Chapter 3

Current Electricity

CBSE · 12 subtopics

Formula SheetEvery formula this chapter can be examined on, in one table.

Electric current

Formula	Symbols, units and conditions
$I = \dfrac{q}{t}$	$q$ is the net charge crossing a chosen cross-section in time $t$ . Steady currents only. SI unit: ampere (A), an SI base unit, $1\ \text{A} = 1\ \text{C s}^{-1}$ .
$I(t) = \lim_{\Delta t \to 0} \dfrac{\Delta Q}{\Delta t}$	Instantaneous current, for currents that are not steady. $\Delta Q$ crosses the section in the interval $\Delta t$ .
$q = \int I\,\mathrm{d}t = \text{area under the } I\text{–}t \text{ graph}$	Use this whenever the question gives an $I$ – $t$ plot instead of a number.
$N = \dfrac{q}{e} = \dfrac{It}{e}$	$N$ is the number of electrons that cross the section, $e = 1.6 \times 10^{-19}\ \text{C}$ .

Ohm's law, resistance and resistivity

Formula	Symbols, units and conditions
$V = IR$	Ohm's law for a conductor at constant temperature: $V$ in volt, $I$ in ampere, $R$ in ohm. $1\ \Omega = 1\ \text{V A}^{-1}$ .
$R = \dfrac{\rho l}{A}$	$l$ = length (m), $A$ = area of cross-section (m $^2$ ), $\rho$ = resistivity of the material ( $\Omega$ m). Holds for a uniform conductor.
$\rho = \dfrac{RA}{l}, \qquad \sigma = \dfrac{1}{\rho}$	$\sigma$ = conductivity, SI unit $\Omega^{-1}\text{m}^{-1}$ (also S m $^{-1}$ or mho m $^{-1}$ ).
$j = \dfrac{I}{A}, \qquad E = \dfrac{V}{l}$	$j$ = current density in A m $^{-2}$ (taken normal to the current); $E$ = uniform field inside a conductor of length $l$ in V m $^{-1}$ .
$\mathbf{E} = \rho\,\mathbf{j} \qquad \text{or} \qquad \mathbf{j} = \sigma \mathbf{E}$	Microscopic (point) form of Ohm's law. $\mathbf{j}$ is a vector parallel to $\mathbf{E}$ ; current $I$ itself is a scalar.
$R' = n^{2} R \quad \text{on stretching a wire to } n \text{ times its length}$	Volume stays constant, so $A$ falls by $n$ while $l$ rises by $n$ . A wire cut lengthwise into two and rejoined end to end gives length $2l$ , area $A/2$ , so $R' = 4R$ .

Drift of electrons, conductivity and mobility

Formula	Symbols, units and conditions
$\mathbf{a} = \dfrac{-e\mathbf{E}}{m}$	Acceleration of a free electron of charge $-e$ and mass $m = 9.1 \times 10^{-31}\ \text{kg}$ in the field $\mathbf{E}$ inside the conductor.
$\mathbf{v}_d = \dfrac{-e\mathbf{E}\tau}{m}, \qquad \lvert v_d \rvert = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}$	$\tau$ = relaxation time, the average time between two successive collisions (s). The minus sign says the drift is opposite to $\mathbf{E}$ . Typical $v_d \sim 1\ \text{mm s}^{-1}$ .
$I = neAv_d$	$n$ = number density of free electrons (m $^{-3}$ ), $A$ = area of cross-section. For copper $n \approx 8.5 \times 10^{28}\ \text{m}^{-3}$ .
$j = \dfrac{I}{A} = nev_d = \dfrac{ne^{2}\tau}{m}E$	Current density in terms of relaxation time; comparing with $j = \sigma E$ gives the conductivity.
$\sigma = \dfrac{ne^{2}\tau}{m}, \qquad \rho = \dfrac{m}{ne^{2}\tau}, \qquad R = \dfrac{ml}{ne^{2}\tau A}$	The origin of resistivity: $\rho$ falls as $n$ or $\tau$ rises. This is the form CBSE marks in the derivation question.
$\mu = \dfrac{\lvert v_d \rvert}{E} = \dfrac{e\tau}{m}, \qquad \sigma = ne\mu$	Mobility: drift speed per unit field. SI unit $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ . Mobility is always positive and does not depend on $V$ or $E$ .
$t = \dfrac{l}{v_d} = \dfrac{neAl}{I}$	Time for one electron to drift the whole length $l$ of a wire.

Resistivity of various materials

Formula	Symbols, units and conditions
$\rho_{\text{metal}} \approx 10^{-8}\ \text{to}\ 10^{-6}\ \Omega\,\text{m}$	Conductors have the lowest resistivities; semiconductors lie in between; insulators are about $10^{18}$ times (NCERT Summary: $10^{22}$ – $10^{24}$ times for glass and rubber) larger than metals.

