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Current Electricity

01
CONCEPT

Electric Current & Drift Velocity

In one line
Electric current is the rate of flow of charge through any cross-section of a conductor.
I = qt   ·   I = dqdt   ·   q = ∫ I dt

SI unit ampere: 1 A = 1 C s⁻¹. Current is a scalar — it has a direction but adds by ordinary (scalar) addition. The charge that flows is the area under the I–t graph.

−−−−−−−electrons driftE, II = q/t = neAvd
Electrons drift opposite to E
Conventional current
direction of flow of positive charge (+ to − outside the cell)
Electronic current
direction of flow of electrons — opposite to the conventional current
CARRIERS

Who carries the current?

MediumCharge carriers
Metals (conductors)free electrons
Semiconductorsfree electrons and holes
Liquids (electrolytes)positive and negative ions
Gases (ionised)positive ions and electrons
DRIFT VELOCITY

Slow average speed of free electrons

In one line
Drift velocity vd is the average velocity with which free electrons drift towards the positive end under an applied field.
vd = eEτm   ·   I = neAvd

τ = relaxation time (mean time between collisions, ~10⁻¹⁴ s); n = free electrons per m³. vd is only ~mm/s. For a steady current in a wire of changing area, I is the same everywhere, so vd ∝ 1/A. As temperature rises, τ falls, so vd falls.

Practise PYQs on this topicElectric Current, Drift Velocity & Current Density · 35 questions with solutions
02
CONCEPT

Mobility & Current Density

In one line
Mobility is the drift velocity per unit electric field.
μ = vdE = eτm

SI unit m² V⁻¹ s⁻¹. Electrons are more mobile than holes (μe > μh) because electrons move freely while holes move from bond to bond.

+++applied field Eaverage drift vd
Random collisions, slow net drift
CURRENT DENSITY

Current per unit area

In one line
Current density J at a point is the current per unit area held normal to the current. J is a vector along the current.
J = IA = nevd   ·   I = ∫ J · dA

SI unit A m⁻², dimension [L⁻² A]. If the area makes angle θ with the normal to the current: I = JA cosθ.

JdAθI = ∫ J · dA = JA cosθ
Only the normal part of the area counts
OHM’S LAW (MICROSCOPIC)

Linking J, σ and E

J = σE
vector form of Ohm’s law
σ = 1ρ = ne²τm
conductivity from n and τ
A wire carries 5 A. Its area is 5 mm² and n = 10²⁹ m⁻³. Find vd.
STEP 1vd = I/(neA)STEP 2vd = 5/(10²⁹ × 1.6 × 10⁻¹⁹ × 5 × 10⁻⁶) = 6.25 × 10⁻⁵ m s⁻¹
Practise PYQs on this topicElectric Current, Drift Velocity & Current Density · 35 questions with solutions
03
CONCEPT

Ohm’s Law, Resistance & Resistivity

In one line
If the physical state of a conductor (temperature, strain) stays the same, the current through it is proportional to the p.d. across it.
V ∝ I  ⇒  V = IR

R is the resistance: 1 Ω = 1 V A⁻¹, dimension [M L² T⁻³ A⁻²]. The V–I graph of an ohmic conductor is a straight line through the origin; slope of V–I = R.

Aρlength lR = ρl/A
R depends on l, A and the material
Ohmic
metals at constant temperature — straight-line V–I graph
Non-ohmic
diode, transistor, vacuum tube, electrolyte, discharge tube — curved V–I graph
RESISTIVITY

A property of the material only

R = ρlA   ·   ρ = RAl

ρ (Ω m) does not depend on the shape or size of the body, only on the material and temperature. Conductivity σ = 1/ρ (S m⁻¹); conductance G = 1/R (siemens, mho). Order: ρinsulator > ρalloy > ρsemiconductor > ρconductor.

