Current Electricity — NEET notes

Electric Current

Electric current is the rate of flow of electric charge through a conductor. Electrons drift slowly in response to an electric field, but the effect travels at nearly the speed of light — exactly like a long pipe full of water.

I = dQ dt

Electric Current

Current is not electrons zipping through a wire at light speed. It is the net charge crossing a cross-section per second — and the direction matters.

Electric current I through a surface is defined as the net charge dQ flowing through it in time dt . If the charge flow is steady, I = Q/t ; for varying flow, we use the instantaneous form.

I = dQ dt (NCERT Eq. 3.1)

• SI unit of current → Ampere (A) = 1 C/s
• Current is a scalar quantity (it has no direction in 3D space)
• The direction of conventional current is the direction of positive charge flow
• Electrons flow from negative to positive terminal, but conventional current flows from positive to negative
• A 1 A current means 6.25 10^ 18 electrons cross the section every second

Drift Velocity

Free electrons in a metal move chaotically at ~10⁶ m/s in all directions. An electric field adds a tiny bias — a 'drift' — that is the real engine of current.

Without an electric field, free electrons collide with lattice ions every seconds (called the relaxation time or mean free time) and move randomly — net displacement is zero. When a field E is applied, each electron picks up a velocity kick eE m in the direction opposite to E before the next collision resets it.

v_d = eE m (NCERT Eq. 3.16)

Here e is the electron charge ( 1.6 10^ -19 C), m is electron mass ( 9.1 10^ -31 kg), E is the applied electric field, and is the average time between collisions (~ 10^ -14 s in copper).

I = nAev_d (NCERT Eq. 3.18)

• n = free electron number density (electrons per m³) — a property of the material
• A = cross-sectional area of the wire
• e = magnitude of electron charge
• v_d = drift velocity
• This equation directly links microscopic drift to macroscopic current — the most important relation in this chapter

Quick recall

If the cross-sectional area of a wire is doubled but current stays the same, what happens to drift velocity?

From I = nAev_d : v_d = I/(nAe) . If A doubles, v_d halves. Thicker wire → same current carried by slower-drifting electrons.

Current Density

Current density J zooms in on 'how crowded' the current flow is at any point inside the conductor — unlike I, it is a vector.

While current I is a bulk property (total charge through a wire), current density J describes flow at every point inside the conductor. It is the amount of current per unit cross-sectional area, directed along the current flow.

J = I A J = nev_d E

Current density is connected to the electric field through conductivity . This is the microscopic (field-level) form of Ohm's Law, valid at every point inside the material.

J = E (NCERT Eq. 3.23, microscopic Ohm's Law)

• J is a vector — direction same as conventional current (opposite to electron drift)
• SI unit of J → A/m²
• SI unit of (conductivity) → S/m or ^ -1 m^ -1
• = ne^2 m — conductivity depends only on material properties, not geometry
• Resistivity = 1/ (they are exact reciprocals)

Mobility

Mobility μ is a single number that tells you how productive each unit of electric field is at generating drift velocity.

Mobility is defined as the magnitude of drift velocity per unit electric field. A high-mobility material produces a large drift velocity even with a small applied field — it's a 'slippery' conductor.

= |v_d| E = e m (NCERT Eq. 3.20)

• SI unit of → m²/(V·s) or equivalently m²V⁻¹s⁻¹
• Mobility is always positive (it's a magnitude ratio)
• Conductivity relates to mobility: = ne
• Higher temperature → more lattice vibrations → smaller → smaller → metals become worse conductors at high temp
• Semiconductors have much lower n but can have higher than metals at room temperature

= ne = ne

Quick recall

If temperature increases, what happens to mobility of a metallic conductor?

Temperature ↑ → lattice vibrations ↑ → collision frequency ↑ → ↓ → = e /m decreases. Metal becomes a poorer conductor at higher temperature.

Ohm's Law (Microscopic Form)

The familiar V = IR is just the macroscopic shadow of the deeper law J = σE — a material property that holds point-by-point inside any conductor.

The microscopic Ohm's Law J = E states that at every point in a conductor, current density is proportional to electric field. This is a statement about the material itself — is the proportionality constant called electrical conductivity.

J = E E = J (NCERT Eq. 3.23)

For a wire of length L and area A : substituting E = V/L and J = I/A into J = E gives V = IR with R = L/A . So V = IR is a consequence of J = E for uniform geometry.

R = L A = 1 L A (NCERT Eq. 3.26)

• Conductivity = ne^2 m — depends on electron density and relaxation time
• Resistivity = m ne^2 — increases with temperature for metals (more collisions)
• Ohmic material: J E graph is a straight line through origin
• Non-ohmic materials (diode, electrolytes, thyristors): J – E relation is nonlinear
• V = IR is macroscopic Ohm's Law; J = E is microscopic — NCERT treats both

Quick recall

A copper wire is stretched to double its length (same volume). How does resistance change?

Volume constant → AL = A'L' → A' = A/2 . New R' = (2L)/(A/2) = 4 L/A = 4R . Resistance becomes 4 times. Classic NEET trap!

This sub-concept is the microscopic root of the ENTIRE Current Electricity chapter. Drift velocity → J = nev_d → J = E → V = IR . Understand this chain and everything else follows logically.

Ohm's Law

At constant temperature, the current through a conductor is directly proportional to the potential difference across it — V = IR. This linear relationship holds for ohmic conductors but breaks down for devices like diodes and filament bulbs.

V = IR

Ohm's Law

Georg Ohm's 1827 discovery: for many materials at constant temperature, doubling the voltage exactly doubles the current. Simple, powerful, and the foundation of all circuit analysis.

Ohm's Law states that the potential difference V across a conductor is directly proportional to the current I through it, provided physical conditions (mainly temperature) remain constant. The proportionality constant is called resistance R .

V = IR R = V I (unit: ohm, )

• Resistance R is the property of the conductor that opposes current — its value depends on material, length, area, and temperature
• SI unit of resistance: ohm ( ), named after Georg Simon Ohm (1789–1854)
• Ohm's Law holds when temperature and other physical conditions are constant
• 1 = 1 V/A — one ohm is the resistance that allows exactly 1 A when 1 V is applied

Quick recall

A bulb has resistance 100 Ω and is connected to a 220 V supply. What is the current through it?

I = V/R = 220/100 = 2.2 A . Straightforward Ohm's Law — this is a NEET-level numerical.