Temperature dependence of resistivity

Formula	Symbols, units and conditions
$\rho_T = \rho_0\left[1 + \alpha (T - T_0)\right]$	$\rho_T$ = resistivity at $T$ , $\rho_0$ = resistivity at the reference temperature $T_0$ , $\alpha$ = temperature coefficient of resistivity. Valid over a limited range of $T$ around $T_0$ .
$R_T = R_0\left[1 + \alpha (T - T_0)\right]$	Same relation for resistance when the dimensions do not change appreciably — the form used in every numerical.
$\alpha = \dfrac{\rho_T - \rho_0}{\rho_0 (T - T_0)} = \dfrac{R_T - R_0}{R_0 (T - T_0)}$	Fractional change in resistivity per unit rise in temperature. Dimension (temperature) $^{-1}$ ; unit K $^{-1}$ or $^\circ$ C $^{-1}$ . Positive for metals, negative for semiconductors and insulators, very small for alloys.
$t = \dfrac{R_t - R_0}{R_{100} - R_0} \times 100\ ^\circ\text{C}$	Platinum resistance thermometer: $R_0$ , $R_{100}$ and $R_t$ are the resistances at the ice point, the steam point and the unknown temperature.

Electrical energy and power

Formula	Symbols, units and conditions
$\Delta W = VI\,\Delta t, \qquad P = \dfrac{\Delta W}{\Delta t} = VI$	Energy dissipated as heat in a conductor carrying $I$ across a potential difference $V$ . SI unit of power: watt (W); of energy: joule (J).
$P = I^{2}R = \dfrac{V^{2}}{R}$	Ohmic (Joule) loss. Use $P = I^2R$ when the current is common (series) and $P = V^2/R$ when the voltage is common (parallel).
$H = I^{2}Rt$	Joule's heating, in joule. $1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}$ .
$P_c = I^{2}R_c = \dfrac{P^{2}R_c}{V^{2}}$	Power wasted in transmission cables of resistance $R_c$ while delivering power $P$ at voltage $V$ : hence transmission at high $V$ .
$R = \dfrac{V^{2}}{P}\ \text{(rated)}, \qquad P' = \dfrac{V'^{2}}{R}$	Resistance of a bulb from its rating $(P, V)$ , and the power it actually draws at a different supply voltage $V'$ .

Combination of resistors

Formula	Symbols, units and conditions
$R_s = R_1 + R_2 + R_3 + \dots$	Series: same current through each, potential differences add. $R_s$ is larger than the largest resistor.
$\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots, \qquad R_p = \dfrac{R_1R_2}{R_1+R_2}\ \text{(two resistors)}$	Parallel: same potential difference across each, currents add. $R_p$ is smaller than the smallest resistor, and $I \propto 1/R$ in the branches.
$\dfrac{1}{P_s} = \dfrac{1}{P_1} + \dfrac{1}{P_2}, \qquad P_p = P_1 + P_2$	Effective power of two appliances rated $(P_1, V)$ and $(P_2, V)$ run off the same supply $V$ , in series and in parallel.

Cell, emf and internal resistance

Formula	Symbols, units and conditions
$\varepsilon = V_+ + V_-$	Emf of a cell: the potential difference between its terminals on open circuit (no current drawn). Unit: volt. Emf is a potential difference, not a force.
$V = \varepsilon - Ir$	Terminal potential difference while the cell is discharging; $r$ = internal resistance ( $\Omega$ ). $V < \varepsilon$ whenever current is drawn.
$V = \varepsilon + Ir$	Terminal potential difference while the cell is being charged (current driven into the positive terminal).
$I = \dfrac{\varepsilon}{R + r}, \qquad V = IR = \dfrac{\varepsilon R}{R + r}$	Single cell of emf $\varepsilon$ , internal resistance $r$ , across an external resistance $R$ .
$r = \dfrac{\varepsilon - V}{I} = R\left(\dfrac{\varepsilon - V}{V}\right), \qquad I_{\max} = \dfrac{\varepsilon}{r}$	Internal resistance from a measured terminal voltage; maximum current is drawn when $R = 0$ (short circuit).

Cells in series and in parallel

Formula	Symbols, units and conditions
$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2, \qquad r_{eq} = r_1 + r_2$	Series, cells supporting each other (current leaves each cell from its positive electrode). If one cell is reversed its emf enters with a minus sign: $\varepsilon_{eq} = \varepsilon_1 - \varepsilon_2$ .
$I = \dfrac{n\varepsilon}{R + nr}$	$n$ identical cells in series across $R$ .
$\dfrac{1}{r_{eq}} = \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dots, \qquad \dfrac{\varepsilon_{eq}}{r_{eq}} = \dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2} + \dots$	Parallel combination, general form for any number of cells.
$\varepsilon_{eq} = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}, \qquad r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$	Two cells in parallel. If the second cell is reversed, put $\varepsilon_2 \to -\varepsilon_2$ .
$I = \dfrac{\varepsilon}{R + r/n}$	$n$ identical cells in parallel across $R$ . Series and parallel give the same current when $R = r$ .

