NeetCrackers Physics Notes
Complete Notes

Current Electricity

Every topic explained in depth, with multiple practice questions and answers after each one.

📑 Topics Covered

  1. Electric Current & Drift Velocity
  2. Ohm's Law
  3. Resistivity & Conductivity
  4. Temperature Dependence of Resistance
  5. Series & Parallel Combination of Resistors
  6. EMF, Internal Resistance & Terminal Voltage
  7. Kirchhoff's Laws
  8. Wheatstone Bridge
  9. Metre Bridge
  10. Electrical Energy & Power
  11. Potentiometer

Electric Current & Drift Velocity

Electric Current

Electric current is the rate of flow of charge through a conductor: $I = Q/t$ (for steady current), measured in amperes (A). By convention, current direction is taken as the direction of flow of positive charge, even though in metals it's actually electrons (negative charge) that physically move, in the opposite direction.

Drift Velocity

Free electrons in a conductor move randomly at high speeds due to thermal motion, but this random motion averages to zero net displacement. When an electric field is applied, electrons acquire a small, net average velocity superimposed on this random motion, called the drift velocity ($v_d$) — it is this slow drift, not the electrons' actual (much faster) random thermal speed, that constitutes the electric current.

Current in Terms of Drift Velocity $$I = nAev_d$$

where $n$ = number density of free electrons, $A$ = cross-sectional area, $e$ = charge of an electron.

Solved Example

A copper wire of cross-sectional area $1\times10^{-6}\ \text{m}^2$ carries a current of 2 A. If the free electron density is $8.5\times10^{28}\ \text{m}^{-3}$, calculate the drift velocity of the electrons. ($e = 1.6\times10^{-19}$ C)

$v_d = \dfrac{I}{nAe} = \dfrac{2}{(8.5\times10^{28})(1\times10^{-6})(1.6\times10^{-19})}$

$= \dfrac{2}{1.36\times10^{4}} \approx 1.47\times10^{-4}\ \text{m/s}$ — an extremely slow speed, much slower than the electrons' actual random thermal motion.

Q. Why does a light bulb turn on almost instantly when a switch is flipped, even though the drift velocity of electrons is extremely slow (a fraction of a millimetre per second)?
Ans: When the switch is closed, the electric field is established throughout the entire circuit almost instantaneously (at a speed close to the speed of light), causing every free electron throughout the wire — not just those near the switch — to begin drifting simultaneously. Since electrons are already present all along the wire, including right at the bulb's filament, the bulb lights up as soon as this field reaches it, without needing to wait for any individual electron to physically travel the entire length of wire from the switch to the bulb.
Q. Why is conventional current direction taken as opposite to the actual direction of electron flow in a metal conductor?
Ans: The convention for current direction (direction of positive charge flow) was established historically, before the electron was even discovered, based on early experiments with current-carrying effects. When electrons were later identified as the actual mobile charge carriers in metals, it was found they move in the direction opposite to this already-established conventional current direction — rather than changing the long-standing convention, physics simply retained it, so conventional current and actual electron flow point in opposite directions in a metallic conductor.
Relaxation Time & Mobility

The relaxation time (τ) is the average time between successive collisions of an electron with the lattice ions. Drift velocity can be expressed as $v_d = \dfrac{eE\tau}{m}$, showing it depends directly on the applied field and relaxation time. Mobility ($\mu$) is defined as drift velocity per unit electric field: $\mu = v_d/E = e\tau/m$ — a measure of how easily charge carriers move through a specific material under a given field.

Solved Example

Explain why increasing temperature reduces the drift velocity of electrons in a metal for the same applied electric field.

As temperature increases, the metal's lattice ions vibrate more vigorously, causing more frequent collisions between these ions and the drifting electrons — this reduces the average relaxation time (τ) between collisions. Since $v_d = eE\tau/m$, a smaller τ (for the same field E) directly results in a smaller drift velocity, explaining why metallic resistance increases with temperature (electrons drift less efficiently for the same applied field, requiring a larger field/voltage to maintain the same current).