V-I Characteristics

Plot voltage on the Y-axis and current on the X-axis: an ohmic conductor gives a perfect straight line through the origin; non-ohmic devices give curves that reveal their interesting (and useful) physics.

A V-I characteristic is a graph of voltage V (Y-axis) versus current I (X-axis) for a device. The slope of this graph equals resistance: R = V/I = slope . A straight line through the origin = constant resistance = ohmic conductor.

• Nichrome wire at room temperature: nearly perfect ohmic conductor — used in heaters and lab resistors
• p-n junction diode: almost zero current in reverse bias; exponential rise in forward bias — classic non-ohmic
• Filament bulb: resistance increases as it heats up (tungsten's R rises with temperature) — slope of V-I increases
• For any device at any point on its V-I curve, R = V/I still gives the static resistance at that point

Limitations of Ohm's Law

Ohm's Law is not a universal law — it's an empirical approximation that holds for a class of materials under specific conditions. NCERT explicitly lists three situations where it fails.

Ohm's Law ( V I ) is not a fundamental law of nature like Newton's Laws — it's an experimental rule that many (but not all) materials follow. NCERT Class 12 Chapter 3 explicitly states three breakdowns.

• V-I relationship is non-linear: In diodes and transistors, I does not increase proportionally with V — the graph curves. Resistance is not constant
• Direction matters (polarity-dependent): In a p-n junction diode, the current is very different for the same magnitude of forward vs reverse voltage — I depends on the sign of V , not just its magnitude
• Relation is not unique: In devices like GaAs (gallium arsenide) operated in certain conditions, the same voltage can correspond to two or three different currents — the V-I curve doubles back on itself

NCERT explicitly says: 'Ohm's law is not a fundamental law. It is obeyed by many substances, and these are called ohmic conductors.' This one-liner has appeared almost verbatim in NEET assertion-reason questions.

Resistance & Resistivity

Resistance is a conductor's opposition to current flow (R = V/I); resistivity is an intrinsic material property (R = ρL/A). Temperature affects both — metals and semiconductors behave opposite to each other. A NEET essential.

R = L A

Resistance

Defined by Ohm's law as R = V/I. Originates from collisions between free electrons and the lattice ions of the conductor.

When a potential difference V is applied across a conductor and current I flows, the conductor's resistance R is defined as the ratio of voltage to current. This is Ohm's law in its simplest form.

R = V I (SI unit: ohm, = V/A)

At the atomic level, resistance arises because free electrons drift through a lattice of positive ions. They constantly collide with these ions (especially at imperfections and thermal vibrations), losing kinetic energy with each collision — this is what resists the flow.

• R = V/I — valid for ohmic conductors where R is constant
• Unit: ohm (Ω), named after Georg Simon Ohm
• Resistance is a property of a specific conductor — depends on its material, geometry, and temperature
• For a non-ohmic device (like a diode), R = V/I still gives resistance at that operating point — but R is not constant

Resistivity

R = ρL/A shows that resistance depends on length, cross-sectional area, and the material constant ρ (resistivity). Resistivity is a pure material property.

The resistance of a uniform conductor depends on three things: its length L, its cross-sectional area A, and what it's made of. The material factor is called resistivity ρ (rho).

R = L A

where is resistivity (SI unit: ·m)

• R L — double the length → double the resistance
• R 1 A — double the cross-section area → half the resistance
• Resistivity depends ONLY on the material and temperature — NOT on the wire's shape or size
• Unit of : ·m (ohm-metre)
• Copper: 1.7 10^ -8 ·m (excellent conductor). Glass: 10^ 10 ·m (insulator)

Quick recall

A wire has resistance R. It is bent into a circle. What is the resistance between the two ends of the diameter?

The wire is split into two equal halves (each R/2), connected in parallel. Equivalent resistance = (R/2 × R/2) / (R/2 + R/2) = R/4.

Temperature Dependence

For metals, resistivity increases with temperature (α > 0). For semiconductors and insulators, resistivity decreases with temperature (α < 0). A NEET favourite distinction.

Resistivity changes with temperature. For metals, the relationship is approximately linear over a moderate temperature range:

= _0 (1 + T )

where _0 = resistivity at reference temperature T_0, = temperature coefficient of resistivity

• Metals: > 0 → resistivity increases with temperature (more vibrations, more collisions)
• Semiconductors/insulators: < 0 → resistivity decreases with temperature (more charge carriers liberated)
• Alloys (Nichrome, Manganin): 0 → nearly temperature-independent resistivity — used in standard resistors and heating elements
• Superconductors: resistivity drops to exactly zero below a critical temperature T_c

Resistor Color Code

A 4-band resistor encodes its resistance value in colored bands. The mnemonic 'BB ROY of Great Britain has Very Good Wife' maps colors to digits 0–9.

Carbon resistors (the small cylindrical ones) have 4 colored bands printed on them. Reading left to right: Band 1 = first significant digit, Band 2 = second significant digit, Band 3 = multiplier (power of 10), Band 4 = tolerance (gold/silver).

Multiplier (Band 3): Same color → same digit → multiply by 10^ digit . Gold = 0.1 , Silver = 0.01 . Tolerance (Band 4): Gold = ±5%, Silver = ±10%, No band = ±20%.

• 4-band code: [Digit 1] [Digit 2] [Multiplier] [Tolerance]
• Gold tolerance band is always on the right — use this to identify which end to read from
• Multiplier for Red = ×100, Orange = ×1000, Yellow = ×10,000
• Gold multiplier (×0.1) and Silver multiplier (×0.01) are used for resistances below 10 Ω

Quick recall

Bands: Orange – Orange – Brown – Gold. What is the resistance?

Orange=3, Orange=3, Brown=×10, Gold=±5% → 33 × 10 = 330 Ω ± 5%.

Conductivity & Conductance

Conductance G = 1/R (unit: siemens) and conductivity σ = 1/ρ (unit: S/m) are reciprocals. They describe how easily current flows rather than how much it is opposed.

Everything about resistance has a mirror-image quantity that measures how easily current flows, rather than how much it is blocked.

G = 1 R (Conductance, unit: siemens S = ^ -1 )

= 1 (Conductivity, unit: S/m or ^ -1 m^ -1 )

• Conductance G = 1/R , unit: siemens (S), also written as mho (℧)
• Conductivity = 1/ , unit: S/m (siemens per metre)
• J = E — current density is proportional to electric field via conductivity
• Metals: 10^7 S/m (very high). Semiconductors: 10^ -4 to 10^4 S/m . Insulators: < 10^ -10 S/m (extremely low)
• Mobility = v_d E — drift velocity per unit electric field, unit: m²/(V·s)

In NEET, conductance and conductivity questions are less frequent than resistance/resistivity questions — but they appear every 2–3 years, usually in a 1-mark definition or unit-identification question. Worth 30 seconds to memorize.