Kirchhoff's rules

Formula	Symbols, units and conditions
$\Sigma I = 0 \quad \text{at any junction}$	Junction rule: total current entering a junction equals total current leaving it. Follows from conservation of charge.
$\Sigma \varepsilon = \Sigma IR \quad \text{around any closed loop}$	Loop rule: the algebraic sum of changes of potential around a closed loop is zero. Follows from conservation of energy.

Wheatstone bridge and metre bridge

Formula	Symbols, units and conditions
$\dfrac{P}{Q} = \dfrac{R}{S} \qquad \text{i.e.} \qquad \dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}, \qquad R_1R_4 = R_2R_3$	Balance (null) condition: no current through the galvanometer. $P = R_1$ (arm AB) and $Q = R_2$ (arm BC) form one branch, $R = R_3$ (arm AD) and $S = R_4$ (arm DC) the other, with G across BD and the cell across AC.
$S = Q\,\dfrac{R}{P} \qquad \text{or} \qquad R_4 = R_3\,\dfrac{R_2}{R_1}$	Unknown arm found from the other three at balance.
$\dfrac{R}{S} = \dfrac{l}{100 - l}$	Metre bridge: $l$ is the balancing length in cm from the left end, $R$ the unknown, $S$ the resistance box.
$\rho = \dfrac{RA}{L} = \dfrac{R\,\pi d^{2}}{4L}$	Specific resistance of the material of the unknown wire, of length $L$ and diameter $d$ , once $R$ is known.

Constants and data CBSE expects

Quantity Symbol Value
Electronic charge $e$ $1.6 \times 10^{-19}\ \text{C}$
Electron mass $m$ $9.1 \times 10^{-31}\ \text{kg}$
Free-electron density in copper $n$ $8.5 \times 10^{28}\ \text{m}^{-3}$
Typical drift speed $v_d$ $\sim 1\ \text{mm s}^{-1}$
Thermal speed of atoms at 300 K — $\sim 2 \times 10^{2}\ \text{m s}^{-1}$
Speed at which the field is set up $c$ $3 \times 10^{8}\ \text{m s}^{-1}$
Unit of energy on a bill kWh $3.6 \times 10^{6}\ \text{J}$
Transition temperature of mercury $T_c$ $\approx 4.2\ \text{K}$

Short NotesThe chapter subtopic by subtopic, in the textbook's own order.12 subtopics

The whole chapter in brief

Electric current

Charges in motion constitute an electric current. For a steady current, $I = q/t$ ; in general $I = \lim_{\Delta t \to 0} (\Delta Q / \Delta t)$ .
SI unit: ampere (A), an SI base unit, defined through the magnetic effect of currents. $1\ \text{A} = 1\ \text{C s}^{-1}$ .
Conventional current is in the direction of flow of positive charge; in a metal the electrons drift the opposite way.
Current is a scalar even though it is drawn with an arrow — currents add algebraically at a junction, not by the parallelogram law. Formally $I = \mathbf{j}\cdot\Delta\mathbf{S}$ , a scalar product of two vectors.
Orders of magnitude: domestic appliances a few A, lightning tens of thousands of A, nerve currents in $\mu$ A.
Charge transferred = area under the $I$ – $t$ graph. Number of electrons $N = q/e$ .

Electric currents in conductors

In metals some electrons are practically free; the positive ions are fixed. In electrolytes both positive and negative ions carry the current. In insulators the electrons stay bound.
With no field, electrons move randomly, collide with the ions and emerge with the same speed in a random direction; the average velocity is zero, so there is no net current.
With a field, a small drift is superposed on the random motion and a current flows. The random motion itself is unchanged in character.
NCERT's cylinder with $+Q$ and $-Q$ discs at its ends carries a current only for a short while — the charges neutralise. A steady current needs a cell or battery to replenish the charge and maintain a steady field.

Ohm's law

Statement: at constant temperature the current through a conductor is proportional to the potential difference across its ends, $V \propto I$ , i.e. $V = IR$ .
$R = V/I$ merely defines resistance and applies to every device; Ohm's law is the extra assertion that $R$ is independent of $V$ , i.e. that the $I$ – $V$ plot is a straight line through the origin.
Geometry: doubling the length doubles $R$ ; halving the area doubles $R$ . Hence $R = \rho l / A$ .
Point form: $E = \rho j$ , or $j = \sigma E$ — a material obeys Ohm's law when $\rho$ does not depend on the magnitude or direction of the applied field.
$\rho$ and $\sigma$ are properties of the material (and its temperature) alone; $R$ depends also on shape and size.