Q. How does mobility relate to relaxation time, and what does a higher mobility value indicate about a material?
Ans: Mobility μ = eτ/m, directly proportional to relaxation time τ. A higher mobility value indicates that charge carriers in that material can achieve a larger drift velocity for the same applied electric field — meaning the material offers less impedance to charge carrier motion (longer average time between collisions), making it a generally better conductor of electricity for a given carrier density.

Ohm's Law

Statement

At constant temperature, the current through a conductor is directly proportional to the potential difference across it: $V = IR$, where $R$ (resistance) is a constant of proportionality for that conductor under those conditions.

Ohmic vs Non-Ohmic Conductors

Materials that obey Ohm's law (giving a straight-line V-I graph through the origin) are called ohmic (e.g., most metals, at constant temperature). Materials that don't follow this simple linear relationship — like diodes, transistors, or a filament bulb (where resistance itself changes significantly with current/temperature) — are called non-ohmic.

Solved Example

A resistor carries a current of 0.5 A when a potential difference of 10 V is applied across it. Calculate its resistance.

$R = \dfrac{V}{I} = \dfrac{10}{0.5} = 20\ \Omega$

Q. Why does a filament light bulb not obey Ohm's law accurately over its full operating range?
Ans: As current through the filament increases, it heats up significantly (due to its own resistive heating), and the filament's resistance itself increases substantially with this rising temperature — since Ohm's law assumes resistance remains constant, but here R changes considerably with the very current flowing through it, the V-I relationship for a filament bulb becomes noticeably non-linear, making it a non-ohmic device.
Q. A conductor obeying Ohm's law has a resistance of 15 Ω. Calculate the current through it when a voltage of 30 V is applied.
Ans: I = V/R = 30/15 = 2 A.

Resistivity & Conductivity

Resistivity

Resistivity (ρ) is a material-specific property (independent of the conductor's actual shape or size) describing its intrinsic resistance to current flow: $R = \rho\dfrac{L}{A}$, where $L$ = length, $A$ = cross-sectional area.

Conductivity

Conductivity ($\sigma$) is simply the reciprocal of resistivity: $\sigma = 1/\rho$ — materials with high conductivity (like copper, silver) are good conductors; materials with very low conductivity (high resistivity) are good insulators.

Solved Example

Calculate the resistance of a copper wire of length 2 m and cross-sectional area $1\times10^{-6}\ \text{m}^2$, given the resistivity of copper is $1.7\times10^{-8}\ \Omega\text{m}$.

$R = \rho\dfrac{L}{A} = (1.7\times10^{-8})\dfrac{2}{1\times10^{-6}} = (1.7\times10^{-8})(2\times10^6) = 0.034\ \Omega$

Q. Why is resistivity considered a material property, while resistance is not?
Ans: Resistivity depends only on the intrinsic nature of the material itself (and temperature), remaining the same regardless of the specific dimensions of a sample — e.g., copper always has the same resistivity, whether shaped into a thin wire or a thick rod. Resistance, however, explicitly depends on the geometry (length and cross-sectional area) of a specific conductor, via R = ρL/A — the same material can have very different resistance values depending on how it's shaped, so resistance is a property of a specific object, not of the material in general.
Q. If a wire's length is doubled and its cross-sectional area is halved, how does its resistance change (resistivity unchanged)?
Ans: Since R = ρL/A, doubling L and halving A both act to increase resistance — combined, R increases by a factor of 2 × 2 = 4.
Solved Example

A wire of resistivity $2\times10^{-8}\ \Omega\text{m}$ and length 5 m has a resistance of 0.5 Ω. Calculate its cross-sectional area.

