Current Electricity: Class 12 Physics · Chapter 3

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 26 Sep 2026 · Updated 28 Sep 2026

130 min
3 simulations
17 practice questions
43 MCQs

What you will learn in Current Electricity

Define electric current and current density, and relate current to the drift of charge carriers
Derive the drift velocity, Ohm's law and the conductivity of a metal from a simple collision model
Use R = ρl/A and the temperature dependence of resistivity to find resistances and temperatures
Recognise non-ohmic behaviour from current–voltage graphs
Calculate electrical power and energy, and explain why power is transmitted at high voltage
Work with the emf and internal resistance of cells, singly and in series or parallel
Apply Kirchhoff's rules to find the currents in a network, including the Wheatstone bridge

Original companion text aligned to NCERT Physics Class 12, Chapter 3, pp. 81–106. Not affiliated with or endorsed by NCERT.

3.1 Introduction

Plug a phone into its charger and a small, steady stream of charge flows into the battery for an hour or more. A lightning strike moves charge too, but in a violent burst lasting a fraction of a second. In the last two chapters every charge sat still. Now we set charges moving and ask what keeps them moving, what holds them back and where their energy goes.

This chapter deals with steady currents, like those that run a TV remote, a pocket calculator or a solar lantern. We will define current precisely, see why metals conduct, derive Ohm's law from the motion of electrons, and learn how resistance depends on material, shape and temperature. We then follow the energy, look inside a cell, and finish with two rules that can solve any circuit, plus a clever circuit for measuring resistance.

3.2 Electric current

To measure a current we need somewhere to count charge. Pick a surface that the moving charges pass through, such as the cross-section of a wire, and decide which way across it counts as forward. In a salt solution carrying a current, positive ions and negative ions travel in opposite directions, and a few of each wander the wrong way. Positive charge carried forward and negative charge carried backward have the same effect: both leave the forward side more positive. So over a time $t$ we record $q_{+}$ , the positive charge carried forward minus that carried back, and $q_{-}$ , the same count for negative charge. The net charge transferred forward is $q = q_{+} - q_{-}$ .

For a steady flow, doubling the time doubles $q$ , so the ratio $q / t$ does not depend on how long we count. This ratio is the electric current (विद्युत धारा) through the surface:

Steady current

I = \frac{q}{t}

(3.1)

where

$q$: net charge crossing forward
$t$: time taken

3/3The current is $q$ divided by $t$ , in the direction positive charge moves.

Fig. 3.1Current counts the net charge crossing a surface per unit time. Negative charges moving backward add to the forward current.

A horizontal band represents a conducting region, with a dashed oval window drawn across its middle and an arrow marking the forward direction. In the first step four positive and four negative carriers move in random directions, so equal numbers cross each way and the net charge is zero. In the second step the positive carriers all move forward and the negative carriers all move backward, and a label reads q equals q plus minus q minus, the net forward charge in time t. In the third step a thick arrow below the band shows the current, in the same direction as the positive carriers, labelled I equals q over t.

Many currents change from moment to moment; the current in a mobile phone speaker, for example, reverses thousands of times a second. To capture such a current at one instant, count the net charge $Δ Q$ that crosses in a very short interval $Δ t$ starting at time $t$ , and let the interval shrink:

Instantaneous current

I (t) = \lim_{Δ t \to 0} \frac{Δ Q}{Δ t} = \frac{d Q}{d t}

(3.2)

If $I$ comes out negative, the net transfer of positive charge is backward. The SI unit of current is the ampere (A), one of the seven SI base units. Since 2019 it has been fixed by giving the elementary charge the exact value $e = 1.602176634 \times 10^{- 19} C$ , so a current of $1 A$ carries about $6.24 \times 10^{18}$ elementary charges past a point every second. Chapter 4 describes the magnetic effect that was used to define the ampere earlier.

Table 3.1Currents in everyday life span more than ten powers of ten.

Situation	Typical current
Signal in a nerve fibre	a few microamperes
LED indicator on a charger	about 10 mA
Phone charging cable	1 to 3 A
Room heater on a 230 V supply	about 8 A
Car starter motor	100 to 300 A
Lightning stroke	10 000 to 100 000 A for microseconds

Quick check

1 of 3

A phone charger delivers a steady

2.0 A

for

30

minutes. How much charge flows through the cable?

Show answer

Answer 3600 C

q = I t = 2.0 A \times (30 \times 60) s = 3600 C

. Remember to convert minutes to seconds.

3.3 Electric currents in conductors

A field pushes on every charge, but only charges that are free to travel can carry a current. Free charges are more common than you might think. Salt water and the acid in a car battery hold mobile ions, the ionosphere high above us is full of free electrons and ions, and every metal contains a sea of free electrons.

Metals owe their conductivity to a sharing arrangement. Each atom in a metal releases one or more of its outer electrons, which then belong to the solid as a whole and can roam anywhere in it. These conduction electrons carry the current, while the positive ions they leave behind stay locked in the lattice. Glass and rubber have no such shared electrons: every electron stays tied to its own atom or bond, which is why they insulate. The rest of this chapter concentrates on metals.

Switch off all fields and look closely at a copper wire lying on a table. Its conduction electrons are anything but still: they dart about at enormous speeds, bouncing off the vibrating ions and flying off in a new direction after every collision. Yet the wire carries no current. For every electron that crosses a cross-section one way, another crosses the other way, so the net flow is zero.

Now try to make a current flow using stored charge alone. Take two metal plates carrying charges $+ Q$ and $- Q$ , and join them with a wire through a small bulb and a switch (Fig. 3.2). When the switch closes, a field appears along the wire, pointing away from the positive plate, and electrons are pulled towards that plate. The bulb flashes. But every electron that arrives cancels a little of the plate's charge, so the field weakens, and within a fraction of a second the plates are neutral and the current has stopped.

4/4A cell keeps restoring the charges, so the field and the current stay steady.

Fig. 3.2Charged plates give only a brief current. A cell keeps resupplying charge, so the field and the current stay steady.NCERT Fig. 3.1 · p. 83

On the left, a plate marked plus Q and a plate marked minus Q are joined across the top by a wire with a switch and a bulb. On the right is a graph of current against time. In the first step the switch is open, the bulb is dark and the graph is empty. In the second step the switch is closed, arrows show current flowing from the positive plate through the bulb to the negative plate, the bulb glows, and the graph shows a large current at the start. In the third step the plate charges have faded, the bulb is dark and the graph curve has fallen to zero. In the fourth step a cell joins the plates from below, the bulb glows steadily and the graph shows a flat line of steady current.

A lasting current therefore needs a device that keeps restoring the charges at the ends, so that the field inside the wire never collapses. A cell does this by chemistry and a generator by magnetism; either can keep a current going for hours. All the currents in the rest of this chapter are of this steady kind.

3.4 Ohm's law

In the 1820s, long before anyone knew about electrons, Georg Simon Ohm measured how the current in metal wires changed with the voltage across them. For a given wire kept at a fixed temperature, doubling the potential difference doubled the current.

LawOhm's law

For a metallic conductor at constant temperature, the current $I$ through it is proportional to the potential difference $V$ across it: $V \propto I$ .

Ohm's law

V = R I

(3.3)

where

$V$: potential difference across the conductor
$I$: current through it
$R$: its resistance

Definition

Resistance

The ratio $R = V / I$ of the potential difference across a conductor to the current through it. Its SI unit is the ohm, $1 Ω = 1 V / A$ .

ExampleA heater that draws $4.0 A$ from a $230 V$ supply has a resistance of $57.5 Ω$ .

Resistance (प्रतिरोध) is not fixed by the metal alone: a long thin copper wire and a short thick copper bar have very different resistances. Try it yourself in the simulation, which offers two wires. The slope of each $I$ – $V$ line tells you $1 / R$ .

SimulationOhm's law bench

A cell with a voltage slider drives a current through one of two resistors, called wire A and wire B, with an ammeter and a voltmeter. Each setting of the slider adds a point to a current against voltage graph beside the circuit, and a readout shows the ratio V divided by I.

Fig. 3.3Change the cell voltage and record points on the

I

–

V

graph for each wire.

Try this

Record at least five points for wire A by moving the cell slider. Do they lie on a straight line through the origin?
Switch to wire B and repeat. Which line is steeper, and what does the steepness tell you about resistance?
Check the V ÷ I readout at every point. Does it change as the voltage changes?

To find how $R$ depends on size, start with a uniform wire whose resistance is $R$ , whose length is $l$ and whose cross-section has area $A$ . Now imagine joining identical pieces together in two ways (Fig. 3.4).

4/4So $R \propto l$ and $R \propto 1 / A$ , giving $R = ρ l / A$ .