Drift of electrons and the origin of resistivity

Drift velocity: the average velocity acquired by the free electrons under an applied field, superposed on their random thermal motion, directed opposite to $\mathbf{E}$ .
Relaxation time $\tau$ : the average time between two successive collisions of a free electron with the lattice ions.
Results to memorise: $v_d = eE\tau/m = eV\tau/ml$ ; $I = neAv_d$ ; $j = nev_d$ ; $\sigma = ne^2\tau/m$ ; $\rho = m/ne^2\tau$ ; $R = ml/ne^2\tau A$ .
Ohm's law follows: $I = (ne^2A\tau/ml)V$ , so $V/I$ is a constant at fixed temperature.
Drift speed is tiny ( $\sim 1\ \text{mm s}^{-1}$ ), about $10^{-5}$ of the thermal speed and $10^{-11}$ of the speed at which the field is set up. The current starts almost instantly because the field is established throughout the circuit at nearly the speed of light, not because electrons travel from one end to the other.
Large currents are possible in spite of the tiny drift speed because $n$ is enormous, $\sim 10^{29}\ \text{m}^{-3}$ .
Between collisions the paths are straight lines when there is no field, and curved (parabolic) when there is one.

Mobility

Mobility $\mu$ is the magnitude of the drift velocity per unit electric field: $\mu = \lvert v_d \rvert / E = e\tau/m$ .
SI unit $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ ; practical unit $\text{cm}^2\text{V}^{-1}\text{s}^{-1}$ , and $1\ \text{m}^2\text{V}^{-1}\text{s}^{-1} = 10^{4}\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$ .
Mobility is always positive; it depends only on $\tau$ (hence on temperature and the material), never on the applied voltage, the field or the geometry.
Conductivity in terms of mobility: $\sigma = ne\mu$ .
Charge carriers: electrons in metals; electrons and positive ions in an ionised gas; positive and negative ions in an electrolyte.

Limitations of Ohm's law

Ohm's law is not a fundamental law of nature. Deviations are of three kinds.
(a) $V$ is not proportional to $I$ — the $V$ – $I$ curve bends away from the straight line at large currents (for instance when $\rho$ rises with $I$ ).
(b) The relation depends on the sign of $V$ : reversing the polarity does not give the same current in the opposite direction. A junction diode behaves this way.
(c) The relation is not unique: more than one value of $V$ gives the same $I$ . Gallium arsenide (GaAs) shows this.
Negative resistance: the region of a characteristic where the current falls as the voltage rises, $\mathrm{d}I/\mathrm{d}V < 0$ . On the GaAs curve this is the falling portion (labelled DE in CBSE's figure); the straight portion through the origin (AB) is where Ohm's law holds.

Resistivity of various materials

Class	Resistivity at room temperature	Behaviour with rising temperature
Conductors (metals)	$10^{-8}$ to $10^{-6}\ \Omega\,\text{m}$	$\rho$ increases, $\alpha$ positive
Alloys (nichrome, manganin, constantan)	high, of order $10^{-6}\ \Omega\,\text{m}$	$\rho$ almost unchanged, $\alpha$ very small
Semiconductors (Si, Ge)	intermediate	$\rho$ decreases, $\alpha$ negative
Insulators (glass, rubber, ceramic, plastic)	about $10^{18}$ times that of metals or more	$\rho$ decreases

Adding a small amount of a suitable impurity (doping) lowers the resistivity of a semiconductor — the basis of electronic devices.
Standard resistors are wire-wound from alloys because they have a very low temperature coefficient of resistivity (the value hardly drifts as they warm up) together with a high specific resistance (a short length gives a large resistance).

Temperature dependence of resistivity

$\rho_T = \rho_0[1 + \alpha(T - T_0)]$ , and the same form for $R$ . $\alpha$ is the fractional change in resistivity per degree rise, unit K $^{-1}$ .
Why a metal's resistivity rises: $\rho = m/ne^2\tau$ . Heating makes the ions vibrate more, collisions get more frequent, $\tau$ falls; $n$ is practically unchanged, so $\rho$ rises.
Why a semiconductor's resistivity falls: $n$ increases sharply with temperature and more than compensates the fall in $\tau$ .
Graphs to be able to sketch: copper — nearly a straight line of small positive slope over the ordinary range, curving and falling steeply towards a small value well below $0\ ^\circ$ C; nichrome — a straight line of very small slope; a semiconductor — a curve falling steeply as $T$ rises.
Superconductivity: the resistance of mercury falls smoothly on cooling and then drops abruptly to zero at $T_c \approx 4.2\ \text{K}$ ; below $T_c$ it offers no resistance at all.
Reading an $I$ – $V$ graph: $R$ is the reciprocal of the slope, so the steeper line is the smaller resistance. For a metal, the steeper line is at the lower temperature; for two wires of equal length and diameter, the less steep line is the higher resistivity.

Electrical energy and power

In time $\Delta t$ the charge $I\Delta t$ falls through $V$ , so the energy delivered is $\Delta W = VI\Delta t$ . Because collisions pass this energy to the lattice, it appears as heat.
$P = VI = I^2R = V^2/R$ ; heat $H = I^2Rt$ . $1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$ .
Series: the current is common, so $P \propto R$ — the larger resistance heats up more (nichrome heats more than copper). Parallel: the voltage is common, so $P \propto 1/R$ — the smaller resistance dissipates more.
Bulb rated $(P, V)$ has $R = V^2/P$ ; at another supply voltage $V'$ it draws $P' = V'^2/R$ .
Transmission: to deliver power $P$ at voltage $V$ through cables of resistance $R_c$ , the wasted power is $P_c = P^2R_c/V^2$ . It falls as $1/V^2$ , which is why transmission is at very high voltage, stepped down by a transformer at the far end.