STRETCHING

Volume stays constant

beforel₁, A₁after stretchingl₂ > l₁, A₂ < A₁A₁l₁ = A₂l₂ ⇒ R ∝ l² ∝ 1/r⁴
Wire stretched so thatNew resistanceSmall % change
length becomes n timesR′ = n²RΔR/R = 2 Δl/l
radius becomes 1/n timesR′ = n⁴RΔR/R = 4 Δr/r
cut into n equal parts, joined in parallelR′ = R/n²—
Practise PYQs on this topicOhm's Law, Resistivity & Temperature · 80 questions with solutions
04
CONCEPT

Temperature Dependence & Colour Code

In one line
For metals the resistance rises with temperature because ions vibrate more and electrons collide more often (τ falls).
RT = R₀[1 + α(T − T₀)]
α = R₂ − R₁R₁(T₂ − T₁)
ρTMetal (α +ve)ρTManganin/alloyρTSemiconductor
ρ–T graphs
MaterialαR on heating
Metalspositiveincreases
Alloys (manganin, constantan, nichrome)very small +vealmost constant
Semiconductors, insulators, electrolytesnegativedecreases

Alloys with small α and high ρ are used for standard resistors. Superconductors: below a critical temperature the resistance falls to zero (used in MRI magnets, loss-free cables). Thermistors: NTC (R falls on heating) are temperature sensors; PTC (R rises on heating) act as resettable fuses.

COLOUR CODE

Carbon resistors: B B R O Y G B V G W

ColourDigit / multiplierColourDigit / multiplier
Black0 / 10⁰Green5 / 10⁵
Brown1 / 10¹Blue6 / 10⁶
Red2 / 10²Violet7 / 10⁷
Orange3 / 10³Grey8 / 10⁸
Yellow4 / 10⁴White9 / 10⁹

4th band (tolerance): gold ±5%, silver ±10%, no band ±20%. Gold and silver as 3rd band mean ×10⁻¹ and ×10⁻².

ABCDAB × 10C ± D%
Orange-Orange-Red-Gold = 33 × 10² Ω ± 5%
Practise PYQs on this topicOhm's Law, Resistivity & Temperature · 80 questions with solutions
05
CONCEPT

Combination of Resistors

In one line
In series the same current flows through each resistor; in parallel each resistor has the same p.d.
Series: Rs = R₁ + R₂ + R₃
Parallel: 1Rp = 1R₁ + 1R₂ + 1R₃

Rs is larger than the largest resistor; Rp is smaller than the smallest one.

Series: same IR₁R₂R₃IParallel: same VR₁R₂R₃
Series and parallel
V₁ = VR₁R₁ + R₂
voltage divider (series)
I₁ = IR₂R₁ + R₂
current divider (two in parallel)
SHORTCUTS

Asked again and again

SituationResult
n identical resistors R in series / in parallelnR / R/n  (ratio Rs/Rp = n²)
Wire R cut into n equal parts, all in parallelR/n²
Wire R bent into a circle: across a diameterR/4
Arc θ of the circle against the restR1R2/(R1 + R2) with R1 = Rθ/2π
Wire R bent into a triangle: across any side2R/9
Bridge network with P/Q = R/Sremove the middle arm (no current)
A 12 Ω wire is bent into a circle. Find the resistance between two points a quarter-circle apart.
STEP 1Arcs: 3 Ω and 9 Ω in parallelSTEP 2R = 3 × 9/(3 + 9) = 2.25 Ω
Practise PYQs on this topicCombination of Resistors · 88 questions with solutions
06
CONCEPT

Cells: EMF & Internal Resistance

In one line
EMF E is the p.d. across the terminals of a cell when no current is drawn. When current flows, part of E is lost inside the cell.
I = ER + r  ·  V = E − Ir
r = (EV − 1) R

Internal resistance r depends on the nature and separation of the electrodes, the electrolyte (nature, concentration, temperature) and the area of the electrodes dipped in it.