Fig. 3.4Doubling the length doubles the resistance. Doubling the area halves it. Together,

R = ρ l / A

.NCERT Fig. 3.2 · p. 83

Rows of rectangular conductor pieces, each with a current arrow entering from the left and a double arrow below showing the potential difference. In the first step one piece carries current I with potential difference V, so R equals V over I. In the second step a row with two pieces joined end to end appears: the same current I needs 2V, so the resistance is 2R, labelled twice as long. In the third step a row with two pieces stacked side by side appears: with the same V the current is 2I, so the resistance is R over 2, labelled twice the area. In the fourth step a box shows the combined result R equals rho l over A.

DerivationHow resistance depends on length and area

Two pieces end to end. The same current $I$ passes through each piece, and each needs a potential difference $V = I R$ . The total potential difference is $2 V$ , so the combination has resistance $2 V / I = 2 R$ . A wire twice as long has twice the resistance, so $R \propto l$ .

Two pieces side by side. Join the pieces at both ends so that both have the same potential difference $V$ . Each carries a current $I$ , so the total is $2 I$ and the combination has resistance $V / (2 I) = R / 2$ . This is one conductor with twice the cross-section, so $R \propto 1 / A$ .

Putting the two results together, $R \propto l / A$ . The constant of proportionality depends only on the material and its temperature:

Resistance of a uniform conductor

R = ρ \frac{l}{A}

(3.9)

where

$ρ$: resistivity of the material
$l$: length along the current
$A$: cross-sectional area

Definition

Resistivity

The property $ρ$ of a material defined by $R = ρ l / A$ . It is the resistance of a piece of unit length and unit cross-section, measured in ohm metre ( $Ω m$ ). It depends on the material and its temperature but not on the shape.

ExampleCopper at room temperature has $ρ \approx 1.7 \times 10^{- 8} Ω m$ .

Resistivity (प्रतिरोधकता) lets us write Ohm's law in a form that holds at each point inside a material. Define the current density $j = I / A$ , the current per unit area of cross-section (unit $A / m^{2}$ ). In a uniform wire the field is $E = V / l$ . Substituting $V = E l$ and $R = ρ l / A$ into $V = I R$ gives $E l = (j A) (ρ l / A)$ , that is, $E = ρ j$ .

Current density is a vector that points along $𝐄$ in an ordinary conductor, so we can write the result in vector form. The reciprocal of resistivity, $σ = 1 / ρ$ , is the conductivity (चालकता), measured in siemens per metre ( $1 S = 1 Ω^{- 1}$ ).

Ohm's law at a point

𝐄 = ρ 𝐣

(3.12)

Ohm's law in terms of conductivity

𝐣 = σ 𝐄

(3.13)

Example

Resistance, current density and field in a copper wire

A copper wire is $50 m$ long and $1.0 m m$ in diameter. Copper has resistivity $1.7 \times 10^{- 8} Ω m$ . (a) Find the resistance of the wire. (b) If it carries $5.0 A$ , find the current density and the electric field inside it. (c) The wire is drawn out to twice its length, keeping its volume fixed. What is its new resistance?

Try it yourself first. Then reveal the solution one step at a time.

Try this

A resistor you can draw

You’ll need

a soft pencil (6B or 8B)
a sheet of thick paper
a ruler
a digital multimeter set to measure resistance
two metal paper clips

Predict first

You measure the resistance of a pencil strip

2 c m

long, then of

8 c m

of the same strip. What do you expect?

About four times the resistanceAbout the same resistanceAbout one quarter of the resistanceAbout sixteen times the resistance

Shade a strip about $1 c m$ wide and $10 c m$ long, going over it several times so the graphite layer is thick and even.
Clip one probe to the start of the strip. Press the other probe onto the strip at $2$ , $4$ , $6$ and $8 c m$ and note the resistance each time.
Shade a second strip twice as wide, with the same pressure and number of passes, and measure $8 c m$ of it.

Your readings

Length (cm)	Resistance (kΩ)
2
4
6
8
8 (double width)

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 3.3

Quick check

1 of 4

A metal wire is drawn out to three times its original length. Its volume stays the same. By what factor does its resistance change?

It triplesIt stays the sameIt becomes one thirdIt becomes nine times as large

Show answer

Answer (D) It becomes nine times as large

Length becomes

3 l

and, at constant volume, area becomes

A / 3

. So

R = ρ l / A

grows by

3 \times 3 = 9

. Tripling alone ignores the thinner cross-section.

3.5 Drift of electrons and the origin of resistivity

Why should current be proportional to voltage at all? A simple model of electrons bouncing through a lattice of ions answers this and tells us what resistivity depends on.

Start with no field. Each electron flies in a straight line until it hits an ion, then leaves with about the same speed in a random new direction. At any instant, then, the electrons' velocities point every which way. Number the $N$ conduction electrons $i = 1, 2, \dots, N$ and call their velocities at that instant $𝐯_{i}$ ; the random directions make the average vanish:

No field: zero average velocity

\frac{1}{N} \sum_{i = 1}^{N} 𝐯_{i} = 0

(3.14)

Now switch on a field $𝐄$ . Every electron, of charge $- e$ and mass $m$ , gets the same acceleration between collisions:

Acceleration of an electron

𝐚 = - \frac{e 𝐄}{m}

(3.15)

DerivationDrift velocity from collisions

Give every electron an imaginary stopwatch that resets to zero at each collision. At some instant, electron $i$ reads $t_{i}$ on its stopwatch, and it left its last collision with velocity $𝐯_{i}$ . Throughout that time it has felt the constant acceleration (3.15), so its velocity now is

𝐕_{i} = 𝐯_{i} - \frac{e 𝐄}{m} t_{i}

(3.16)

To find the drift, average this over all $N$ electrons at the same instant. The first term averages to zero: an electron leaves each collision in a random direction, so (3.14) applies to these velocities too. The stopwatches show different readings, because collisions happen at random moments, but their average is a fixed time $τ$ , the relaxation time, whatever instant we pick. So the average velocity is

Drift velocity

𝐯_{d} = ⟨ 𝐕_{i} ⟩ = - \frac{e 𝐄}{m} τ

(3.17)

Notice what this says. Every electron is accelerating, yet the average velocity does not change with time. Each collision wipes out what an electron has just gained from the field, so the crowd settles into a steady, slow drift (अपवाह वेग) opposite to $𝐄$ .

3/3After many collisions the electron has drifted a distance $v_{d} t$ against the field.

Fig. 3.5Between collisions the field bends every flight the same way. The random part averages out and a slow drift opposite to

𝐄

remains.NCERT Fig. 3.3 · p. 85

A regular grid of positive ions fills the picture. In the first step, with no field, one electron follows a zigzag path of straight flights between collisions, starting at an open circle and ending at a filled dot close to where it began. In the second step a field arrow labelled E points to the right; the old path is shown dashed and a new blue path curves slightly to the left in every flight, so it ends well to the left of the start. In the third step a thick arrow from the start to the new end point, pointing opposite to the field, is labelled v d t to show the net drift.

From drift to current. Picture a flat surface of area $A$ across the wire, at right angles to $𝐄$ (Fig. 3.6). The electrons approach it in a slow procession at speed $∣ 𝐯_{d} ∣$ . Those that will reach it within the next short time $Δ t$ are the ones now in a thin slice of length $∣ 𝐯_{d} ∣ Δ t$ beside it, on the side they come from. The slice has volume $A ∣ 𝐯_{d} ∣ Δ t$ , so with $n$ conduction electrons per unit volume it holds $n A ∣ 𝐯_{d} ∣ Δ t$ electrons, each with charge of size $e$ . The charge crossing in $Δ t$ is $I Δ t$ , so

3/3 $n A v_{d} Δ t$ electrons, each carrying charge $e$ , cross in $Δ t$ .

Fig. 3.6In time

Δ t

every electron in the shaded slice crosses the surface, giving

I = n e A v_{d}

.NCERT Fig. 3.4 · p. 86

A horizontal cylindrical wire is filled with blue dots for electrons, with a dashed oval cross-section of area A in the middle. A field arrow E points right above the wire and a drift arrow v d points left below it. In the second step a slice of the wire of length v d delta t just to the right of the cross-section is shaded and the electrons inside it are circled. In the third step all the electrons have shifted left, the circled ones are now just past the cross-section, a current arrow I points right, and the result I equals n e A v d is shown.

Current from drift speed

I = n e A ∣ 𝐯_{d} ∣

(3.18)

DerivationOhm's law from the drift model

Divide (3.18) by $A$ to get the current density, $j = n e ∣ 𝐯_{d} ∣$ , and substitute the drift speed $∣ 𝐯_{d} ∣ = e τ E / m$ from (3.17). The current density points along $𝐄$ (the electrons drift the other way, but they are negative), so

𝐣 = \frac{n e^{2} τ}{m} 𝐄

(3.22)

This has exactly the form $𝐣 = σ 𝐄$ of equation (3.13). The collision model therefore predicts Ohm's law and gives the conductivity in terms of microscopic quantities:

Conductivity of a metal

σ = \frac{n e^{2} τ}{m}

(3.23)

The prediction holds as long as $n$ and $τ$ do not depend on the field. For metals this is true over an enormous range of fields, which is why metal wires obey Ohm's law so well.