$R = \rho\dfrac{L}{A} \Rightarrow A = \dfrac{\rho L}{R} = \dfrac{(2\times10^{-8})(5)}{0.5} = \dfrac{1\times10^{-7}}{0.5} = 2\times10^{-7}\ \text{m}^2$

Q. Two wires of the same material have the same length, but wire A has twice the diameter of wire B. Compare their resistances.
Ans: Since cross-sectional area is proportional to diameter squared, wire A (twice the diameter) has 4 times the cross-sectional area of wire B. Since R ∝ 1/A (for the same ρ and L), wire A has 1/4 the resistance of wire B — the thicker wire offers considerably less resistance.

Temperature Dependence of Resistance

Metals — Resistance Increases with Temperature

For most metals, resistance increases with rising temperature, since increased thermal vibration of the metal's lattice ions causes more frequent collisions with the flowing electrons, impeding their drift: $R_T = R_0(1+\alpha\Delta T)$, where $\alpha$ is the temperature coefficient of resistance.

Semiconductors — Resistance Decreases with Temperature

For semiconductors, increasing temperature provides more thermal energy to free additional charge carriers (electrons and holes) from the material's covalent bonds — this increase in the number of available charge carriers dominates over any increased collision effects, causing semiconductor resistance to decrease with rising temperature, unlike the behaviour of metals.

Solved Example

A metal wire has a resistance of 10 Ω at 20°C. If its temperature coefficient of resistance is 0.004 per °C, calculate its resistance at 70°C.

$R_T = R_0(1+\alpha\Delta T) = 10(1+(0.004)(70-20)) = 10(1+0.2) = 10(1.2) = 12\ \Omega$

Q. Why does the resistance of a semiconductor decrease with increasing temperature, opposite to the behaviour of a typical metal?
Ans: In a semiconductor, raising temperature provides enough thermal energy to break additional covalent bonds, releasing more free electrons (and creating more holes) available to carry current — this significant increase in the number of charge carriers outweighs any increased lattice vibration/collision effects, causing overall resistance to fall. In a metal, the number of free electrons is already essentially fixed (all readily available) regardless of temperature, so only the increased lattice vibration/collision effect matters, causing resistance to rise instead.
Q. What is meant by a material having a "negative temperature coefficient of resistance," and which class of materials typically shows this?
Ans: It means the material's resistance decreases as temperature increases (opposite to the usual positive coefficient seen in metals) — semiconductors (and insulators) typically show this negative temperature coefficient behaviour.

Series & Parallel Combination of Resistors

Series Combination

Resistors in series carry the same current, while the total voltage divides among them. Equivalent resistance is the sum: $R_{eq} = R_1+R_2+\cdots$ — always larger than the largest individual resistance.

Parallel Combination

Resistors in parallel have the same voltage across each, while the total current divides among them. Equivalent resistance: $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots$ — always smaller than the smallest individual resistance.

Solved Example

Two resistors of 6 Ω and 3 Ω are connected in parallel. Calculate the equivalent resistance.

$\dfrac{1}{R_{eq}} = \dfrac{1}{6}+\dfrac{1}{3} = \dfrac{1}{6}+\dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2} \Rightarrow R_{eq} = 2\ \Omega$

Q. Why does adding more resistors in parallel always decrease the overall (equivalent) resistance, while adding more resistors in series always increases it?
Ans: In series, each additional resistor adds directly to the total path length current must travel through, increasing overall opposition to flow. In parallel, each additional resistor provides an extra, independent path for current to flow through simultaneously — more available paths make it collectively easier for charge to flow overall, decreasing the equivalent resistance, similar to how adding more lanes to a road reduces overall traffic congestion despite each lane individually still having its own resistance.
Q. Three resistors of 2 Ω, 3 Ω, and 5 Ω are connected in series. Calculate the equivalent resistance.
Ans: R_eq = 2 + 3 + 5 = 10 Ω.