Electrical Energy & Power

Power is the rate of doing electrical work. It dissipates as heat in resistors (Joule heating), gets spent inside the cell (internal loss), and is measured commercially in kWh — the unit on your electricity bill.

P = VI = I^2R = V^2 R

Electrical Power

Power is energy per unit time. In a circuit, it equals the product of voltage and current — with three equivalent forms depending on what's known.

When a charge dq moves through a potential difference V in time dt , the work done is dW = V dq . Power is P = dW dt = V dq dt = VI . This is the most general form — valid for any circuit element.

P = VI = I^2R = V^2 R (unit: watt, W)

• 1 watt = 1 joule per second = 1 J/s
• Power depends on the square of current (or voltage) — doubling I quadruples power loss
• P = I^2R → higher resistance means more heat at the same current
• P = V^2/R → higher resistance means less heat at the same voltage (opposite!)

Joule Heating

Electrical energy dissipated in a resistor appears entirely as heat. This is Joule's law — the basis of heaters, toasters, and why wires get warm.

When current flows through a resistor, electrical energy is irreversibly converted to heat. The total energy (heat) produced in time t is just power multiplied by time.

W = Pt = VIt = I^2Rt = V^2 R t (unit: joule, J)

Joule's law: The heat produced in a conductor is proportional to (1) the square of the current, (2) the resistance, and (3) the time for which current flows. Double the current → four times the heat.

• Joule heating is always positive and irreversible — energy flows heat → surroundings, never back
• This is why long-distance power lines use very HIGH voltage (low current → minimal I^2R loss)
• Electric heater, iron, geyser, fuse — all work on Joule heating principle
• 1 calorie = 4.18 J (useful when questions ask heat produced in calories)

Power from a Cell

A cell converts chemical energy to electrical energy at rate εI. Part is delivered to the external circuit, part is lost internally.

A cell of EMF and internal resistance r connected to external resistance R drives current I = R+r . The cell converts chemical energy to electrical energy at total rate P_ cell = I .

P_ total = I (total power supplied by cell)

P_ external = I^2R = I - I^2r (useful power to external circuit)

P_ internal = I^2r (power wasted inside cell as heat)

• Total power by cell: P = I (source of all energy in the circuit)
• Internal loss: P_r = I^2 r — this is why old batteries feel warm
• Useful (external) power: P_R = I - I^2 r
• Efficiency = P_ ext P_ total = R R+r — higher R → better efficiency

Maximum Power Transfer

There's a trade-off: to get maximum power from a source you match the load to the internal resistance — but exactly half the power is then wasted inside.

External power is P = I^2 R = ^2 R (R+r)^2 . To find the maximum, differentiate with respect to R and set to zero. The result is elegant: P is maximum when R = r .

P_ max = ^2 4r when R = r

• Maximum power condition: R = r (load = internal resistance)
• Maximum power: P_ max = ^2 4r
• Efficiency at max power = 50% (always — a fundamental result)
• For maximum efficiency (not max power): use R r
• At R = r : terminal voltage = 2 (half the EMF appears across load)

Quick recall

A cell has EMF 12V and internal resistance 2Ω. What is the maximum power it can deliver to an external load?

Maximum power when R = r = 2 . P_ max = ^2 4r = 144 8 = 18 W .

Practical Applications

Real-world electrical power: how your bill is calculated, what a fuse protects against, and why bulb brightness depends on circuit configuration.

The commercial unit of electrical energy is the kilowatt-hour (kWh), not the joule. 1 kWh is the energy consumed by a 1 kW appliance in 1 hour.

1 kWh = 1000 W 3600 s = 3.6 10^6 J = 3.6 MJ

A fuse is a thin wire with a low melting point placed in series. If current exceeds the rated value, I^2 R heating melts it — breaking the circuit before wires overheat and start a fire.

• 1 kWh = 3.6 MJ (memorize this — direct NEET MCQ)
• Higher wattage bulb has LOWER resistance (it draws more current at 220V)
• In parallel (same V): brighter = higher wattage = lower R
• In series (same I): brighter = higher resistance = LOWER wattage (flip!)
• Fuse protects against overcurrent via Joule heating — always in series

Quick recall

A household uses a 2kW AC for 6 hours and a 100W TV for 8 hours daily. What is the daily energy consumption in kWh?

AC: 2 6 = 12 kWh. TV: 0.1 8 = 0.8 kWh. Total = 12.8 kWh per day.

The series vs parallel bulb question appears in NEET almost every alternate year. The key is: identify what stays constant (V in parallel, I in series), then choose the matching P formula.

Combinations of Resistors

Resistors can be joined in series (one after another, same current) or parallel (side by side, same voltage). Every real circuit is some mix of the two — learn to identify and simplify step by step.

R_ series = R_1 + R_2 + 1 R_ parallel = 1 R_1 + 1 R_2 +

Series Combination

Resistors in series behave like a single longer resistor. The equivalent resistance is just the arithmetic sum — and the voltage splits proportionally.

In series, resistors are connected end-to-end so there is only a single conducting path. The same current I passes through every element. The total voltage V is shared across them.

V = V_1 + V_2 + V_3 = IR_1 + IR_2 + IR_3 = I(R_1 + R_2 + R_3)

R_ eq = R_1 + R_2 + R_3 + + R_n

• Current is the same through all resistors: I_1 = I_2 = I_3 = I
• Voltage adds up: V = V_1 + V_2 +
• R_ eq is always greater than the largest individual resistor
• Adding more resistors in series → total resistance increases, current decreases

Voltage Divider: In series, each resistor 'grabs' a fraction of the total voltage proportional to its resistance. This is the voltage divider rule — a fundamental tool in electronics.

V_1 = V R_1 R_1 + R_2 , V_2 = V R_2 R_1 + R_2

Parallel Combination

Resistors in parallel share the same two terminals. Current splits among them. The equivalent resistance is always smaller than the smallest individual resistor.

In parallel, all resistors share the same two terminals, so the same voltage V appears across each. The total current I from the source splits into branches — each branch carries current I_k = V/R_k .