Let us put in numbers for copper: $n \approx 8.5 \times 10^{28} m^{- 3}$ and $ρ = 1 / σ \approx 1.7 \times 10^{- 8} Ω m$ . Solving (3.23) for $τ$ gives $τ = m / (n e^{2} ρ) \approx 2.5 \times 10^{- 14} s$ . An electron in copper collides with the lattice roughly forty million million times a second.

SimulationElectrons drifting in a copper wire

The top of the stage shows a circuit with a cell, a switch and a copper wire. Below it, a magnified piece of the wire shows electrons darting about and colliding, with two tracer electrons leaving trails. Sliders set the cell voltage, the load resistance and the wire's cross-sectional area, and readouts give the current, the drift speed in millimetres per second and the time to drift one metre.

Fig. 3.7The electron cloud jiggles fast but drifts slowly. The readout uses

v_{d} = I / (n e A)

for copper.

Try this

Raise the current and watch the tracer trails. How does the drift speed readout change?
Keep the current fixed and give the wire four times the cross-sectional area. What happens to the drift speed, and why?
Open the switch. Do the electrons stop moving, or only stop drifting?

Example 3.1NCERT Example 3.1 · p. 87

How slow is the drift?

A house wiring circuit for a geyser uses aluminium wire of cross-section $4.0 m m^{2}$ and carries $16 A$ . Aluminium has density $2.70 \times 10^{3} k g / m^{3}$ and molar mass $27.0 g / m o l$ , and each atom contributes three conduction electrons. (a) Find the drift speed of the electrons. (b) Compare it with (i) the typical random thermal speed of an electron at $300 K$ , estimated classically as $\sqrt{3 k_{B} T / m}$ , and (ii) the speed at which the electric signal travels along the wire, close to $3 \times 10^{8} m / s$ .

Try it yourself first. Then reveal the solution one step at a time.

The electron speed in the random motion is huge; quantum theory makes it larger still, about $10^{6} m / s$ in metals. The drift superimposed on it is tiny. Yet the current can be large because $n$ is enormous: about $10^{29}$ carriers in every cubic metre.

Example 3.2NCERT Example 3.2 · p. 88

Five claims about drifting electrons

A classmate makes five claims. Decide whether each is right, and explain. (a) “Since electrons drift only a fraction of a millimetre per second, a bulb 5 m from the switch should light up only after hours.” (b) “A steady force on an electron must give it an ever-increasing speed, so the drift speed should keep growing.” (c) “With such a small drift speed and such a small charge per electron, a large current is impossible.” (d) “When electrons drift towards higher potential, every free electron in the wire moves that way.” (e) “The path of an electron between two collisions is always a straight line.”

Try it yourself first. Then reveal the solution one step at a time.

(a) Wrong. A wire is always full of free electrons, like a hose that is already full of water. Open the tap and water pours from the far end at once, long before the water that entered at the tap arrives. Closing the switch sends a change in the field along the circuit at nearly the speed of light, so the electrons in a filament $5 m$ away are pushed within about $2 \times 10^{- 8} s$ . The current then needs only a very short time to settle to its steady value.
(b) Wrong, because it ignores collisions. An electron gains velocity along $- 𝐄$ only during its short free flight; the next collision scrambles its direction and wipes out the gain. Averaged over many flights, the extra velocity is a fixed amount, $e E τ / m$ . A marble rolling down a board studded with pegs behaves the same way: it keeps speeding up and being knocked about, and ends up with a steady average speed downhill.

Quick check

1 of 2

The electric field in a metal wire is doubled, with the temperature kept the same. What happens to the drift speed of the electrons?

It stays the sameIt doublesIt becomes four times as largeIt keeps increasing with time

Show answer

Answer (B) It doubles

v_{d} = e τ E / m

with

τ

unchanged, so

v_{d} \propto E

. It does not grow with time, because collisions keep resetting the gained velocity.

3.5.1 Mobility

Every conductor needs mobile charge carriers, but the carriers differ. A metal has electrons; a gas discharge, such as a neon sign, has electrons and positive ions; the acid in a car battery has positive and negative ions. To compare how freely different carriers respond to a field, we use their mobility (गतिशीलता), the drift speed produced by unit field:

Mobility

μ = \frac{∣ 𝐯_{d} ∣}{E}

(3.24)

Mobility is measured in $m^{2} V^{- 1} s^{- 1}$ and is always quoted as a positive number, whatever the sign of the carrier. Semiconductor data tables usually give it in $c m^{2} V^{- 1} s^{- 1}$ , which is $10^{- 4}$ of the SI unit. For electrons, (3.17) gives

Mobility and relaxation time

μ = \frac{e τ}{m}

(3.25)

Combining with (3.23) gives a compact result, $σ = n e μ$ : conductivity is the carrier density times the charge times the mobility. For copper, $μ = e τ / m \approx 4.3 \times 10^{- 3} m^{2} V^{- 1} s^{- 1}$ . Electrons in pure silicon are about thirty times more mobile, yet silicon conducts far worse because it has vastly fewer carriers.

Example

Two routes to the same drift speed

Return to the copper wire of the earlier example, which carries $5.0 A$ with a field of $0.108 V / m$ inside. Take $n = 8.5 \times 10^{28} m^{- 3}$ and $ρ = 1.7 \times 10^{- 8} Ω m$ for copper. Find the relaxation time and the mobility, and use the mobility to find the drift speed. Check the answer with $v_{d} = I / (n e A)$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A copper wire of cross-section

1.5 m m^{2}

carries

3.0 A

. With

n = 8.5 \times 10^{28} m^{- 3}

, what is the drift speed in millimetres per second?

mm/s

Show answer

Answer 0.147 mm/s

v_{d} = I / (n e A) = 3.0 / [(8.5 \times 10^{28}) (1.6 \times 10^{- 19}) (1.5 \times 10^{- 6})] = 1.47 \times 10^{- 4} m / s = 0.147 m m / s

3.6 Limitations of Ohm's law

Ohm's law describes metals and many other materials well, but plenty of useful devices break it. The departures come in three kinds, each easy to spot on a graph of current against voltage (Fig. 3.8).

$I$ is not proportional to $V$ . The graph curves. A filament lamp is the everyday example: as the current grows the filament heats up, its resistance rises, and the graph bends towards the voltage axis.
The response depends on the sign of $V$ . Reversing the voltage does not simply reverse the current. A diode passes a large current one way and almost none the other way; this one-way behaviour is used in rectifiers, which you will meet in Chapter 14.
One current, several voltages. Some materials, such as gallium arsenide (GaAs), have a graph that rises, falls and rises again, so the same current occurs at more than one voltage.

4/4A GaAs-type curve: the same current occurs at three different voltages.

Fig. 3.8An ohmic conductor gives a straight line through the origin. The other curves show the three ways a device can break Ohm's law.NCERT Figs. 3.5–3.7 · p. 89

Axes show voltage V horizontally and current I vertically, with the origin near the middle-left so negative values are visible. In the first step a green straight line passes through the origin, labelled ohmic. In the second step the straight line becomes a dashed reference and an orange curve bends towards the voltage axis at larger voltages, labelled filament lamp. In the third step a vermillion diode curve stays near zero for negative voltages and rises steeply for positive ones. In the fourth step a purple curve rises, dips and rises again; a dashed horizontal line meets it at three points, labelled one current, three voltages.

Devices that break Ohm's law are not faults to be avoided. Diodes, transistors, LEDs and solar cells all depend on non-ohmic behaviour. For the rest of this chapter, however, and for several chapters after it, every resistor is assumed to be ohmic.

3.7 Resistivity of various materials

Resistivity spans more powers of ten than almost any other property of matter: about 24 of them between silver and the best insulators. On a logarithmic scale, materials fall into three broad groups with wide gaps between them (Fig. 3.9).

Conductors. Metals and their alloys have resistivities between about $10^{- 8}$ and $10^{- 6} Ω m$ .
Semiconductors. Germanium, silicon and similar materials lie in between. Their resistivity falls as the temperature rises, and adding a tiny trace of a suitable impurity (doping) can lower it by many powers of ten. Every chip in your phone relies on this: its transistors are made by doping chosen regions of a silicon crystal.
Insulators. Glass, rubber, ceramics and plastics have resistivities from about $10^{10}$ to beyond $10^{16} Ω m$ : roughly $10^{18}$ to $10^{24}$ times that of a metal.