cellErRIV = E − Ir = IR
Terminal voltage V < E on discharge
Discharging: V = E − Ir (V < E)
Charging: V = E + Ir (V > E)
Open circuit: I = 0, V = E
Short circuit: R = 0, I = E/r, V = 0
GROUPING

Cells in series and in parallel

n identical cellsEeqreqCurrent in RBest when
SeriesnEnrnE/(R + nr)R ≫ r
ParallelEr/nnE/(nR + r)R ≪ r
Mixed (n in series, m rows)nEnr/mmnE/(mR + nr)R = nr/m

One cell reversed in a series row reduces the emf by 2E. Two cells of equal emf in series give zero p.d. across cell 1 when R = r₁ − r₂.

P is maximum when R = r
Pmax = E²4r
Practise PYQs on this topicCells: EMF, Internal Resistance & Combinations · 55 questions with solutions
07
CONCEPT

Kirchhoff’s Laws & Measuring Instruments

In one line
KCL: the algebraic sum of currents meeting at a junction is zero (current in = current out).
Σ I = 0  — conservation of charge
In one line
KVL: the algebraic sum of all p.d.s round any closed loop is zero.
Σ ΔV = 0  — conservation of energy
nodeI₁I₂I₃I₄I₁ + I₂ = I₃ + I₄
Charge cannot pile up at a node
SIGN RULES

Going round a loop

Element crossedChange in potential
Resistor along the current− IR (drop)
Resistor against the current+ IR (rise)
Cell from − to ++ E
Cell from + to −− E
METERS

Galvanometer → ammeter / voltmeter

Ammeter: S = IgGI − Ig  (shunt in parallel)
Voltmeter: R = VIg − G  (in series)

Ammeter: RA = GS/(G + S), low — ideal 0, joined in series. Voltmeter: RV = G + R, high — ideal ∞, joined in parallel. To raise a voltmeter range from V to nV add (n − 1)RV in series; a larger range means a larger resistance. A real voltmeter draws current, so it reads less than the true p.d.

Ammeter: low S in parallelGSVoltmeter: high R in seriesGR
Conversions of a galvanometer
Practise PYQs on this topicKirchhoff's Laws & Measuring Instruments · 210 questions with solutions
08
CONCEPT

Wheatstone Bridge & Metre Bridge

In one line
A Wheatstone bridge is four resistors with a galvanometer across one diagonal. When the galvanometer shows no deflection, VB = VD and the bridge is balanced.
PQ = RS

At balance the cell and galvanometer can be interchanged without disturbing balance. The bridge is most sensitive when all four resistances are of the same order. A balanced bridge’s middle arm can be removed when finding equivalent resistance.

PQRSGABCDBalanced: P/Q = R/S, Ig = 0
Balanced when Ig = 0
METRE BRIDGE

Slide-wire form of the Wheatstone bridge

ACRSGJl100 − lS = R(100 − l)/l
R in the left gap, unknown S in the right gap
RS = l100 − l  ⇒  S = R(100 − l)l

The 1 m wire is of manganin/constantan (high ρ, low α). If R and S are interchanged, the null point moves from l to 100 − l. Specific resistance of the wire in a gap: ρ = Sπr²/L.

In a metre bridge, 15 Ω is in the left gap and 45 Ω in the right gap. Where is the null point?
STEP 1l/(100 − l) = 15/45 = 1/3STEP 2l = 25 cm from the left end (25 cm from the centre)
Practise PYQs on this topicWheatstone Bridge & Metre Bridge · 87 questions with solutions
09
CONCEPT

Potentiometer

In one line
When a steady current flows through a uniform wire, the p.d. across any length is proportional to that length.
V = Kl  ·  K = VL = IRL = IρA

K is the potential gradient (V m⁻¹). At the null point no current is drawn from the cell, so the potentiometer measures the true emf: E = Kl.