EMF, Internal Resistance & Terminal Voltage

EMF (Electromotive Force)

The EMF of a cell/battery is the total energy provided per unit charge by the source, including the energy needed to overcome its own internal resistance ($r$) — it represents the maximum possible potential difference the cell could provide, measured when no current is drawn (open circuit).

Terminal Voltage $$V = \varepsilon - Ir$$

where $\varepsilon$ = EMF, $I$ = current drawn from the cell, $r$ = internal resistance. Terminal voltage is always less than EMF whenever current flows (except in the idealised case of zero internal resistance).

Solved Example

A cell of EMF 12 V and internal resistance 0.5 Ω delivers a current of 2 A to an external circuit. Calculate its terminal voltage.

$V = \varepsilon - Ir = 12 - (2)(0.5) = 12 - 1 = 11\ \text{V}$

Q. Why does a battery's terminal voltage drop below its EMF as more current is drawn from it?
Ans: Some of the EMF is used up internally to push current through the battery's own internal resistance (a voltage drop of Ir occurs within the battery itself) — the terminal voltage available to the external circuit is what remains after this internal voltage drop is subtracted from the total EMF, so as current increases, this internal voltage drop (Ir) increases, further reducing the terminal voltage below the fixed EMF value.
Q. A cell has an EMF of 6 V. When short-circuited (external resistance ≈ 0), it delivers a current of 3 A. Calculate the cell's internal resistance.
Ans: When short-circuited, essentially all EMF drops across internal resistance: ε = Ir → 6 = (3)r → r = 2 Ω.
Cells in Series

For $n$ identical cells (each of EMF $\varepsilon$ and internal resistance $r$) connected in series: total EMF $= n\varepsilon$, total internal resistance $= nr$ — useful for obtaining a higher voltage.

Cells in Parallel

For $n$ identical cells connected in parallel: total EMF remains $\varepsilon$ (unchanged), but total internal resistance $= r/n$ (reduced) — useful for supplying a larger current while keeping the same voltage, since the reduced internal resistance allows more current to be drawn without as much internal voltage drop.

Solved Example

5 identical cells, each of EMF 2 V and internal resistance 0.2 Ω, are connected in series. Calculate the total EMF and total internal resistance of this combination.

Total EMF $= n\varepsilon = 5 \times 2 = 10\ \text{V}$

Total internal resistance $= nr = 5 \times 0.2 = 1\ \Omega$

Q. Why is connecting cells in parallel (rather than series) more advantageous when a circuit requires a large current, while keeping the same voltage as a single cell?
Ans: Connecting cells in parallel keeps the overall EMF the same as a single cell (unlike series, which multiplies EMF) but reduces the total internal resistance to r/n — this smaller internal resistance means less voltage is lost internally for a given current drawn, allowing the parallel combination to supply a larger current more efficiently (with a smaller internal voltage drop) than a single cell alone could provide.

Kirchhoff's Laws

Kirchhoff's Current Law (KCL / Junction Rule)

The sum of currents entering any junction in a circuit equals the sum of currents leaving that junction — a direct consequence of conservation of charge (charge cannot accumulate indefinitely at a junction under steady-state conditions).

Kirchhoff's Voltage Law (KVL / Loop Rule)

The sum of all potential differences (EMFs and IR drops) around any closed loop in a circuit is exactly zero — a direct consequence of conservation of energy (a charge returning to its starting point after going around any closed loop must have zero net change in potential energy).

Solved Example

At a circuit junction, currents of 3 A and 2 A flow in, while a single current I flows out. Using KCL, find I.