3/3Glass and rubber: $10^{10} Ω m$ and far beyond.

Fig. 3.9On a logarithmic scale the three groups are far apart: a rubber band resists current about

10^{23}

times more strongly than copper of the same shape.

A horizontal logarithmic axis of resistivity in ohm metre runs from ten to the minus eight to ten to the sixteen, with tick labels every four powers of ten. In the first step a narrow orange band at the far left is labelled metals, with markers for copper and nichrome. In the second step a wide green band in the middle is labelled semiconductors, with markers for germanium near one ohm metre and silicon near two thousand. In the third step a blue band at the right is labelled insulators, with markers for glass near ten to the twelve and rubber near ten to the fifteen.

Table 3.2Approximate values near room temperature. Alloys combine a fairly high resistivity with a very small

α

Material	Resistivity ρ at 20 °C (Ω m)	Temperature coefficient α (per °C)	Typical use
Silver	$1.6 \times 10^{- 8}$	$3.8 \times 10^{- 3}$	contacts, high-end connectors
Copper	$1.7 \times 10^{- 8}$	$3.9 \times 10^{- 3}$	house wiring, motor windings
Aluminium	$2.7 \times 10^{- 8}$	$3.9 \times 10^{- 3}$	overhead power lines
Tungsten	$5.6 \times 10^{- 8}$	$4.5 \times 10^{- 3}$	lamp filaments
Manganin (alloy)	$4.8 \times 10^{- 7}$	about $2 \times 10^{- 6}$	standard resistors
Nichrome (alloy)	$1.1 \times 10^{- 6}$	about $1.7 \times 10^{- 4}$	heating elements
Germanium (pure)	$0.46$	negative, large	early transistors, detectors
Silicon (pure)	about $2 \times 10^{3}$	negative, large	chips, solar cells
Glass	$10^{10}$ to $10^{14}$	—	insulating supports
Hard rubber	$10^{13}$ to $10^{16}$	—	cable insulation, gloves

Quick check

1 of 3

Arrange these materials in order of increasing resistivity.

nichrome
germanium
glass
copper

Show answer

Answer

copper
nichrome
germanium
glass

Copper (

\sim 10^{- 8} Ω m

) < nichrome (

\sim 10^{- 6}

) < germanium (

\sim 0.5

) < glass (

10^{10}

or more). An alloy is still a conductor, far below any semiconductor.

3.8 Temperature dependence of resistivity

Measure the resistance of a torch bulb's filament with a multimeter and you get a small value. Calculate it from the bulb's rating when it glows and you get a much larger one. Temperature is the reason, and materials differ greatly in how strongly, and even in which direction, it changes their resistivity.

Metals are the simplest case. Warm a metal wire through a moderate range, say a few hundred degrees, and its resistivity rises almost in proportion to the temperature rise:

Resistivity and temperature

ρ_{T} = ρ_{0} [1 + α (T - T_{0})]

(3.26)

where

$ρ_{0}$: resistivity at a reference temperature $T_{0}$
$ρ_{T}$: resistivity at temperature $T$
$α$: temperature coefficient of resistivity

Definition

Temperature coefficient of resistivity

The fractional change in resistivity per degree rise in temperature, $α = \frac{ρ_{T} - ρ_{0}}{ρ_{0} (T - T_{0})}$ . Its unit is $K^{- 1}$ (or $^{\circ} C^{- 1}$ , the same size). It is positive for metals.

ExampleFor copper near room temperature, $α \approx 3.9 \times 10^{- 3}^{\circ} C^{- 1}$ : its resistivity rises by about $0.4 %$ per degree.

Look at the copper curve in Fig. 3.10. Above about $100 K$ it is nearly straight, which is what (3.26) assumes. At lower temperatures it bends over and levels off at a small residual resistivity, set by impurities and defects in the crystal rather than by the vibrating ions. So treat (3.26) as a local fit: choose $T_{0}$ close to the temperatures you care about, and the straight line serves well.

4/4A semiconductor's resistivity falls as the temperature rises.

Fig. 3.10Metals:

ρ

rises with

T

. Alloys such as nichrome: almost no change. Semiconductors:

ρ

falls steeply as

T

rises.NCERT Figs. 3.8–3.10 · p. 90

The vertical axis shows resistivity divided by its value at 300 kelvin, from 0 to about 2.5, and the horizontal axis shows temperature from 0 to 600 kelvin. In the first step an orange copper curve starts nearly flat near zero, then rises almost linearly, reaching about 2 at 600 kelvin. In the second step a dashed straight line touches the copper curve at 300 kelvin, where a point is marked rho zero, and the formula rho T equals rho zero times one plus alpha times T minus T zero is shown; the line separates from the curve below about 100 kelvin. In the third step a blue nearly flat line for nichrome is added, staying close to 1. In the fourth step a purple semiconductor curve falls steeply from high values at low temperature to near zero at 600 kelvin.

The other two curves in the figure behave quite differently. An alloy such as nichrome, manganin or constantan barely changes: its $α$ is tens to thousands of times smaller than copper's, which is why precision resistors are wound from these alloys. A semiconductor goes the other way, its resistivity dropping steeply as it warms. The thermistor in a digital fever thermometer uses exactly this effect to sense temperature.

Why such different behaviour? Turn (3.23) upside down:

Resistivity from the drift model

ρ = \frac{1}{σ} = \frac{m}{n e^{2} τ}

(3.27)

Two quantities sit in the denominator: the carrier density $n$ and the relaxation time $τ$ . Heating shortens $τ$ in every solid, because the more violently the ions vibrate, the sooner an electron is knocked off course. What decides the outcome is $n$ . A metal already has its full set of conduction electrons, so $n$ hardly changes and the shorter $τ$ makes $ρ$ grow. In a semiconductor or an insulator, warmth shakes loose extra carriers, and $n$ rises so steeply that it swamps the fall in $τ$ : $ρ$ drops.

For a wire whose dimensions barely change, the resistance follows the same rule, $R_{T} = R_{0} [1 + α (T - T_{0})]$ . Measuring a resistance therefore measures a temperature. A platinum resistance thermometer is calibrated at the ice point and the steam point, and a reading $R_{t}$ in between gives the temperature $t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100^{\circ} C$ .

Example 3.3NCERT Example 3.3 · p. 91

How hot is a heater coil?

A room heater's nichrome coil reads $40.0 Ω$ on a multimeter in a $25.0^{\circ} C$ room; the meter's tiny current does not warm it. Switched on at $230 V$ , the heater first draws a larger current, which falls over a few seconds and settles at $5.10 A$ once the coil glows. Use $α = 1.70 \times 10^{- 4}^{\circ} C^{- 1}$ for nichrome. How hot is the glowing coil?

Try it yourself first. Then reveal the solution one step at a time.

Example 3.4NCERT Example 3.4 · p. 92

A probe for hot food

A school canteen checks that dal is served hot using a platinum probe (a Pt100 sensor). Its resistance is $100.00 Ω$ in melting ice and $138.50 Ω$ in boiling water at standard pressure. Dipped into a pot of dal, it reads $127.08 Ω$ . What is the temperature of the dal? What is the temperature coefficient of platinum?

Try it yourself first. Then reveal the solution one step at a time.

Try this

Cold filament, hot filament

You’ll need

a 12 V, 21 W bike or car bulb (or a 6 V torch bulb)
a digital multimeter
a 12 V battery or adapter (or a 6 V battery for the torch bulb)
connecting wires

Predict first

The bulb's rating tells you its hot resistance,

R = V^{2} / P

. How will the cold resistance you measure with a multimeter compare with it?

About the sameMuch largerMuch smallerExactly zero

Work out the hot resistance from the rating: for a $12 V$ , $21 W$ bulb, $R = 12^{2} / 21 \approx 6.9 Ω$ .
With the bulb disconnected and cold, measure its resistance with the multimeter.
If a meter with a current range is available, connect the bulb to the low-voltage source and let it glow. Record the current (meter in series) and the potential difference (meter across the bulb), then work out $V / I$ .

Filament resistance

Condition	Resistance (Ω)
Cold (multimeter)
Hot (from rating)
Hot (from V ÷ I)

Your entries are saved automatically.

Quick check

1 of 3

A copper coil has a resistance of

10.0 Ω

20^{\circ} C

. With

α = 3.9 \times 10^{- 3}^{\circ} C^{- 1}

, what is its resistance at

70^{\circ} C

Show answer

Answer 11.95 Ω

R = 10.0 [1 + (3.9 \times 10^{- 3}) (50)] = 10.0 \times 1.195 = 11.95 Ω

3.9 Electrical energy, power

A current warms every wire it passes through, sometimes usefully (a kettle), sometimes wastefully (a charger that gets hot). Where does this heat come from?

