current between the cells.

Mistakes to avoid

In a parallel group the equivalent resistance is smaller than the smallest one; in series larger than the largest.
Stretching a wire changes both l and A; R becomes n²R, not nR.
In a mixed grouping the EMF of a row is nE, not mnE.
Maximum power in the load (R = r) is not the maximum efficiency (which is for a large R).
In the meter bridge check which gap holds the unknown before writing X = R l/(100 − l).
The voltage across a cell is E − Ir only while it delivers current; it is E + Ir when it is charged.
Drift velocity is tiny, but the current starts at once everywhere in the wire.
In Kirchhoff’s loop rule, count the sign of the cell and the direction of the current in each element.

Simulations for Current Electricity

Cells in series and parallel — Combine n identical cells in series or in parallel and compare the current through a load.
Ohm’s law circuit — Change battery voltage and resistance to control the current in a closed circuit.
Electron drift in a wire — Change wire thickness and current to see why electron drift is so slow.
A real cell under load — Connect a load to a cell and separate its emf from the voltage delivered at its terminals.
Kirchhoff’s laws: two-loop circuit — Two cells drive currents through three resistors. Solve the loops and watch each branch current.
Potentiometer: comparing EMFs — Slide the jockey along a 1 m wire until the galvanometer reads zero. The balancing length measures the…
Resistance and temperature — Heat a resistor and watch its resistance change. Metals rise; carbon falls.
Series and parallel resistors — Compare equivalent resistance for two connection types.
Wheatstone bridge — Adjust four resistors and bring the galvanometer reading to zero.

More: Class 12 Physics simulations · all JEE Main chapters.