I = I_1 + I_2 + I_3 = V R_1 + V R_2 + V R_3 = V ( 1 R_1 + 1 R_2 + 1 R_3 )

1 R_ eq = 1 R_1 + 1 R_2 + 1 R_3 + + 1 R_n

• Voltage is the same across all resistors: V_1 = V_2 = V_3 = V
• Current adds up: I = I_1 + I_2 +
• R_ eq is always less than the smallest individual resistor
• Adding more resistors in parallel → total resistance decreases, total current increases

Current Divider: For two resistors in parallel, each branch carries current inversely proportional to its resistance. The smaller resistor carries more current (easier path).

I_1 = I R_2 R_1 + R_2 , I_2 = I R_1 R_1 + R_2

Quick recall

A 6 Ω and 3 Ω resistor are in parallel. Total current = 6 A. How much current flows through 6 Ω?

I_ 6 = 6 3 6+3 = 6 1 3 = 2 A . The smaller 3 Ω branch carries 4 A — more current takes the easier path.

Series vs Parallel

The most tested comparison in Current Electricity. Nail every row.

Mixed Circuits

Real circuits combine series and parallel sections. The trick is to reduce the circuit from the innermost group outward, one step at a time.

Most NEET circuit problems are neither pure series nor pure parallel — they mix both. The method is always the same: identify the innermost independent group, replace it with its equivalent, and repeat until one resistor remains.

• Step 1: Redraw the circuit clearly (label nodes A, B, C…)
• Step 2: Find resistors that are definitely in series (carry the same current — no junction between them)
• Step 3: Find resistors that are definitely in parallel (share the same two nodes)
• Step 4: Replace each group with its equivalent R_ eq
• Step 5: Repeat until a single equivalent resistance remains

Example: Three resistors — R_1 = 2 in series with the parallel combination of R_2 = 6 and R_3 = 3 , across a 12 V battery.

Step 1: R_ 23 = R_2 R_3 R_2 + R_3 = 6 3 6 + 3 = 18 9 = 2

Step 2: R_ eq = R_1 + R_ 23 = 2 + 2 = 4

Step 3: I_ total = V R_ eq = 12 4 = 3 A

Step 4: V_ 23 = I_ total R_ 23 = 3 2 = 6 V

I_2 = 6 6 = 1 A, I_3 = 6 3 = 2 A (check: 1+2 = 3 A )

Special Cases

Four patterns appear so often in NEET that you should have the answer in 5 seconds without writing a single step.

Case 1: Two equal resistors in parallel. When R_1 = R_2 = R :

R_ eq = R R R + R = R^2 2R = R 2

Case 2: n identical resistors in series.

R_ eq = nR

Case 3: Two unequal resistors in parallel (product over sum). Only works for exactly two resistors:

R_ eq = R_1 R_2 R_1 + R_2

Case 4: Infinite resistor ladder. A semi-infinite chain of resistors R in series and R in parallel (repeating). Let the equivalent resistance be x . Because the ladder is infinite, adding one more unit doesn't change x :

x = R + Rx R + x

x^2 - Rx - R^2 = 0 x = R(1 + 5 ) 2 1.618 R

• n resistors R in series: R_ eq = nR
• n resistors R in parallel: R_ eq = R/n
• Two resistors R_1, R_2 in parallel: R_ eq = R_1R_2/(R_1+R_2) — product over sum
• Infinite ladder (series R + parallel R ): R_ eq = R where = (1+ 5 )/2
• Short circuit across a resistor: R_ eq = 0 (that resistor is bypassed completely)
• Open circuit (broken wire) in series: R_ eq , current = 0

Quick recall

You have 6 identical 12 Ω resistors. How do you connect them to get 4 Ω equivalent resistance?

Connect 3 of the 6 resistors in parallel: R_ eq = 12/3 = 4 . Only 3 resistors are needed; the remaining 3 are unused. No arrangement using all 6 identical 12 Ω resistors gives exactly 4 Ω.

NCERT Exercise 3.7 and 3.8 are based on series-parallel combinations — solve them fully. Questions from these exercises appeared verbatim in NEET 2018 and 2022.

Cells, EMF & Internal Resistance

Every real battery has an internal resistance that causes the terminal voltage to drop under load. EMF is the work done per unit charge by non-electrostatic forces inside the cell — not an actual force.

V = - Ir

EMF of a Cell

EMF (electromotive force) is a misnomer — it is not a force. It is the energy supplied per coulomb by the cell's internal chemistry to drive charge around the circuit.

Inside a cell, chemical reactions create a non-electrostatic force (sometimes called the 'chemical force') that pushes positive charges from the negative terminal to the positive terminal — against the electric field. The work done by this chemical force per unit positive charge moved from the negative to positive terminal (through the electrolyte) is defined as the EMF of the cell.

= W q

where W = work done by non-electrostatic (chemical) force to move charge q from negative to positive terminal inside the cell. SI unit: volt (V) = joule per coulomb.

• EMF stands for electromotive force but it is NOT a force — it is energy per unit charge (unit: volt)
• EMF is produced by non-electrostatic forces: chemical (batteries), magnetic (generators), thermal (thermocouples)
• EMF = potential difference across terminals only when no current flows (open circuit)
• Symbol: (epsilon) or sometimes E

Internal Resistance

The electrolyte and electrodes inside a cell have resistance. This internal resistance r causes energy loss inside the cell itself, reducing the voltage available to the external circuit.

A real cell is modelled as an ideal EMF source in series with a small resistor r — the internal resistance. This r is inside the cell; you cannot remove it. When current I flows, the voltage drop Ir occurs inside the cell, and only the remaining voltage - Ir appears across the external terminals.

• Internal resistance r depends on: electrolyte concentration (higher concentration → lower r ), electrode separation (larger separation → higher r ), and electrode area (larger area → lower r )
• For an ideal cell: r = 0 , so V = always
• As a cell ages or discharges, electrolyte weakens → r increases → terminal voltage drops more under load
• Unit of r : ohm ( ) — same as any resistance

NCERT explicitly lists three factors affecting r : nature and concentration of electrolyte, separation between electrodes, area of electrodes. These are common 1-mark NEET questions.

Terminal Voltage

Terminal voltage is the potential difference across the external terminals of the cell. It is less than EMF during discharge and more than EMF during charging.