DerivationWhere the energy goes

Let a steady current $I$ pass through a resistor. The end where it enters is at a higher potential than the end where it leaves, by an amount $V$ . Over a short time $Δ t$ , a charge $Δ Q = I Δ t$ goes in at the high end and an equal charge comes out at the low end; nothing piles up inside. The net effect is the same as carrying $Δ Q$ straight down the potential drop $V$ , so its electric potential energy changes by

Δ U_{pot} = - Δ Q V = - I V Δ t < 0

(3.28)

If nothing held the carriers back, this lost potential energy would reappear as kinetic energy, $Δ K = I V Δ t$ , and the carriers would pick up speed along the wire. They do not: they drift at a steady $v_{d}$ . At every collision an electron hands the energy it has just gained to the ion it hits, so the lattice vibrates harder and the conductor warms up. This is Joule heating, and the heat produced in $Δ t$ is $Δ W = I V Δ t$ .

Dividing by $Δ t$ gives the power converted into heat. Using $V = I R$ gives two other useful forms:

Electrical power

P = I V

(3.32)

Power in a resistor

P = I^{2} R = \frac{V^{2}}{R}

(3.33)

This is the power that heats a toaster coil or makes a filament glow white-hot. A resistor only converts energy, so something else must supply it: the chemicals in a cell, or the generator at a power station. Which form of (3.33) is handier depends on what stays fixed. For resistors joined in series (same $I$ ), $P = I^{2} R$ shows that the largest resistance gets hottest. For appliances on the same mains supply (same $V$ ), $P = V^{2} / R$ shows that the smallest resistance draws the most power.

The SI unit of power is the watt ( $1 W = 1 J / s$ ). Electricity boards bill energy in kilowatt hours, the “units” on your bill: $1 k W h = 1000 W \times 3600 s = 3.6 \times 10^{6} J$ .

Example

The geyser and the electricity bill

A $1.5 k W$ , $230 V$ geyser runs for $40$ minutes every day. (a) What current does it draw, and what is the resistance of its element when hot? (b) How much energy does it use in a $30$ -day month, and what does that cost at ₹7 per unit?

Try it yourself first. Then reveal the solution one step at a time.

Sending power over long distances. A city's electricity may come from a coal plant or a hydroelectric dam hundreds of kilometres away, and the cables in between have resistance. Say the load at the far end must receive a power $P$ at voltage $V$ , and the cables have total resistance $R_{c}$ . They carry the same current as the load, $I = P / V$ , so the power they turn into useless heat is

Loss in transmission cables

P_{c} = I^{2} R_{c} = \frac{P^{2} R_{c}}{V^{2}}

(3.35)

For the same power delivered, the loss scales as $1 / V^{2}$ : double the voltage and the loss drops to a quarter; raise it tenfold and the loss drops a hundredfold. India's long-distance grid therefore runs at $220$ , $400$ and $765 k V$ . Such voltages would be lethal inside a home, so transformers step the voltage up as power leaves the station and back down to $230 V$ before it reaches your wiring (Fig. 3.11). Chapter 7 explains how transformers work.

3/3Near the town the voltage is stepped down to 230 V for homes.

Fig. 3.11The same power at a higher voltage needs a smaller current, so far less power is lost as heat in the lines.

A power plant stands on the left and two houses on the right, joined by a line carried on three lattice towers; the line resistance is labelled R c and the formula P loss equals I squared R c is shown at the top. In the first step there are no transformers, the line is drawn thick and hot with heat ripples above it, and the caption reads low voltage, big current, hot wires. In the second step a step-up transformer appears beside the plant, the line becomes a thin normal wire and the heat ripples vanish: high voltage, small current, tiny loss. In the third step a step-down transformer appears near the houses and their windows light up, stepped down to 230 volts for homes.

Example

Why the feeder line runs at 11 kV

A village substation must deliver $100 k W$ through a feeder line of total resistance $5.0 Ω$ . Compare the power lost in the line if the power is sent at (a) $440 V$ and (b) $11 k V$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

The current through a resistor is doubled. By what factor does the rate of heating change?

42½It does not change

Show answer

Answer (A) 4

P = I^{2} R

with

R

fixed, so doubling

I

multiplies

P

by 4. Doubling it only would treat

P

as proportional to

I

3.10 Cells, emf, internal resistance

A cell is a chemical device that keeps the charges at the ends of a circuit topped up. Two electrodes, marked positive (P) and negative (N), stand in an electrolyte. At each electrode a chemical reaction moves charge between the metal and the liquid until the electrode settles at a definite potential relative to the electrolyte touching it.

Let the positive electrode settle a potential $V_{+}$ above the electrolyte beside it, and the negative electrode a potential $V_{-}$ below the electrolyte beside it (both numbers positive). With no current there is no drop inside the electrolyte, so the only changes on the way from N to P are the two jumps: up by $V_{-}$ , then up by $V_{+}$ . Hence

Emf of a cell

ε = V_{+} + V_{-}

(3.36)

Definition

Electromotive force (emf)

The potential difference $ε$ between the terminals of a source when no current flows. It equals the work done by the source's non-electrostatic (for a cell, chemical) action in taking unit positive charge from its negative terminal to its positive terminal.

ExampleA fresh alkaline AA cell has an emf of about $1.5 V$ .

The name emf (विद्युत वाहक बल), short for electromotive force, survives from the early 1800s. An emf is really a potential difference, measured in volts, and has nothing to do with the newton.

Close the circuit with a resistor $R$ . Outside the cell, current runs from P through $R$ to N. To complete the loop it must also cross the electrolyte, from N back to P, and the electrolyte resists this like any other conductor. Its resistance $r$ is the cell's internal resistance (आंतरिक प्रतिरोध). Crossing it in the direction of the current costs a potential drop $I r$ (Fig. 3.12), so the potential difference left between the terminals, the terminal voltage, is

2/3With a load $R$ , the potential falls by $I r$ across the electrolyte, so $V = ε - I r$ .

Fig. 3.12The chemical jumps at the electrodes add up to

ε

. When current flows, the electrolyte takes away

I r

, leaving

V = ε - I r

at the terminals.NCERT Fig. 3.12 · p. 94

At the top, a beaker of electrolyte holds a blue negative electrode N on the left and an orange positive electrode P on the right. Below it, a graph shows potential against position through the cell, lined up under the electrodes. In the first step there is no external circuit: the potential jumps up by V minus at N, stays flat across the electrolyte, then jumps up by V plus at P, so P is epsilon above N. In the second step a resistor R joins the electrodes, arrows show current leaving P through R and crossing the electrolyte from N to P; the flat part now slopes down by I r, so P is only V equals epsilon minus I r above N, shown against a dashed line at epsilon. In the third step the resistor is replaced by a plain wire; the drop across the electrolyte equals epsilon and the terminal potential difference is zero.

Terminal voltage while supplying current

V = ε - I r

(3.38)

The same $V$ is the potential difference across $R$ , so $V = I R$ as well. Setting $I R = ε - I r$ gives the current:

Current from a cell

I = \frac{ε}{R + r}

(3.40)

Three consequences follow. First, a voltmeter that draws almost no current reads the emf itself. Second, the terminal voltage sags as the load draws more current. Third, joining the terminals with a thick wire ( $R \approx 0$ ) gives the largest current the cell can drive, $I_{\max} = ε / r$ . Such a short circuit overheats the cell and can ruin it, and large batteries can start fires this way, so makers specify a safe maximum current far below $ε / r$ .

If an external source pushes current into the positive terminal, as when a battery is being charged, the drop across $r$ is reversed and the terminal voltage is larger than the emf:

Terminal voltage while charging

V = ε + I r

(3.60)

Internal resistance can be tiny or large. A lead-acid car battery has around $0.01 Ω$ , so it can push well over a hundred amperes through a starter motor. A small dry cell has far more, and its $r$ grows as it ages. That is why an old torch cell can still read $1.5 V$ on a voltmeter yet light a bulb only dimly: once current flows, most of its emf is lost across its own $r$ .

SimulationCell with internal resistance

A cell of emf 6 volts with an internal resistance of 1 ohm drives a network in which R1 is in series with R2 and R3 in parallel. An ammeter shows the total current, and a voltmeter can be placed across the cell or across any part of the network. Moving markers show the current in each branch, and sliders change the emf and each resistance.

Fig. 3.13A

6 V

cell with

r = 1 Ω

drives

R_{1}

in series with

R_{2}

and

R_{3}

in parallel.

Try this

Put the voltmeter across the cell. Open the switch and read the emf, then close it. How much does the terminal voltage drop?
Lower R1 step by step. What happens to the current and to the terminal voltage? Compare with $V = ε - I r$ .
Watch the markers at the junction after R1. Is the current in R1 equal to the currents in R2 and R3 added together?