Combination of resistors

Feature	Series	Parallel
Common quantity	current $I$	potential difference $V$
Equivalent	$R_s = R_1 + R_2 + \dots$	$\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$
Compared with the parts	larger than the largest	smaller than the smallest
Sharing	$V \propto R$	$I \propto 1/R$
Power	$P \propto R$	$P \propto 1/R$

Two resistors in parallel: $R_p = R_1R_2/(R_1+R_2)$ . $n$ equal resistors $R$ in parallel: $R/n$ .
Points joined by a resistanceless wire are at the same potential; a resistor across two such points carries no current and may be ignored.

Cells, emf and internal resistance

Emf $\varepsilon$	Terminal voltage $V$
Potential difference between the terminals on open circuit	Potential difference between the terminals on closed circuit
No current is drawn	Current $I$ flows
Fixed for a given cell, set by its chemistry	Depends on the current and the external resistance
Cause	Effect
$\varepsilon = V + Ir$	$V = \varepsilon - Ir$ , less than $\varepsilon$ while discharging

Internal resistance $r$ is the resistance of the electrolyte between the electrodes. Dry cells have a much higher $r$ than ordinary electrolytic cells. $r$ may be neglected when $\varepsilon \gg Ir$ .
$I = \varepsilon/(R+r)$ ; largest current when $R = 0$ , $I_{\max} = \varepsilon/r$ (usually far above the safe limit of a real cell).
While charging, current is pushed into the positive terminal and $V = \varepsilon + Ir > \varepsilon$ .
Graph $V$ against $I$ : a straight line, intercept $\varepsilon$ on the $V$ -axis, slope $-r$ , meeting the $I$ -axis at $\varepsilon/r$ . The intercept gives the emf and the magnitude of the slope gives the internal resistance.
Graph $V$ against $R$ : a curve from the origin, rising steeply then flattening, approaching $V = \varepsilon$ asymptotically as $R \to \infty$ .
Direction of current: from negative to positive terminal inside the cell, from positive to negative outside it.

Cells in series and in parallel

Series	Parallel
$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$	$\dfrac{\varepsilon_{eq}}{r_{eq}} = \dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2}$
$r_{eq} = r_1 + r_2$	$\dfrac{1}{r_{eq}} = \dfrac{1}{r_1} + \dfrac{1}{r_2}$
Reversed cell enters with a minus sign	Reversed cell: put $\varepsilon_2 \to -\varepsilon_2$
$n$ identical cells: $I = \dfrac{n\varepsilon}{R + nr}$	$n$ identical cells: $I = \dfrac{\varepsilon}{R + r/n}$

For two cells in parallel, $\varepsilon_{eq} = (\varepsilon_1r_2 + \varepsilon_2r_1)/(r_1+r_2)$ and $r_{eq} = r_1r_2/(r_1+r_2)$ .
Series is worth using when $R \gg r$ , parallel when $R \ll r$ ; the two give the same current when $R = r$ .
If identical cells are in parallel, the equivalent emf is that of one cell and the equivalent internal resistance is $r/n$ .

Kirchhoff's rules

Junction rule: at any junction the sum of the currents entering equals the sum of the currents leaving, $\Sigma I = 0$ . Based on conservation of charge — no charge piles up at a junction in the steady state.
Loop rule: the algebraic sum of the changes in potential round any closed loop is zero, $\Sigma \varepsilon = \Sigma IR$ . Based on conservation of energy — potential is a function of position, so returning to the same point must give zero net change.
Signs: going through a resistor along the assumed current, the change is $-IR$ ; against it, $+IR$ . Going through a cell from $-$ to $+$ , the change is $+\varepsilon$ ; from $+$ to $-$ , it is $-\varepsilon$ .
A negative answer for a current simply means the real current is opposite to the arrow you drew.
Bending or re-routing a wire does not affect the junction rule.

Wheatstone bridge

Four resistors $P, Q, R, S$ form a quadrilateral ABCD; the cell is across the diagonal AC (battery arm) and a galvanometer across BD (galvanometer arm).
Balance condition: the galvanometer shows no deflection when $\dfrac{P}{Q} = \dfrac{R}{S}$ , i.e. $R_1R_4 = R_2R_3$ .
At balance B and D are at the same potential; the galvanometer arm carries no current and may be removed or replaced by a wire without changing anything.
A null method: the balance condition contains only the four resistances, so the result does not depend on the emf of the cell or the sensitivity of the galvanometer — which is why it is accurate.
Metre bridge is a practical Wheatstone bridge: two of the arms are the two parts of a uniform one-metre wire on either side of the jockey, so $\dfrac{R}{S} = \dfrac{l}{100-l}$ , and then $\rho = RA/L$ for the material of the unknown wire.