Apply Kirchhoff's voltage law around the full circuit (including the cell's inside). Going from negative terminal to positive terminal through the cell: the chemical EMF acts as a source, but the internal resistance r causes a drop Ir . The net terminal voltage across A and B is:

V = - Ir (discharging — cell delivering current)

V = + Ir (charging — current forced into cell from outside)

• Open circuit ( I = 0 ): V = — voltmeter reads full EMF
• Discharging ( I > 0 ): V = - Ir < — terminal voltage drops
• Charging ( I < 0 by convention, i.e., current into positive terminal): V = + Ir > — terminal voltage rises
• The V - I graph is a straight line: slope = -r , y-intercept =

Quick recall

A cell of EMF 2 V and internal resistance 0.5 Ω delivers 2 A. What is the terminal voltage?

V = - Ir = 2 - (2)(0.5) = 2 - 1 = 1 V. Half the EMF is lost inside the cell!

Maximum Current

When external resistance R → 0 (short circuit), current reaches its maximum value ε/r. All power is dissipated internally — nothing reaches the load.

For a cell with EMF , internal resistance r , and external resistance R , the current in the circuit is:

I = R + r

As R increases, I decreases. As R decreases toward zero (short circuit), I approaches its maximum value:

I_ = r (when R = 0 , short circuit)

• General current: I = R + r — both R and r limit the current
• Maximum current occurs at short circuit ( R = 0 ): I_ = r
• At short circuit: terminal voltage V = 0 (all EMF dropped across r )
• Open circuit ( R ): I 0 , terminal voltage V =
• Power delivered to external R : P = I^2 R = ^2 R (R+r)^2

Quick recall

A cell of EMF 6 V and internal resistance 2 Ω is short-circuited. Find (a) current, (b) terminal voltage.

(a) I_ = /r = 6/2 = 3 A. (b) V = - I_ r = 6 - 3 2 = 0 V. Terminal voltage is zero — all 6 V dropped across internal resistance.

Power to external load: P_ ext = I^2 R = ^2 R (R+r)^2

Cells in Series & Parallel

Multiple cells can be combined in series (voltages add, internal resistances add) or parallel (voltage stays same, internal resistance divides) to match the needs of a circuit. Mixed grouping optimises current delivery. When cells oppose each other, net EMF subtracts.

I_ max = mn mR + nr (mixed grouping: m rows, n cells each)

Cells in Series

Cells connected end-to-end (positive terminal of one touching negative of the next) combine their EMFs and their internal resistances both add up.

For n identical cells (each EMF , internal resistance r ) in series driving current through external resistance R :

_ eq = n r_ eq = nr

I = n R + nr

For non-identical cells (EMFs _1, _2, and internal resistances r_1, r_2, ), simply sum them individually, provided all cells are correctly oriented (aiding).

_ eq = _1 + _2 + + _n r_ eq = r_1 + r_2 + + r_n

• Net EMF increases with each cell added in series — great for high-voltage needs
• Net internal resistance also increases — a large r_ eq drops terminal voltage under load
• Series combination is most useful when R nr (external resistance dominates internal)
• If one cell is reversed, its EMF subtracts: _ eq = _1 - _2 (for that cell)

Quick recall

4 cells, each of EMF 2 V and internal resistance 0.5 Ω, are in series with R = 2 Ω. Find current.

_ eq = 4 2 = 8 V, r_ eq = 4 0.5 = 2 Ω. I = 8 2 + 2 = 2 A.

Cells in Parallel

Cells connected with all positive terminals joined together and all negative terminals joined together maintain the same EMF as a single cell, but the effective internal resistance drops — allowing more current.

For n identical cells (EMF , internal resistance r each) in parallel:

_ eq = r_ eq = r n

I = R + r/n

For n non-identical cells in parallel (general formula using reciprocals):

1 r_ eq = 1 r_1 + 1 r_2 + + 1 r_n

_ eq = r_ eq ( _1 r_1 + _2 r_2 + + _n r_n )

• Net EMF does not increase in parallel — it stays equal to one cell's EMF (for identical cells)
• Net internal resistance decreases — each added cell provides another current path
• Parallel combination is most useful when R r (small external load, like a motor starter)
• All cells must have the same polarity when connected in parallel — reversed cell causes circulating current and wastes energy

Quick recall

6 identical cells, EMF 1.5 V each, r = 3 Ω each, in parallel with R = 0.5 Ω. Find current.

r_ eq = 3/6 = 0.5 Ω, _ eq = 1.5 V. I = 1.5/(0.5 + 0.5) = 1.5 A.

Mixed Grouping

When neither pure series nor pure parallel is optimal, cells are arranged in m rows with n cells per row. The optimal grouping condition for maximum current delivery has a beautiful symmetry.

Suppose you have mn identical cells (each EMF , internal resistance r ), arranged in m parallel rows, each row containing n cells in series.

_ eq = n r_ eq = nr m

I = n R + nr/m = mn mR + nr

To find the value of n (or m ) that gives maximum current, treat I as a function of n and minimise the denominator mR + nr subject to fixed total cells. Applying AM-GM inequality:

I_ max when mR = nr R = nr m = r_ eq

• Condition mR = nr means R_ ext = r_ eq — external equals internal
• If R is large, use more cells in series (increase n ) to match
• If R is small, use more rows in parallel (increase m ) to reduce r_ eq
• Maximum current: I_ max = mn 2nr = m 2r (substituting mR = nr )

Quick recall

24 cells, ε = 2 V, r = 1 Ω each. External R = 2 Ω. Find the best grouping (m rows, n per row) and max current.

Total cells: mn = 24 . Condition: mR = nr → 2m = n . With mn = 24 : m 2m = 24 → m^2 = 12 (not integer). Take m = 3, n = 8 : check mR = 3 2 = 6 , nr = 8 1 = 8 (close but not exact). Best practical: m = 4, n = 6 : mR = 8 , nr = 6 . Or try m = 2, n = 12 : I = (2 12 2)/(2 2 + 12 1) = 48/16 = 3 A. Try m = 4, n = 6 : I = 48/(8+6) = 3.43 A. Try m = 6, n = 4 : I = 48/(12+4) = 3 A. Maximum is at m = 4, n = 6 I 3.43 A.

Mixed grouping is a NEET favourite every 2–3 years. The formula for I_ max and the condition mR = nr are must-memorise.

Cells in Opposition

If two cells are connected with the same polarity facing each other (positive-to-positive), their EMFs oppose. The stronger cell drives current through the weaker one, potentially charging it.

When two cells of EMFs _1 > _2 are connected in opposition (positive terminals facing each other), the net EMF drives current in the direction of the stronger cell.