Example

Emf and internal resistance from two readings

A voltmeter connected across a cell reads $1.20 V$ when the cell drives a $4.0 Ω$ resistor and $1.35 V$ when it drives a $9.0 Ω$ resistor. Find the emf and the internal resistance of the cell. Assume the voltmeter draws no current.

Try it yourself first. Then reveal the solution one step at a time.

Try this

Find the emf and internal resistance of a cell

You’ll need

a used AA cell in a holder
a digital multimeter
resistors of about 10 Ω, 4.7 Ω and 2.2 Ω (rated 2 W or more)
a push switch and connecting wires

Predict first

Compared with its reading on open circuit, what will the voltmeter show when the cell drives the

2.2 Ω

resistor?

A higher voltageThe same voltageZeroA lower voltage

Measure the voltage across the cell with nothing else connected. This is very nearly its emf.
Connect one resistor across the cell through the push switch. Press the switch, read the voltage quickly and release it.
Repeat with the other two resistors. Work out each current as $I = V / R$ .
Plot $V$ against $I$ . The intercept on the $V$ axis is $ε$ ; the slope is $- r$ .

Cell under load

Load R (Ω)	Voltage V (V)	Current I = V/R (A)
none (open)		0
10
4.7
2.2

Your entries are saved automatically.

Can’t do it now? Try the simulation · Fig. 3.13

3.11 Cells in series and in parallel

A torch uses two cells end to end; a solar lantern may use several cells side by side. Just as resistors combine into one equivalent resistor, a group of cells can be replaced by one equivalent cell with emf $ε_{e q}$ and internal resistance $r_{e q}$ , as far as the rest of the circuit is concerned (Fig. 3.14).

DerivationTwo cells in series

Join the negative terminal of cell 1 (emf $ε_{1}$ , internal resistance $r_{1}$ ) to the positive terminal of cell 2 ( $ε_{2}$ , $r_{2}$ ). Call the free positive terminal A, the joint B and the free negative terminal C. With a current $I$ flowing out of A through an external circuit, (3.38) applied to each cell gives $V (A) - V (B) = ε_{1} - I r_{1}$ and $V (B) - V (C) = ε_{2} - I r_{2}$ . Adding,

V_{A C} = (ε_{1} + ε_{2}) - I (r_{1} + r_{2})

(3.43)

A single cell between A and C would give $V_{A C} = ε_{e q} - I r_{e q}$ . Comparing the two expressions for every value of $I$ :

Cells in series

ε_{e q} = ε_{1} + ε_{2}, r_{e q} = r_{1} + r_{2}

(3.45)

Now turn cell 2 round, so that the two negative terminals meet at B. The current then passes through cell 2 from its positive terminal to its negative one, which gives $V (B) - V (C) = - ε_{2} - I r_{2}$ and hence $V_{A C} = (ε_{1} - ε_{2}) - I (r_{1} + r_{2})$ . Comparing with one equivalent cell as before:

ε_{e q} = ε_{1} - ε_{2} (ε_{1} > ε_{2}), r_{e q} = r_{1} + r_{2}

(3.47)

The same bookkeeping works for a string of any length. Walk along it in the direction of the current: each cell you cross from − to + adds its emf, each cell you cross from + to − subtracts it, and every internal resistance adds, whichever way its cell faces.

3/3Positive terminals joined: $1 / r_{e q} = 1 / r_{1} + 1 / r_{2}$ and $ε_{e q} / r_{e q} = ε_{1} / r_{1} + ε_{2} / r_{2}$ .

Fig. 3.14In series, emfs add (with signs) and internal resistances add. In parallel, conductances

1 / r

add and so do the ratios

ε / r

.NCERT Figs. 3.13–3.14 · p. 95

Three shaded rows show pairs of cells with the long plate as the positive terminal. In the first step the top row shows cells joined in series between points A, B and C, both facing the same way, with the results epsilon eq equals epsilon 1 plus epsilon 2 and r eq equals r 1 plus r 2. In the second step a second row appears in which the second cell is reversed, giving epsilon eq equals epsilon 1 minus epsilon 2 with r eq still r 1 plus r 2. In the third step a third row shows two cells side by side with their positive terminals joined at point X on the left and their negative terminals joined at point Y on the right, and the results one over r eq equals one over r 1 plus one over r 2 and epsilon eq over r eq equals epsilon 1 over r 1 plus epsilon 2 over r 2.

DerivationTwo cells in parallel

Now join the two positive terminals to a common point X and the two negative terminals to a common point Y (Fig. 3.14, last step). Let $I_{1}$ and $I_{2}$ be the currents leaving the cells' positive terminals, and $I$ the current flowing from X into the external circuit. Charge does not pile up at X, so

I = I_{1} + I_{2}

(3.48)

Both cells have the same potential difference $V = V (X) - V (Y)$ across them, so $V = ε_{1} - I_{1} r_{1} = ε_{2} - I_{2} r_{2}$ . Solve each for its current and add:

I = \frac{ε_{1} - V}{r_{1}} + \frac{ε_{2} - V}{r_{2}} \Rightarrow V = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}} - I \frac{r_{1} r_{2}}{r_{1} + r_{2}}

(3.52)

Compare with $V = ε_{e q} - I r_{e q}$ for one equivalent cell:

ε_{e q} = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}}, r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}}

(3.54)

These are easier to remember in the form

Cells in parallel

\frac{1}{r_{e q}} = \frac{1}{r_{1}} + \frac{1}{r_{2}}, \frac{ε_{e q}}{r_{e q}} = \frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}}

(3.57)

With more cells side by side, each sum simply gains more terms. A cell connected the other way round (its negative terminal on the positive rail) enters with its emf reversed, $ε \to - ε$ .

Two special cases are worth remembering. $n$ identical cells in series give $n ε$ and $n r$ : a higher voltage. $m$ identical cells in parallel give $ε$ and $r / m$ : the same voltage with a lower internal resistance, so they can supply more current and last longer.

Example

A strong cell and a weak cell in parallel

A cell of emf $1.5 V$ and internal resistance $0.50 Ω$ is connected in parallel (positive to positive) with a cell of emf $1.2 V$ and internal resistance $0.30 Ω$ . The pair drives a $2.0 Ω$ resistor. (a) Find the equivalent emf and internal resistance, and the current in the resistor. (b) How much current does each cell supply? (c) What happens if the resistor is removed?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Four cells, each of emf

1.5 V

and internal resistance

0.25 Ω

, are connected in series to a

5.0 Ω

lamp. What current flows?

Show answer

Answer 1 A

ε_{e q} = 6.0 V

and

r_{e q} = 1.0 Ω

, so

I = 6.0 / (5.0 + 1.0) = 1.0 A

3.12 Kirchhoff's rules

Series and parallel rules take you only so far. Put cells in two different branches, or join resistors in a bridge, and no amount of combining will simplify the network. Two rules stated by Gustav Kirchhoff, Kirchhoff's rules (किरखोफ़ के नियम), handle every such case.

Before using them, set up your labels. Mark the + and − terminals of every cell. Then give each branch a current symbol with an arrow. The arrow is only a guess at the direction, and the algebra will correct it: a current that comes out negative is simply flowing against its arrow. With the labels in place, these sign rules give the potential change across each element as you walk through it:

Through a resistor along the arrow of its current, the potential falls by $I R$ ; walking against the arrow, it rises by $I R$ .
Through a cell from its negative to its positive terminal, the potential rises by $ε$ ; from positive to negative, it falls by $ε$ .
A cell's internal resistance is treated like any other resistor. If the current $I$ in the cell flows from N to P, then $V (P) - V (N) = ε - I r$ , as in (3.38). If it flows from P to N, $V (P) - V (N) = ε + I r$ , as in (3.60).

LawJunction rule

At any junction, the total current flowing in equals the total current flowing out.

Suppose more charge entered a junction each second than left it. Charge would build up there, the potential of the junction would keep rising, and the currents would change: the circuit would not be steady. In a steady state, therefore, charge leaves every junction exactly as fast as it arrives, so the junction rule is conservation of charge. It holds at any point along a single wire too, however the wire bends: the current flowing into the point equals the current flowing out.

2/3 $3 + 2 = 4 + I$ , so the unknown current is $1 A$ .

Fig. 3.15Charge does not collect at a junction, so the currents in must add up to the currents out. The same holds at any point on a wire.NCERT Fig. 3.15 · p. 98

Four wires meet at a node in the middle of the picture. Arrows show 3 amperes flowing in from the left and 2 amperes flowing in from above, 4 amperes flowing out to the right, and an unknown current I flowing out along a wire to the lower right. In the second step the node is highlighted, the equation 3 plus 2 equals 4 plus I is shown, and the unknown current is revealed as 1 ampere. In the third step the junction fades and a single bent wire is shown with a marked point on it; the same current I flows into and out of the point.