ABE₀RhEGJl
Primary (E₀, Rh) and secondary (E, G) circuits
USES

Three standard experiments

ExperimentResult
Compare emfs of two cellsE₁/E₂ = l₁/l₂
Sum and difference method(E₁ + E₂)/(E₁ − E₂) = l₁/l₂
Internal resistance (cell shunted by R)r = R(l₁/l₂ − 1)
Driving cell E₀ with series Rs, wire resistance RwK = E₀Rw/[(Rw + Rs + r₀)L]

Sensitivity is higher when K is smaller: use a longer wire or reduce the current (increase Rh). With a fixed p.d. across a longer wire, K falls and the null point moves further. No balance is possible if the p.d. across the whole wire is less than the emf being measured.

VoltmeterPotentiometer
High but finite resistance; draws currentInfinite resistance at balance; draws no current
Deflection method, reads less than true p.d.Null method, reads the true emf; not direct reading
Practise PYQs on this topicPotentiometer · 90 questions with solutions
10
CONCEPT

Electrical Power & Heating Effect

In one line
Electric power is the rate at which electrical energy is used; the energy appears as heat in a resistor (Joule heating).
P = VI = I²R = V²R
H = I²Rt = V²tR J = I²Rt4.2 cal

1 kWh (1 unit) = 1000 W × 3600 s = 3.6 × 10⁶ J. Units used = (watt × hour)/1000. 1 HP = 746 W.

Series: same IP₁P₂lower-watt bulb glows brighterParallel: same Vhigher-watt bulb glows brighter
Bulbs in series and in parallel
RATED APPLIANCES

Same R, different supply

R = Vrated²Prated
P = Prated(VVrated)²
Series (same I)Parallel (same V)
Total power of bulbs1/P = 1/P₁ + 1/P₂ + …P = P₁ + P₂ + …
Brighter bulblower wattage (higher R)higher wattage
Two heaters boiling watert = t₁ + t₂t = t₁t₂/(t₁ + t₂)
Two identical heaters, same VHparallel : Hseries = 4 : 1

Fuse wire: high resistivity and low melting point (alloy of lead and tin), joined in series. Heater element: high resistivity, high melting point (nichrome). Small change: ΔP/P = 2ΔV/V. Chemical effect: in electrolysis the mass deposited m = ZIt (Faraday’s first law).

Practise PYQs on this topicElectrical Power & Heating Effect · 102 questions with solutions
EXAM-DAY CARD

Just see this before the exam

Current ElectricityI = neAvdJ = σER = ρl/AR = R₀(1 + αΔT)V = E − IrPmax = E²/4rΣI = 0, ΣΔV = 0KirchhoffP/Q = R/Sbridge balanceE = Klpotentiometer
FORMULA SHEET

Whole chapter on one page

Current

I = q/t = dq/dt

vd = eEτ/m

I = neAvd, μ = vd/E

J = I/A = σE

Resistance

V = IR, R = ρl/A

σ = ne²τ/m = 1/ρ

Stretch: R ∝ l² ∝ 1/r⁴

R = R₀(1 + αΔT)

Combinations

Rs = R₁ + R₂ + …

1/Rp = 1/R₁ + 1/R₂ + …

I₁ = IR₂/(R₁ + R₂)

Circle, diameter: R/4

Cells

I = E/(R + r), V = E − Ir

r = (E/V − 1)R

Series nE/(R + nr)

Pmax at R = r

Meters & bridges

S = IgG/(I − Ig)

R = V/Ig − G

P/Q = R/S

S = R(100 − l)/l

Potentiometer & power

K = V/L, E = Kl

r = R(l₁/l₂ − 1)

P = VI = I²R = V²/R

1 kWh = 3.6 × 10⁶ J

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Stretched wire: R ∝ l

Using E for the terminal voltage

Metre bridge: forgetting 100 − l

Bulbs in series: high watt glows more

CORRECT ✓

Volume fixed ⇒ R ∝ l²

V = E − Ir when current flows

R/S = l/(100 − l)

Lower watt (higher R) glows more