By KCL: current in = current out $\Rightarrow 3 + 2 = I \Rightarrow I = 5\ \text{A}$

Q. What fundamental conservation principles underlie Kirchhoff's current law and voltage law respectively?
Ans: Kirchhoff's current law (junction rule) is a direct consequence of conservation of charge — charge cannot be created, destroyed, or accumulate indefinitely at a junction under steady conditions. Kirchhoff's voltage law (loop rule) is a direct consequence of conservation of energy — since electric potential is a well-defined, single-valued quantity at every point, the net change in potential energy for a charge that returns to its exact starting point around a closed loop must be zero.
Q. Why is Kirchhoff's voltage law essential for analysing circuits with multiple loops and EMF sources, where simple series/parallel resistor combination rules are insufficient?
Ans: Simple series/parallel combination rules only apply to networks that can be reduced through repeated straightforward series/parallel simplification — many practical circuits (with multiple batteries, or resistors arranged in complex, non-simplifiable networks) cannot be reduced this way. Kirchhoff's laws, being based on fundamental conservation principles that always hold true regardless of circuit complexity, provide a systematic, universally applicable method (setting up simultaneous equations for currents and voltages) for solving even the most complex circuit configurations.

Wheatstone Bridge

Setup & Balance Condition

A Wheatstone bridge is a four-resistor arrangement, with a galvanometer connected across the bridge's midpoint, used to precisely measure an unknown resistance. The bridge is balanced when the galvanometer shows zero deflection (no current flows through it), which occurs when:

$$\dfrac{P}{Q} = \dfrac{R}{S}$$

where $P, Q, R, S$ are the four resistances forming the bridge's four arms.

Solved Example

In a balanced Wheatstone bridge, P = 4 Ω, Q = 8 Ω, and R = 6 Ω. Calculate the unknown resistance S.

$\dfrac{P}{Q} = \dfrac{R}{S} \Rightarrow \dfrac{4}{8} = \dfrac{6}{S} \Rightarrow S = \dfrac{6\times8}{4} = 12\ \Omega$

Q. Why is a Wheatstone bridge considered a highly sensitive and accurate method for measuring an unknown resistance, compared to simply measuring current and voltage directly?
Ans: The bridge's balance condition depends only on achieving zero current through the galvanometer (a null/balance method), rather than on precisely reading an absolute current or voltage value — null methods are inherently more accurate since they don't depend on the precision or calibration of the measuring meter itself at a non-zero reading, only on reliably detecting the null (zero) point, which a sensitive galvanometer can do very precisely.
Q. In a Wheatstone bridge, why does no current flow through the galvanometer specifically when P/Q = R/S?
Ans: This condition ensures the potential at the two midpoints of the bridge (where the galvanometer connects) are exactly equal — since current only flows between two points when there's a potential difference between them, equal potentials at both galvanometer terminals mean zero potential difference across it, and hence zero current flows through the galvanometer branch.

Metre Bridge

Principle

A metre bridge is a practical application of the Wheatstone bridge principle, using a 1-metre-long uniform resistance wire in place of two of the bridge's arms — a sliding "jockey" contact finds the balance point where the galvanometer shows zero deflection, and the two wire lengths on either side of this point act as two of the bridge's resistance ratio arms (since resistance of a uniform wire is directly proportional to its length).

Metre Bridge Formula $$\dfrac{R}{S} = \dfrac{l_1}{l_2} = \dfrac{l_1}{100-l_1}$$

where $R$ = known resistance, $S$ = unknown resistance, $l_1$ = balance length from one end (in cm), and $100-l_1$ is the remaining length.

Solved Example

In a metre bridge experiment, the balance point is obtained at 40 cm from one end, with a known resistance of 6 Ω connected in the corresponding gap. Calculate the unknown resistance.

$\dfrac{R}{S} = \dfrac{l_1}{100-l_1} \Rightarrow \dfrac{6}{S} = \dfrac{40}{60} \Rightarrow S = \dfrac{6\times60}{40} = 9\ \Omega$

Q. Why must the resistance wire used in a metre bridge be uniform (constant cross-section and material) along its entire length?
Ans: The metre bridge method relies on the assumption that resistance is directly proportional to length along the wire (R = ρL/A, with ρ and A constant throughout) — this proportionality is only valid if the wire has genuinely uniform composition and cross-sectional area throughout its length. Any variation in thickness or material would break this direct length-resistance proportionality, making the calculated balance-point ratios inaccurate for determining the unknown resistance.
Q. In a metre bridge, the balance point is found at 25 cm from the left end, with a known resistance of 5 Ω in the left gap. Calculate the unknown resistance in the right gap.
Ans: R/S = l₁/(100−l₁) → 5/S = 25/75 → S = (5×75)/25 = 15 Ω.