LawLoop rule

Around any closed loop in a circuit, the rises and falls in potential add up to zero.

5/5The wire from d to a has no resistance. The changes add to zero: $+ 10 - 6 - 4 = 0$ .

Fig. 3.16Walking once round the loop: up

10 V

across the cell (its

12 V

emf minus

2 V

lost inside), down

6 V

and

4 V

across the resistors. The total change is zero.NCERT Fig. 3.15 · p. 98

On the left is a rectangular loop with corners a, b, c and d. The left side holds a 12 volt cell with a 1 ohm internal resistance inside a dashed box, the top side a 3 ohm resistor and the right side a 2 ohm resistor; arrows show a clockwise current of 2 amperes. On the right is a graph of potential at points a, b, c, d and back to a. In the first step a dot sits at a and the graph shows only zero volts at a. In the second step the dot moves to b and the graph rises to 12 volts then falls to 10 volts across the internal resistance. In the third step the dot reaches c and the graph falls to 4 volts across the 3 ohm resistor. In the fourth step the dot reaches d and the graph falls to zero across the 2 ohm resistor. In the fifth step the dot is back at a and the sum plus 10 minus 6 minus 4 equals 0 is shown.

Quick check

1 of 2

At a junction, currents of

5 A

and

2 A

flow in, and

4 A

flows out along one wire. What current flows out along the only other wire?

Show answer

Answer 3 A

In = out:

5 + 2 = 4 + I

, so

I = 3 A

A method that always works.

Draw the circuit, label each branch current with an arrow, and use the junction rule straight away so that you introduce as few unknown currents as possible.
Choose enough independent loops: one for each unknown that remains. A loop is independent if it passes through at least one branch not used in the earlier loops.
Walk round each loop in a direction of your choice, add up the potential changes with the sign rules, and set the sum to zero.
Solve the simultaneous equations. A negative current flows against its arrow. Check your answers with a loop you did not use.

Example 3.5NCERT Example 3.5 · p. 98

Six resistors, one clever idea

Six resistors, each of $6.0 Ω$ , connect every pair of the four corners P, Q, R and S of a square: one along each side and one along each diagonal (the diagonals cross without touching). The network is driven by a $9.0 V$ battery, with negligible internal resistance, joined to the neighbouring corners P and Q as shown. Find the equivalent resistance between P and Q and the current in each resistor.

Fig. 3.17Every resistor is

6.0 Ω

. The diagonals cross without touching.NCERT Fig. 3.16 · p. 98

Four corners labelled S and R at the top and P and Q at the bottom are joined by resistors along all four sides and along both diagonals, with the two diagonals crossing at a small hop so they do not touch. A 9.0 volt battery is connected below, between P and Q.

Try it yourself first. Then reveal the solution one step at a time.

That shortcut worked only because the network is so symmetric: all four corners are alike, so the same $3.0 Ω$ appears across any two of them. The next network has two cells and no symmetry at all, so we write out and solve the full set of equations.

Example 3.6NCERT Example 3.6 · p. 99

A network with two cells

Find all the branch currents in the network below. The $12 V$ cell has negligible internal resistance; the $6 V$ cell has an internal resistance of $1 Ω$ . The whole bottom rail E is one node.

Fig. 3.18Assumed directions:

I_{1}

from A to B,

I_{2}

from B to C,

I_{3}

from C to D.NCERT Fig. 3.17 · p. 99

Top nodes A, B, C and D are joined by resistors of 3, 2 and 1 ohm, with current arrows I 1, I 2 and I 3 pointing to the right. From A a 12 volt cell, positive at the top, goes down to the bottom rail E. From B a 6 ohm resistor and from C a 2 ohm resistor go down to the rail. From D a 6 volt cell, positive at the top, with a 1 ohm internal resistance in a dashed box, goes down to the rail.

Try it yourself first. Then reveal the solution one step at a time.

Try this

Check Kirchhoff's rules in the simulation

Open the circuit simulation and note the ammeter reading, the total current.
Find the current through $R_{2}$ and through $R_{3}$ . On a wide screen it is printed beside each resistor; on a phone, tap a resistor to move the voltmeter across it and divide the reading by its resistance. Do the two currents add up to the current through $R_{1}$ ?
Put the voltmeter across $R_{1}$ , then across the parallel pair, then across the cell. Does the terminal voltage equal the sum of the other two readings?

Can’t do it now? Try the simulation · Fig. 3.13

Quick check

1 of 3

Walking round a loop in the direction of its current, you pass through a cell from its positive terminal to its negative terminal. How does the potential change across the cell's emf?

It rises by

ε

It changes by

I r

onlyIt falls by

ε

It does not change

Show answer

Answer (C) It falls by $ε$

From + to − the potential falls by

ε

, whatever the current does. The current only decides the sign of the

I r

term for the internal resistance.

3.13 Wheatstone bridge

Kirchhoff's rules explain one of the most useful circuits in physics, the Wheatstone bridge (व्हीटस्टोन सेतु). It finds an unknown resistance by comparing it with known ones (Fig. 3.19). Four resistors, $P$ from A to B, $Q$ from B to C, $R$ from A to D and $S$ from D to C, make a closed diamond. A cell joins the opposite corners A and C; this branch is the battery arm. A galvanometer G, a sensitive current detector, joins the other two corners, B and D; this branch is the galvanometer arm.The NCERT textbook names the arms $R_{1}$ (AD), $R_{2}$ (AB), $R_{3}$ (DC) and $R_{4}$ (BC) and writes the balance condition as $R_{2} / R_{1} = R_{4} / R_{3}$ . That is the same condition as $P / Q = R / S$ below.

3/3Adjust the resistors until $V_{B} = V_{D}$ : no current in G, and $P / Q = R / S$ .

Fig. 3.19When

P / Q = R / S

, points B and D are at the same potential and the galvanometer reads zero: the bridge is balanced.NCERT Fig. 3.18 · p. 101

Four resistors form a diamond: P from A on the left to B at the top, Q from B to C on the right, R from A to D at the bottom and S from D to C. A galvanometer with a needle is connected from B straight down to D, and a cell below the diamond joins A and C. In the first step the battery arm and galvanometer arm are highlighted and labelled. In the second step arrows show current flowing from B to D through the galvanometer, the needle is deflected to the right and a label reads V B is not equal to V D, current in G. In the third step the needle points straight up at zero and the labels read V B equals V D and P over Q equals R over S, no deflection.

With arbitrary resistors the needle deflects, because some current takes the short cut through G between B and D. By adjusting the resistors, though, we can make the galvanometer current $I_{g}$ exactly zero. That special setting, the balanced bridge, is what makes the circuit useful.

DerivationThe balance condition

Suppose $I_{g} = 0$ . The junction rule at B says the current $I_{1}$ that comes through $P$ carries on, unchanged, through $Q$ . At D, the current $I_{2}$ through $R$ carries on through $S$ . Nothing in the argument depends on the cell, so its emf and internal resistance can be anything.

Apply the loop rule to loop ABDA (through $P$ , then G, then back through $R$ ). There is no drop across G because no current flows in it:

- I_{1} P + 0 + I_{2} R = 0 \Rightarrow I_{1} P = I_{2} R

(3.62)

Now loop BCDB (through $Q$ , then back through $S$ and G):

- I_{1} Q + I_{2} S + 0 = 0 \Rightarrow I_{1} Q = I_{2} S

(3.63)

Dividing the first result by the second removes both currents:

Balance condition

\frac{P}{Q} = \frac{R}{S}

(3.64(a))

In words: the potential drops across $P$ and $R$ are equal, so B and D are at the same potential, and nothing drives a current through G.

In practice the unknown resistance goes in as $S$ . $P$ and $Q$ , the ratio arms, are fixed known resistors, and a variable known resistance $R$ is adjusted until the galvanometer needle rests at zero, the null point. Then

Unknown resistance

S = R \frac{Q}{P}

(3.64(b))

A null method has real advantages. Deciding whether a needle is exactly at zero is far easier than reading a value accurately. The balance condition does not involve the emf of the cell or the resistance of the galvanometer, so neither needs to be known or kept steady. The bridge is most sensitive when all four arms have similar resistances. In the school laboratory the meter bridge uses the same idea: a uniform wire one metre long replaces $P$ and $Q$ , and sliding a contact along it changes their ratio.

Fig. 3.20Choose a mystery resistor and a pair of ratio arms, then adjust

R

until the galvanometer reads zero.

Example 3.7NCERT Example 3.7 · p. 101

Current through an unbalanced bridge

In the bridge shown, $P = 10 Ω$ (AB), $Q = 20 Ω$ (BC), $R = 15 Ω$ (AD) and $S = 20 Ω$ (DC). A galvanometer of resistance $40 Ω$ is connected between B and D, and a $6.0 V$ cell of negligible internal resistance across A and C. Find the current through the galvanometer. What value of $S$ would balance the bridge?