_ net = _1 - _2 r_ net = r_1 + r_2

I = _1 - _2 R + r_1 + r_2

• Net EMF = difference of the two EMFs (subtracts, not adds)
• Net internal resistance = sum of both (still adds, not subtracts)
• Current flows from higher EMF cell; the weaker cell is being charged
• If _1 = _2 , net EMF = 0 but current still circulates if r_1 r_2 (circulating current = 0 since no driving force — actually I = 0 for _1 = _2 )

Quick recall

Two cells: ε₁ = 6 V, r₁ = 1 Ω; ε₂ = 4 V, r₂ = 2 Ω, connected in opposition with R = 3 Ω. Find I and which cell is being charged.

_ net = 6 - 4 = 2 V. r_ eq = 1 + 2 = 3 Ω. I = 2/(3 + 3) = 1/3 A. Current flows in direction of ε₁ (6V cell). The 4V cell (ε₂) is being charged.

Series vs Parallel

A structured side-by-side comparison to lock in the differences and when to use each configuration.

• Series: EMF adds, r adds — use for HIGH external R (voltmeter-like loads)
• Parallel: EMF stays, r divides — use for LOW external R (ammeter-like loads, motors)
• Mixed: compromise — optimize by choosing m and n such that mR = nr
• Opposition: EMF subtracts, r still adds — appears in charging circuits and Kirchhoff problems

This comparison table alone is worth 2–3 NEET marks per year. The 'best when' conditions (R ≫ nr vs R ≪ r) appear as direct MCQs more often than numerical problems.

Kirchhoff's Laws

Two powerful rules — one for junctions (charge conservation) and one for loops (energy conservation) — that let you crack any circuit that Ohm's Law alone cannot handle.

I_ in = I_ out V = 0 (around any loop)

Junction Rule (KCL)

At any junction in a circuit, the total current flowing in equals the total current flowing out. Charge cannot pile up at a node.

A junction (or node) is any point in a circuit where two or more wires meet. Kirchhoff's Current Law (KCL) says: the sum of all currents entering a junction equals the sum of all currents leaving it. Think of it like a UPI transaction — every rupee that arrives must go somewhere. None disappears.

I_ in = I_ out

Equivalently, if we define currents flowing into a junction as positive and currents flowing out as negative, the algebraic sum of all currents at the junction is zero:

_ j I_j = 0 (at any junction)

• KCL is based on conservation of charge — charge doesn't pile up at a node
• Valid for any junction, at any instant, for DC and AC circuits
• If you have n junctions, you get at most n-1 independent KCL equations
• Current directions are assumed initially — a negative answer just means your arrow was backwards

Loop Rule (KVL)

The algebraic sum of all potential differences (EMFs and voltage drops) around any closed loop in a circuit is zero.

Pick any closed loop in a circuit. Start at any point, walk around the loop in any direction, add up every rise and fall in voltage. When you get back to where you started, the total must be zero — because voltage is a state function (like altitude). If you hike around a mountain and end up where you began, your net elevation change is zero.

_ loop V = 0

This can be rewritten as: the sum of all EMFs in a loop equals the sum of all resistive voltage drops in that loop.

= IR

• KVL is based on conservation of energy — energy supplied by EMFs = energy dissipated in resistors
• The loop can be traversed in any direction (clockwise or anticlockwise) — but stick to your chosen direction throughout
• Works for any loop: a real loop in the circuit, or even an imaginary loop you draw
• If you have L independent loops, you get L independent KVL equations

Sign Conventions

Getting signs wrong is the #1 reason students lose marks on Kirchhoff's problems. Two simple rules cover every case.

Sign convention in KVL: as you traverse a loop, every element either adds voltage (rise, positive) or subtracts voltage (drop, negative). The rules below cover resistors and EMF sources — master these and you will never lose a sign again.

• For a resistor: -IR if traversal matches current direction; +IR if opposite
• For an EMF source: + if traversed from - to + ; - if from + to -
• Internal resistance r of a battery is handled exactly like a resistor — apply the same sign rule
• Once you fix a traversal direction, do NOT change it mid-loop

Step-by-Step Circuit Solving

A systematic method to set up and solve any circuit using Kirchhoff's Laws — from labeling unknowns to solving simultaneous equations.

Kirchhoff's Laws give you a system of linear equations. The method below guarantees you get enough equations to solve for all unknowns. Never skip a step — every step has a purpose.

1. Assign branch currents
2. Apply KCL at junctions
3. Apply KVL to independent loops
4. Solve the simultaneous equations

• ALWAYS assign ALL branch currents before writing any equation
• KCL equations reduce unknowns — use them first
• Choose loops that look simple (fewer elements = fewer terms to write)
• An outer loop that encompasses the entire circuit is always valid
• Double-check: the number of equations must equal the number of unknown currents

Worked Example

A classic NCERT-style problem (Example 3.3) with two batteries and a shared resistor, solved step by step using the 4-step method.

This is the circuit type NEET loves most. Two batteries and a common resistor form two loops sharing one branch. Let's solve it completely using mesh currents.

Step 1 — Assign mesh currents: Let I_1 flow clockwise in the left loop (through _1 , r_1 , and R ). Let I_2 flow clockwise in the right loop (through _2 , r_2 , and R ). The shared middle branch carries the net current.

Step 2 — KCL at the top junction: I_1 flows down through the middle branch (from Loop 1), I_2 flows up through it (from Loop 2). Net current through R :

I_3 = I_1 - I_2 (downward through R)

Step 3 — KVL for Loop 1 (traverse clockwise): Rise through _1 : +10 . Drop across r_1 (current I_1 in traversal direction): -I_1(1) . Drop across R (current I_3 = I_1 - I_2 in traversal direction): -(I_1 - I_2)(4) .

10 - I_1(1) - (I_1 - I_2)(4) = 0

10 - I_1 - 4I_1 + 4I_2 = 0 10 = 5I_1 - 4I_2 (1)

KVL for Loop 2 (traverse clockwise): Rise through _2 : +4 . Drop across r_2 (current I_2 in traversal direction): -I_2(1) . Drop across R — but traversal goes upward while I_3 = I_1 - I_2 is downward, so opposite direction → sign flips: -(I_2 - I_1)(4) .