Traps that cost marks

Writing $V = IR$ and calling it Ohm's law. $V = IR$ defines $R$ ; the law is the claim that $R$ does not depend on $V$ .
Saying resistivity depends on the length or the area. It depends only on the material and its temperature; it is $R$ that depends on the dimensions.
Saying mobility or relaxation time changes when the applied voltage changes. Both depend on the material and its temperature, not on $V$ .
In a wire of non-uniform cross-section, only the current is the same everywhere. $j$ , $v_d$ and $E$ all vary as $1/A$ — they are larger where the wire is thinner.
Two wires in parallel across the same battery and of the same material and length have the same current density and the same drift speed, whatever their areas — because they share the same field $E = V/l$ .
Confusing " $\alpha$ is negative" with " $\rho$ is negative". Only the coefficient is negative for a semiconductor; resistivity itself is always positive.
Forgetting to use $R = V^2/P$ before comparing bulbs, and forgetting that the rated power is only drawn at the rated voltage.
Dropping the internal resistance from $I = \varepsilon/(R+r)$ , or writing $V = \varepsilon + Ir$ for a discharging cell (it is $\varepsilon - Ir$ ; the plus sign is for charging).
Adding emfs of cells in parallel as if they were in series. For a parallel combination use $\varepsilon_{eq}/r_{eq} = \Sigma(\varepsilon_i/r_i)$ .
In Kirchhoff problems, writing more loop equations than there are independent loops, or changing the assumed current directions halfway through.
Quoting the bridge condition upside down. Keep the same order: the ratio of the two arms of one branch equals the ratio of the two arms of the other, taken from the same end.
On a metre bridge, using $l$ in metres in $R/S = l/(100-l)$ . The formula is written for $l$ in cm; only the ratio matters, so keep both lengths in the same unit.
Rounding $e$ to $10^{-19}$ C, or forgetting SI units in the final line — CBSE deducts for a missing unit.

Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter.13

1
Derivation 1
Using the concept of drift velocity of charge carriers in a conductor, deduce the relationship between current density and resistivity of the conductor.
Asked in: CBSE Class 12 Physics — SA · Delhi 2015C, An · Section B, 2 marks
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Drift velocity: $v_d = \dfrac{eE\tau}{m}$ Current density: $J = nev_d = \dfrac{ne^2\tau}{m}E = \sigma E$ , where $\sigma = \dfrac{ne^2\tau}{m}$ Since resistivity $\rho = 1/\sigma = \dfrac{m}{ne^2\tau}$ , $J = \dfrac{E}{\rho}$
2
Derivation 2
(a) Define the term 'conductivity' of a metallic wire. Write its SI unit. (b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field $E$ .
Asked in: CBSE Class 12 Physics — SA · 2018 · Section C, 3 marks
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(a) Conductivity ( $\sigma$ ) is the reciprocal of resistivity, $\sigma = 1/\rho$ . SI unit: $\Omega^{-1}\text{m}^{-1}$ . (b) $v_d = \dfrac{eE\tau}{m}$ Current density $J = nev_d = \dfrac{ne^2\tau}{m}E$ Comparing with $J = \sigma E$ : $\sigma = \dfrac{ne^2\tau}{m}$ Hence the relation between current density and applied field: $J = \sigma E$
3
Derivation 3
(a) Deduce the relation between the current $I$ flowing through a conductor and the drift velocity $v_d$ of the electrons. (b) The figure shows a plot of current ' $I$ ' flowing through the cross-section of a wire versus time ' $t$ '. Use the plot to find the charge flowing in 10 s through the wire. [Graph: $I$ rises linearly from 0 to 5 A over $t=0$ to $5$ s, then stays constant at 5 A from $t=5$ s to $10$ s.]
Asked in: CBSE Class 12 Physics — Long Answer · AI 2015C · Section E, 5 marks
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(a) Consider a conductor of length $l$ , area of cross-section $A$ , with $n$ free electrons per unit volume. Total number of free electrons, $N = nAl$ . Time taken for an electron drifting with speed $v_d$ to cross the length $l$ : $t = l/v_d$ . Charge crossing a cross-section in this time: $q = Ne = nAle$ $\therefore I = \dfrac{q}{t} = \dfrac{nAle}{l/v_d} = neAv_d$ (b) Charge = area under the $I$ – $t$ graph: $Q = \dfrac{1}{2}\times 5 \times 5 + (10-5)\times 5 = 12.5 + 25 = 37.5$ C
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Derivation 4
(i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it. (ii) Consider two conducting wires $A$ and $B$ of the same diameter but made of different materials, joined in series across a battery. The number density of electrons in $A$ is 1.5 times that in $B$ . Find the ratio of drift velocity of electrons in wire $A$ to that in wire $B$ .
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2017, Ev · Section E, 5 marks
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(i) The field $E$ exerts a force $F=-eE$ on each electron, giving it an acceleration $a = -eE/m$ between collisions. Since successive collisions randomize the electron's velocity, the average velocity gained (the drift velocity) is $v_d = a\tau = \dfrac{-eE\tau}{m}$ , where $\tau$ is the average time between collisions (relaxation time); the negative sign shows $v_d$ is opposite to $E$ . (ii) $n_A = 1.5\,n_B$ . Since $A$ and $B$ are in series, the current is the same: $I_A = I_B$ . $n_A eA_A v_{dA} = n_B eA_B v_{dB}$ Since the wires have the same diameter, $A_A = A_B$ , so: $\dfrac{v_{dA}}{v_{dB}} = \dfrac{n_B}{n_A} = \dfrac{n_B}{1.5\,n_B} = \dfrac{2}{3}$
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Derivation 5
(i) Derive an expression for the drift velocity of electrons in a conductor. Hence deduce Ohm's law. (ii) A wire whose cross-sectional area is increasing linearly from one end to the other is connected across a battery of $V$ volts. Which of the following quantities remain constant in the wire: drift speed, current density, electric current, electric field? Justify your answer.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2017/2023, Ev · Section E, 5 marks
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(i) As derived above, $v_d = \dfrac{eE\tau}{m}$ . Now $I = neAv_d = \dfrac{ne^2A\tau}{m}E$ , and since $E=V/l$ : $I = \dfrac{ne^2A\tau}{ml}V \ \Rightarrow\ \dfrac{V}{I} = \dfrac{ml}{ne^2A\tau} = \dfrac{\rho l}{A} = R$ So $V = IR$ , which is Ohm's law, with $R = \dfrac{ml}{ne^2A\tau}$ constant for a given conductor at constant temperature. (ii) The rate of flow of charge through a conductor of non-uniform cross-section is the same at every section (steady state), so the electric current remains constant. Since the area of cross-section varies along the wire, the current density $J(=I/A)$ , drift speed $v_d(=J/ne)$ and electric field $E(=J\rho)$ all vary — decreasing as the area increases.
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Derivation 6
Define the term 'drift velocity' of charge carriers in a conductor. Obtain the expression for the current density in terms of relaxation time.
Asked in: CBSE Class 12 Physics — SA · 2/5, Foreign 2014 · Section B, 2 marks
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Drift velocity: the average velocity with which free electrons drift towards the positive end of a conductor under an applied electric field. $v_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}$ Current density: $J = \dfrac{I}{A} = nev_d = \dfrac{ne^2E\tau}{m}$
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Derivation 7
Two conductors, made of the same material, have equal lengths but different cross-sectional areas $A_1$ and $A_2$ ( $A_1>A_2$ ). They are connected in parallel across a cell. Show that the drift velocities of electrons in the two conductors are equal.
Asked in: CBSE Class 12 Physics — SA · 2023 · Section B, 2 marks
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Since the conductors are in parallel across the same cell, the potential difference $V$ across each is the same. $R_1 = \rho L/A_1$ , $R_2 = \rho L/A_2$ $v_{d1} = \dfrac{I_1}{neA_1} = \dfrac{V/R_1}{neA_1} = \dfrac{V}{ne A_1 \cdot \rho L/A_1} = \dfrac{V}{ne\rho L}$ $v_{d2} = \dfrac{I_2}{neA_2} = \dfrac{V}{ne A_2 \cdot \rho L/A_2} = \dfrac{V}{ne\rho L}$ Since neither expression depends on the area, $v_{d1} = v_{d2}$ .
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Derivation 8