Fig. 3.21An unbalanced bridge:

10 / 20 \neq 15 / 20

.NCERT Fig. 3.19 · p. 101

A diamond of four resistors: 10 ohm from A to B, 20 ohm from B to C, 15 ohm from A to D and 20 ohm from D to C. A galvanometer of 40 ohm joins B and D, and a 6.0 volt cell below joins A and C.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

A bridge balances with

P = 10 Ω

Q = 40 Ω

and

R = 25 Ω

. What is the unknown resistance

S

Show answer

Answer 100 Ω

S = R Q / P = 25 \times 40 / 10 = 100 Ω

Table 3.3Quantities at a glance

Quantity	Symbol	SI unit	Dimensions	Defined by
Electric current	$I$	A	$[A]$	SI base quantity
Charge	$q$ , $Q$	C	$[T A]$	$q = I t$
Potential difference, emf	$V$ , $ε$	V	$[M L^{2} T^{- 3} A^{- 1}]$	work per unit charge
Resistance	$R$	Ω	$[M L^{2} T^{- 3} A^{- 2}]$	$R = V / I$
Resistivity	$ρ$	Ω m	$[M L^{3} T^{- 3} A^{- 2}]$	$R = ρ l / A$
Conductivity	$σ$	S/m	$[M^{- 1} L^{- 3} T^{3} A^{2}]$	$σ = 1 / ρ$
Electric field	$E$	V/m	$[M L T^{- 3} A^{- 1}]$	force per unit charge
Current density	$j$	A/m²	$[L^{- 2} A]$	$j = I / A$
Drift speed	$v_{d}$	m/s	$[L T^{- 1}]$	$v_{d} = e E τ / m$
Relaxation time	$τ$	s	$[T]$	mean time between collisions
Mobility	$μ$	m² V⁻¹ s⁻¹	$[M^{- 1} T^{2} A]$	$μ = v_{d} / E$
Power	$P$	W	$[M L^{2} T^{- 3}]$	$P = I V$

Flash cards · 10

Key relations to remember

1 / 10

Recap

Current Electricity: key points and summary

Current is the net charge crossing a surface per unit time, $I = d Q / d t$ ; it is a scalar measured in amperes (Section 3.2).
A steady current needs a closed circuit and a source, such as a cell, that keeps a steady field inside the conductor (Section 3.3).
Ohm's law: at constant temperature, $V = I R$ for a metal. Resistance depends on shape through $R = ρ l / A$ , and locally $𝐣 = σ 𝐄$ (Section 3.4).
Collisions turn the field's push into a steady drift, $𝐯_{d} = - e τ 𝐄 / m$ , giving $I = n e A v_{d}$ and $ρ = m / (n e^{2} τ)$ (Section 3.5).
Drift speeds are tiny (about a millimetre per second or less) but $n$ is enormous; mobility $μ = v_{d} / E = e τ / m$ (Section 3.5.1).
$V = I R$ defines resistance for any device; Ohm's law holds only when $R$ does not depend on $V$ . Lamps, diodes and GaAs devices break it (Section 3.6).
Metals have $ρ \sim 10^{- 8}$ to $10^{- 6} Ω m$ , insulators $10^{18}$ to $10^{24}$ times more, and semiconductors lie in between (Section 3.7).
For metals $ρ_{T} = ρ_{0} [1 + α (T - T_{0})]$ with $α > 0$ ; alloys have tiny $α$ ; semiconductors have $ρ$ falling with temperature (Section 3.8).
Power converted to heat is $P = I V = I^{2} R = V^{2} / R$ ; transmitting at high voltage cuts line losses, which scale as $1 / V^{2}$ (Section 3.9).
A cell's emf is its open-circuit terminal voltage; with internal resistance $r$ , $V = ε - I r$ and $I = ε / (R + r)$ (Section 3.10).
Cells in series: emfs add with sign and internal resistances add. In parallel: $1 / r_{e q} = \sum 1 / r_{i}$ and $ε_{e q} / r_{e q} = \sum ε_{i} / r_{i}$ (Section 3.11).
Junction rule (charge conservation) and loop rule (single-valued potential) solve any circuit (Section 3.12).
A Wheatstone bridge balances when $P / Q = R / S$ , giving an unknown resistance by a null method (Section 3.13).

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 3.1	Practice Q1
NCERT Exercise 3.2	Practice Q2
NCERT Exercise 3.3	Practice Q3
NCERT Exercise 3.4	Practice Q4
NCERT Exercise 3.5	Practice Q5
NCERT Exercise 3.6	Practice Q6
NCERT Exercise 3.7	Practice Q7
NCERT Exercise 3.8	Practice Q8
NCERT Exercise 3.9	Practice Q9

Quick check: MCQs on Current Electricity

3.2 Electric current3 questions
3.3 Electric currents in conductors3 questions
3.4 Ohm's law4 questions
3.5 Drift of electrons and the origin of resistivity2 questions
3.5.1 Mobility3 questions
3.6 Limitations of Ohm's law3 questions
3.7 Resistivity of various materials3 questions
3.8 Temperature dependence of resistivity3 questions
3.9 Electrical energy, power3 questions
3.10 Cells, emf, internal resistance4 questions
3.11 Cells in series and in parallel3 questions
3.12 Kirchhoff's rules2 questions
3.12 Kirchhoff's rules3 questions
3.13 Wheatstone bridge4 questions

Glossary · 14

Key terms (English–Hindi)

conductivityचालकताchaalakta: The reciprocal of resistivity, $σ = 1 / ρ$ , in siemens per metre, with $𝐣 = σ 𝐄$ .First seen in the chapter →
current densityधारा घनत्वdhaara ghanatva: Current per unit area of a surface normal to the flow, $j = I / A$ , in $A / m^{2}$ . It is a vector.First seen in the chapter →

drift velocityअपवाह वेगapavaah veg: The small average velocity of charge carriers in a field, $𝐯_{d} = - e τ 𝐄 / m$ for electrons, on top of their random motion.First seen in the chapter →

electric currentविद्युत धाराvidyut dhaara: The net charge crossing a surface per unit time, $I = d Q / d t$ , measured in amperes.First seen in the chapter →
emfविद्युत वाहक बलvidyut vaahak bal: The terminal potential difference of a source on open circuit: the work its chemical (or other) action does per unit charge. Measured in volts.First seen in the chapter →

internal resistanceआंतरिक प्रतिरोधaantarik pratirodh: The resistance $r$ inside a cell, such as that of its electrolyte, which causes a drop $I r$ when a current flows.First seen in the chapter →

Kirchhoff's rules: Two circuit rules: the currents into a junction equal those out, and the potential changes round any closed loop add to zero.First seen in the chapter →

mobilityगतिशीलताgatisheelta: Drift speed per unit field, $μ = v_{d} / E = e τ / m$ , in $m^{2} V^{- 1} s^{- 1}$ .First seen in the chapter →

relaxation time: The average time $τ$ since the last collision of the conduction electrons; about $10^{- 14} s$ in metals.First seen in the chapter →
resistanceप्रतिरोधpratirodh: The ratio $R = V / I$ for a conductor, measured in ohms ( $Ω$ ).First seen in the chapter →
resistivityप्रतिरोधकताpratirodhakta: The material property $ρ$ in $R = ρ l / A$ , in $Ω m$ . It depends on the material and temperature, not on shape.First seen in the chapter →

temperature coefficient of resistivityप्रतिरोधकता ताप गुणांकpratirodhakta taap gunaank: The fractional change in resistivity per degree, $α$ in $ρ_{T} = ρ_{0} [1 + α (T - T_{0})]$ . Positive for metals.
terminal voltage: The potential difference across a source while it supplies a current $I$ : $V = ε - I r$ .First seen in the chapter →

Wheatstone bridgeव्हीटस्टोन सेतुwheatstone setu: Four resistors in a diamond, with a cell across one diagonal and a galvanometer across the other. It balances when $P / Q = R / S$ .First seen in the chapter →

Current Electricity: Class 12 Physics · Chapter 3

What you will learn in Current Electricity

3.1 Introduction

3.2 Electric current

3.3 Electric currents in conductors

3.4 Ohm's law

3.5 Drift of electrons and the origin of resistivity

3.5.1 Mobility

3.6 Limitations of Ohm's law

3.7 Resistivity of various materials

3.8 Temperature dependence of resistivity

3.9 Electrical energy, power

3.10 Cells, emf, internal resistance

3.11 Cells in series and in parallel

3.12 Kirchhoff's rules

3.13 Wheatstone bridge

Current Electricity: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Current Electricity

Key terms (English–Hindi)

You reached the end