Electrical Energy & Power

Electrical Power

The rate at which electrical energy is converted to other forms (heat, light, mechanical work) in a circuit element:

$$P = VI = I^2R = \dfrac{V^2}{R}$$
Electrical Energy & Household Billing

Electrical energy consumed is $E = Pt$, commonly measured in kilowatt-hours (kWh) for household billing purposes — 1 kWh (also called 1 "unit" of electricity) is the energy consumed by a 1000 W appliance running for 1 hour.

Solved Example

A 100 W bulb is used for 5 hours a day. Calculate the electrical energy consumed in one day, in kWh.

$E = Pt = (0.1\ \text{kW})(5\ \text{h}) = 0.5\ \text{kWh}$

Q. Two resistors, 4 Ω and 8 Ω, are connected in series to a battery. Which resistor dissipates more power, and why?
Ans: The 8 Ω resistor dissipates more power. Since they're in series, both carry the same current I, and power dissipated is P = I²R — with the same I for both, the resistor with the larger R (8 Ω) dissipates proportionally more power than the smaller one (4 Ω).
Q. Calculate the power dissipated in a 20 Ω resistor carrying a current of 3 A.
Ans: P = I²R = (3)²(20) = (9)(20) = 180 W.

Potentiometer

Principle

A potentiometer is a long, uniform resistance wire used to measure an unknown EMF or potential difference by comparison, again using a null (balance point) method — this makes it more accurate than a simple voltmeter, since at the balance point, no current is drawn from the source being measured at all, avoiding any loading effect or voltage drop due to internal resistance.

Comparing Two EMFs $$\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}$$

where $l_1$, $l_2$ are the balance lengths obtained for each EMF source in turn, using the same potentiometer wire and driving current.

Solved Example

A potentiometer gives a balance length of 60 cm for a standard cell of EMF 1.5 V, and 90 cm for an unknown cell. Calculate the EMF of the unknown cell.

$\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2} \Rightarrow \dfrac{1.5}{\varepsilon_2} = \dfrac{60}{90} \Rightarrow \varepsilon_2 = \dfrac{1.5\times90}{60} = 2.25\ \text{V}$

Q. Why is a potentiometer considered more accurate than an ordinary voltmeter for measuring EMF?
Ans: At its balance point, a potentiometer draws exactly zero current from the cell/source whose EMF is being measured — this means there is no voltage drop across that source's internal resistance during the measurement, allowing the potentiometer to measure the true EMF directly. An ordinary voltmeter, by contrast, always draws at least some small current to operate, causing a small voltage drop across the source's internal resistance, meaning it measures a slightly reduced terminal voltage rather than the true EMF.
Q. A potentiometer wire of length 100 cm carries a standard cell balance point at 50 cm for a 2 V reference. If an unknown cell gives a balance point at 70 cm, calculate its EMF.
Ans: ε₁/ε₂ = l₁/l₂ → 2/ε₂ = 50/70 → ε₂ = (2×70)/50 = 2.8 V.

⚡ Formula Cheat-Sheet

Current from drift velocityI = nAev_d
Ohm's lawV = IR
Resistance from resistivityR = ρL/A
Series resistorsR_eq = ΣRᵢ
Parallel resistors1/R_eq = Σ(1/Rᵢ)
Terminal voltageV = ε − Ir
Wheatstone bridge balanceP/Q = R/S
Metre bridge formulaR/S = l₁/(100−l₁)
Electrical powerP = VI = I²R = V²/R

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