(a) Two cells of emf $E_1$ and $E_2$ have their internal resistances $r_1$ and $r_2$ respectively. Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance $R$ . Assume that the two cells are supporting each other. (b) In case the two cells are identical, each of emf $E=5$ V and internal resistance $r=2\,\Omega$ , calculate the voltage across the external resistance $R=10\,\Omega$ .
Asked in: CBSE Class 12 Physics — SA · 2020 · Section C, 3 marks
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(a) Let $V$ be the potential difference between the terminals $A,B$ and $I=I_1+I_2$ the total current. For cell $E_1$ : $I_1 = \dfrac{E_1-V}{r_1}$ ; for cell $E_2$ : $I_2 = \dfrac{E_2-V}{r_2}$ $I = \dfrac{E_1-V}{r_1}+\dfrac{E_2-V}{r_2}$ Solving for $V$ : $V = \dfrac{E_1r_2+E_2r_1}{r_1+r_2} - I\left(\dfrac{r_1r_2}{r_1+r_2}\right)$ Comparing with $V=E_{eq}-Ir_{eq}$ : $E_{eq} = \dfrac{E_1r_2+E_2r_1}{r_1+r_2}$ , $r_{eq} = \dfrac{r_1r_2}{r_1+r_2}$ (b) $E_1=E_2=E=5$ V, $r_1=r_2=r=2\,\Omega \Rightarrow E_{eq}=E=5$ V, $r_{eq}=r/2=1\,\Omega$ $I = \dfrac{E_{eq}}{R+r_{eq}} = \dfrac{5}{10+1} = \dfrac{5}{11}$ A Voltage across $R$ : $V = IR = \dfrac{5}{11}\times 10 = \dfrac{50}{11} \approx 4.55$ V
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Derivation 9
Two cells of emf and internal resistance $\varepsilon_1, r_1$ and $\varepsilon_2, r_2$ are connected in parallel. Derive the expressions for the emf and internal resistance of a single cell which can replace this combination.
Asked in: CBSE Class 12 Physics — SA · 2019C · Section C, 3 marks
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By the same derivation as above (equating $V=\varepsilon_{eq}-Ir_{eq}$ to the parallel combination's terminal relation): $\varepsilon_{eq} = \dfrac{\varepsilon_1 r_2+\varepsilon_2 r_1}{r_1+r_2}$ , $r_{eq} = \dfrac{r_1r_2}{r_1+r_2}$
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Derivation 10
(i) State the two Kirchhoff's laws. Explain briefly how these rules are justified. (ii) The current is drawn from a cell of emf $E$ and internal resistance $r$ connected to the network of resistors, each of resistance $r$ , as shown in the figure. Obtain the expression for (a) the current drawn from the cell and (b) the power consumed in the network.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2017 · Section E, 5 marks
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(i) Same as stated in Q69 above: the junction rule follows from conservation of charge, the loop rule from conservation of energy. (ii) The network between $A$ and $B$ can be redrawn as four resistors, $r$ , $2r$ , $2r$ and $r$ , all in parallel: $\dfrac{1}{R_{AB}} = \dfrac{1}{r}+\dfrac{1}{r}+\dfrac{1}{2r}+\dfrac{1}{2r} = \dfrac{3}{r} \Rightarrow R_{AB} = \dfrac{r}{3}$ Net resistance of the circuit (including the cell's own internal resistance $r$ , in series with $R_{AB}$ ): $R = r + R_{AB} = r+\dfrac{r}{3} = \dfrac{4r}{3}$ (a) Current drawn from the cell: $I = \dfrac{E}{R} = \dfrac{3E}{4r}$ (b) Power consumed in the network: $P = I^2R_{AB} = \left(\dfrac{3E}{4r}\right)^2\dfrac{r}{3} = \dfrac{3E^2}{16r}$
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Derivation 11
Derive an expression for the balance condition of a Wheatstone bridge.
Asked in: CBSE Class 12 Physics — SA · 1½/3, 2020 · Section C, 3 marks
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(Same derivation as in Q73.) Applying Kirchhoff's rules to loops $ABDA$ and $BCDB$ and setting the galvanometer current $I_2=0$ at balance gives: $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$
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Derivation 12
(a) Explain the term drift velocity of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of drift velocity. (b) Two cells of emfs $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ respectively are connected in parallel, as shown in the figure. Deduce the expressions for the (i) equivalent emf, (ii) equivalent internal resistance of the combination, and (iii) potential difference between the points $A$ and $B$ .
Asked in: CBSE Class 12 Physics — Long Answer · Section E, 5 marks · CBSE sample question
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(a) Drift velocity is the average velocity acquired by the free electrons of a conductor under an applied electric field, superposed on their random thermal motion. For a conductor of length $l$ , area $A$ , with $n$ free electrons per unit volume, all $N=nAl$ free electrons cross a section in time $t=l/v_d$ , carrying charge $q=Ne=nAle$ . $I = q/t = neAv_d$ (b) Let $V$ = potential difference between $A$ and $B$ , $I=I_1+I_2$ the total current. $I_1 = \dfrac{E_1-V}{r_1}$ , $I_2 = \dfrac{E_2-V}{r_2}$ $I = \dfrac{E_1-V}{r_1}+\dfrac{E_2-V}{r_2}$ , which on solving for $V$ gives $V = \dfrac{E_1r_2+E_2r_1}{r_1+r_2} - I\dfrac{r_1r_2}{r_1+r_2}$ Comparing with $V = E_{eq}-Ir_{eq}$ : (i) $E_{eq} = \dfrac{E_1r_2+E_2r_1}{r_1+r_2}$ (ii) $r_{eq} = \dfrac{r_1r_2}{r_1+r_2}$ (iii) Potential difference between $A$ and $B$ : $V = E_{eq}-Ir_{eq}$ , using the values from (i) and (ii).
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Derivation 13
(a) State Kirchhoff's first rule. Explain briefly how it is justified. State Kirchhoff's second rule and explain how it is justified. (b) Using Kirchhoff's rules, derive the balance condition for a Wheatstone bridge.
Asked in: CBSE Class 12 Physics — Long Answer · Section E, 5 marks · CBSE sample question
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(a) Kirchhoff's first rule (junction rule): the algebraic sum of currents meeting at a junction is zero, $I_1+I_2-I_3+I_4-I_5=0$ for currents at a junction; justified by conservation of charge. Kirchhoff's second rule (loop rule): the algebraic sum of the applied emfs in a closed loop equals the algebraic sum of the $IR$ drops across the resistors of that loop, $\Sigma\varepsilon=\Sigma IR$ ; justified by conservation of energy. (b) (As derived in Q73/Q75 of the board questions.) Applying Kirchhoff's rules to the two loops of the bridge and setting the galvanometer current to zero at balance gives: $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$

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