4 - I_2(1) - (I_2 - I_1)(4) = 0

4 - I_2 - 4I_2 + 4I_1 = 0 4 = -4I_1 + 5I_2 (2)

Step 4 — Solve equations (1) and (2):

From (1): 10 = 5I_1 - 4I_2

From (2): 4 = -4I_1 + 5I_2

Multiply (1) by 4: 40 = 20I_1 - 16I_2

Multiply (2) by 5: 20 = -20I_1 + 25I_2

Add: 60 = 9I_2 I_2 = 20 3 6.67 A

From (1): 5I_1 = 10 + 4 20 3 = 10 + 80 3 = 110 3 I_1 = 22 3 7.33 A

I_3 = I_1 - I_2 = 22 3 - 20 3 = 2 3 0.67 A (downward)

Verification (substitute back): Loop 1: 10 - 22 3 (1) - 2 3 (4) = 10 - 22 3 - 8 3 = 10 - 30 3 = 10 - 10 = 0 ✓ Loop 2: 4 - 20 3 (1) - ( 20 3 - 22 3 )(4) = 4 - 20 3 + 8 3 = 4 - 12 3 = 4 - 4 = 0 ✓

• KCL first: I_3 = I_1 - I_2 reduces 3 unknowns to 2
• Two KVL loops → two equations → solve simultaneously
• Negative current = actual direction is opposite to assumed arrow (didn't happen here, but often does)
• Always verify: substitute solved values back into both KVL equations — both must give zero

Quick recall

In a Kirchhoff's circuit, you solve and get I_2 = -3 A. What does this mean?

The actual current in that branch is 3 A, flowing in the direction OPPOSITE to the arrow you drew in Step 1. The negative sign is information, not an error.

The worked example above follows the NCERT Example 3.3 structure (Chapter 3, Class 12). If you can replicate this solution without notes, you're ready for any Kirchhoff's question NEET can throw at you.

Wheatstone Bridge & Meter Bridge

The Wheatstone Bridge is a null-deflection circuit for finding unknown resistance; the Meter Bridge is just a Wheatstone Bridge built on a 1-metre wire — same formula, physical form.

P Q = R S (balance condition)

Wheatstone Bridge

Four resistors in a diamond (bridge) arrangement. When the galvanometer reads zero, we have the balance condition. That one condition is everything.

The Wheatstone Bridge has four resistors arranged in a diamond: P, Q, R, and S. A battery is connected across vertices A and C; a galvanometer (G) is connected across B and D. When the galvanometer shows zero deflection (null point), the bridge is said to be balanced.

P Q = R S PS = QR

If three resistors are known and one is unknown, adjust until I_G = 0 , then use the balance condition to find the unknown. Typically S is the unknown: S = QR P .

• Null-point method — galvanometer reads zero, NOT maximum
• At balance: P Q = R S — the ratio of adjacent arms are equal
• PS = QR — products of opposite arm pairs are equal (either form is fine)
• S (unknown) can be found as S = QR P
• The battery EMF and its internal resistance do NOT appear in the balance condition

Sensitivity of Wheatstone Bridge

Sensitivity measures how much the galvanometer deflects for a small change in resistance. Maximum sensitivity occurs when all four arms are equal.

A bridge is sensitive if a small change in the unknown resistance S causes a large deflection in the galvanometer. High sensitivity = you can detect small imbalances = more accurate measurement.

• Sensitivity is highest when all four arms are approximately equal in magnitude
• Using a more sensitive galvanometer (lower G ) increases sensitivity
• Using a higher EMF battery increases current and improves sensitivity — but risks overheating the galvanometer
• Very unequal arms (e.g., P = 1 , S = 10000 ) → low sensitivity

Meter Bridge

The Meter Bridge physically implements a Wheatstone Bridge using a 1-metre resistance wire. The balance condition translates into a length ratio.

The 1-metre resistance wire (usually nichrome, uniform cross-section) acts as the two middle arms P and Q of the Wheatstone Bridge. Since resistance is proportional to length ( R l ) and the cross-section is uniform: the arm ratio P/Q = l/(100-l) .

R S = l 100 - l

S = R 100 - l l

Here, l is the balancing length in cm — the length from end A to the jockey position J at which the galvanometer shows zero deflection. R is the known resistance in the left gap.

• l = distance from left end A to jockey J (in cm) at null point
• R is in the left gap, S (unknown) is in the right gap
• The 100 cm wire serves as P (length l ) and Q (length 100-l )
• Formula: S = R(100 - l)/l — find unknown S if R and l are known
• Re-arrange for R: R = Sl/(100-l) if S is known and R is unknown

Quick recall

In a meter bridge, null point is at l = 60 cm with R = 30 . Find S.

S = 30 100-60 60 = 30 40 60 = 20

End Corrections

The copper strips at both ends of the meter bridge wire add small resistances. These 'end errors' shift the effective null point and must be accounted for in precise work.

In a real meter bridge, the resistance wire is soldered to thick copper strips at both ends. These copper strips have small but non-negligible resistance. This means the effective zero of the wire is not exactly at the 0 cm mark — there's a small shift at both ends.

R S = l + (100 - l) +

Here, and are the end corrections (in cm) at the left and right ends respectively. These are determined experimentally by taking two readings with R and S interchanged.

• = left end correction (cm), = right end correction (cm)
• Both are always positive (they represent extra effective wire length)
• To minimize end correction: keep null point between 20 cm and 80 cm — away from both ends
• NCERT recommends taking the average of two readings (R and S swapped) to cancel end errors

Wheatstone vs Meter Bridge

Students often think these are two separate topics. They aren't. The Meter Bridge IS a Wheatstone Bridge — just physically implemented using a wire instead of discrete resistors.

P Q = R S _ Wheatstone l 100-l = R S _ Meter Bridge

• Meter Bridge IS a Wheatstone Bridge — not a separate concept
• The 100 cm wire provides a continuously variable ratio P:Q = l:(100-l)
• Both methods use null deflection — galvanometer reads zero at balance
• Both are independent of battery EMF and internal resistance at balance
• Meter bridge accuracy is limited by: wire uniformity, end corrections, jockey contact resistance

Quick recall

Why is the null-point method more accurate than direct measurement of current/voltage?

At null (zero current through galvanometer), the galvanometer resistance doesn't affect the circuit. No current → no voltage drop across G → no systematic error from galvanometer resistance. This is why null methods are highly precise.

NEET has asked both theory questions ('which condition gives null?') and numerical questions ('find S given l and R'). The comparison table above covers the theory MCQs; the quick-recall covers the numericals.

Potentiometer

A potentiometer measures EMF and internal resistance without drawing any current from the source — the gold standard of voltage measurement. Its principle: potential drop across a wire is proportional to length (V ∝ l). Three NEET-critical applications: comparing EMFs, finding internal resistance, and calibrating instruments.

_1 _2 = l_1 l_2

Principle

The entire potentiometer depends on one simple idea: a uniform wire carrying a steady current has a uniform potential gradient. Slide the jockey until the galvanometer reads zero — that's your null point.

A long uniform wire AB (typically 4–10 m, made of high-resistance alloy like manganin or nichrome) carries a steady current from a driver cell. Because the wire is uniform, resistance per unit length is constant. By Ohm's law, the potential drop per unit length is also constant — this is the potential gradient k .

k = V L = Total potential drop across wire Total length of wire (unit: V/m)

For any length l from end A, the potential at that point relative to A is exactly V = kl . This is V ∝ l — the whole principle in two symbols.

V = kl V l (at constant k)

• Uniform wire → uniform potential gradient k = V/L
• Potential at distance l from end A: V_ Al = kl
• At null point: galvanometer deflection = 0, no current flows through test circuit
• Wire material: Manganin or Nichrome — high resistivity, low temperature coefficient
• Driver cell must have EMF greater than the EMF being measured

Comparing Two EMFs

Connect each cell in turn (using a two-way key), find the null points l₁ and l₂, and the ratio of EMFs follows directly from the ratio of lengths.

With the two-way key switched to cell 1, slide the jockey to get null point at l_1 . At null: _1 = kl_1 . Switch to cell 2, find null point at l_2 . At null: _2 = kl_2 . Divide — k cancels.

_1 = kl_1 and _2 = kl_2

_1 _2 = kl_1 kl_2 = l_1 l_2

• Both cells measured under same k — driver circuit must stay unchanged
• Formula: _1/ _2 = l_1/l_2 — clean and elegant
• The two-way key (or plug key) is used to switch between cells without disturbing the main circuit
• Null point for stronger EMF is at greater length
• If l_1 > l_2 , then _1 > _2 — directly readable from the wire

Quick recall

Two cells give null points at 60 cm and 45 cm. What is ε₁/ε₂?

_1/ _2 = l_1/l_2 = 60/45 = 4/3 . So ε₁ is larger.

Finding Internal Resistance

Connect a resistance box R across the test cell and find two null points: one with R disconnected (open circuit → measures ε) and one with R connected (loaded circuit → measures V). The difference reveals internal resistance.

Step 1: With key K_2 open (no external resistance across test cell), find null point at l_1 . No current flows through test cell at null, so kl_1 = (true EMF).

Step 2: Close K_2 — now resistance R is connected across the cell. Current flows through the cell and R. Slide the jockey to find new null point at l_2 . At this null, kl_2 = V = terminal voltage (not EMF, because current is now flowing).

V = l_1 l_2 V = l_1 l_2

Since = I(R + r) and V = IR , we get /V = (R+r)/R . Equating with the ratio above:

R + r R = l_1 l_2 r = R ( l_1 l_2 - 1 ) = R l_1 - l_2 l_2

• l_1 → null point with K_2 open (measures , open circuit)
• l_2 → null point with K_2 closed (measures V , loaded circuit)
• Always l_1 > l_2 because > V
• Formula: r = R(l_1 - l_2)/l_2
• Larger R → more current → greater drop → larger difference (l_1 - l_2) → more accurate

Quick recall

Null points at l₁ = 75 cm (open) and l₂ = 60 cm (R = 10 Ω connected). Find r.

r = R(l_1 - l_2)/l_2 = 10 (75 - 60)/60 = 10 15/60 = 2.5

Sensitivity

Sensitivity is the ability to detect small differences in EMF. A smaller potential gradient k gives a longer wire per unit voltage — so the null point shifts more noticeably for tiny EMF changes.

Sensitivity of a potentiometer = how precisely it can locate the null point. It depends entirely on the potential gradient k . Lower k → more wire length per unit voltage → null point shifts by more cm for a small voltage change → easier to detect → higher sensitivity.

k = V_ driver L so lower k means longer L (or lower driver voltage)

• Increase wire length L → decreases k → increases sensitivity (most practical method)
• Decrease driver cell voltage → decreases k → increases sensitivity
• Add a rheostat in the driver circuit to reduce current → reduces potential drop per cm
• Sensitivity has nothing to do with the test cell — only the driver circuit and wire matter
• NCERT says: a potentiometer can be made more sensitive by using a longer wire

Potentiometer vs Voltmeter

A voltmeter draws current from the circuit it measures, disturbing it. A potentiometer at null point draws zero current — so it measures true EMF without any disturbance. This is its fundamental advantage.

When you connect a voltmeter across a cell, it draws current I = /(R_ cell + R_ voltmeter ) . This current causes a voltage drop Ir across the internal resistance, so the voltmeter reads terminal voltage V = - Ir , not the true EMF .

At the null point of a potentiometer, galvanometer current = 0, so the cell delivers zero current to the galvanometer branch. Terminal voltage = - I(0) r = . The potentiometer reads true EMF.

• Voltmeter error: V = I r (depends on internal resistance r )
• Potentiometer error: zero (no current at null point)
• Potentiometer can also measure very small EMFs (like thermoelectric EMFs) that voltmeters can't detect
• Potentiometer is passive at null — it doesn't inject energy into the test circuit

Wheatstone Bridge vs Meter Bridge vs Potentiometer

Three precision measurement devices, three distinct purposes, all based on the null-deflection principle. This table is the capstone of Chapter 3 — know it cold for NEET.

• Meter bridge is a practical application of the Wheatstone bridge — same physics, different form
• Meter bridge end errors: contact resistance at junctions A and C (at 0 cm and 100 cm) introduce systematic error
• End error correction: jockey readings near 0 cm and 100 cm are unreliable — keep null point between 20–80 cm
• Potentiometer has no equivalent 'end error' because balance is at a point along a long wire, far from the terminals
• All three: galvanometer current = 0 at null — that's the entire measurement strategy

Quick recall

Which device would you use to find internal resistance of a cell — Wheatstone bridge, meter bridge, or potentiometer?

Potentiometer. Wheatstone and meter bridges measure resistance only. Internal resistance requires measuring true EMF (with no load) and terminal voltage (with load R) — only the potentiometer does this via null method.

Quick recall

Why is a meter bridge used with null point between 20 cm and 80 cm?

To avoid end errors — the contact resistance at 0 cm and 100 cm is significant and unaccounted for. Between 20–80 cm, the wire is most uniform and end effects are negligible.

This is the last sub-concept of Current Electricity (Ch. 3). After this table, you've covered Ohm's Law → resistivity → Kirchhoff's laws → Wheatstone → meter bridge → potentiometer. Every concept builds on the previous one. The null method ties them all together.