Chapter 3 ยท Electricity

3.1 Electric Charge, Current and Potential Difference

Electric current is charge on the move: $I=\frac Qt$, counted in coulombs per second. A cell keeps it moving by doing work on every coulomb โ€” the potential difference $V=\frac WQ$, measured in volts.

A phone charger is labelled "5 V, 2 A"; a torch cell says "1.5 V"; the overhead wire of an electric train carries $25\,000\ \text{V}$. These two numbers โ€” a current in amperes and a potential difference in volts โ€” describe every electrical device you will ever use. The current tells you how much charge flows past each second; the potential difference tells you how much energy each unit of charge carries from the source to the device. Multiply them and, as Section 3.4 will show, you get the power.

In this section we start from the electric charge itself and the tiny electron that carries it in metal wires. We define current and the coulomb, learn the symbols used to draw circuits, define potential difference and the volt, and finish with the practical skill every board practical tests: connecting an ammeter and a voltmeter correctly and reading their scales.

Electric Charge and Electric Current

Rub a plastic comb on dry hair and it picks up bits of paper: it has gained electric charge. Charge comes in two kinds, positive and negative; like charges repel and unlike charges attract. Every atom contains positively charged protons in its nucleus and negatively charged electrons around it. In a metal, some electrons are only loosely held: these free electrons wander through the whole wire. That is what makes metals good conductors. In insulators such as rubber, plastic and glass, almost every electron is tightly bound and charge cannot flow.

When a wire is connected across a cell, the cell pushes all the free electrons along the wire at once. The wire was full of them before the switch closed, so the bulb lights immediately โ€” no electron has to travel from the cell to the bulb. The amount of charge crossing any cross-section of the wire each second is the electric current.

๐Ÿ“˜ Definition 3.1 โ€” Electric current

If a net charge $Q$ flows through any cross-section of a conductor in time $t$, the electric current is the rate of flow of charge:

$$I=\frac Qt .$$

The SI unit of current is the ampere (A): a current of $1\ \text{A}$ is a flow of $1$ coulomb of charge per second, $1\ \text{A}=1\ \text{C/s}$. Small currents are measured in milliamperes ($1\ \text{mA}=10^{-3}\ \text{A}$) and microamperes ($1\ \mu\text{A}=10^{-6}\ \text{A}$). Current is measured with an ammeter.

๐Ÿ“˜ Definition 3.2 โ€” The coulomb and the charge of an electron

The SI unit of charge is the coulomb (C). Charge always comes in whole-number multiples of the charge of one electron, whose size is

$$e=1.6\times10^{-19}\ \text{C}.$$

So a charge $Q$ is carried by $n=\dfrac Qe$ electrons, and $Q=ne$. One coulomb is the charge of $\dfrac{1}{1.6\times10^{-19}}=6.25\times10^{18}$ electrons. The electron's charge is negative; when we only count electrons we use its size, $e$.

Which way does current flow? Long before the electron was discovered, scientists agreed that current flows from the positive terminal of a cell, round the circuit, to the negative terminal. This is the conventional current, and it is the direction we always mark with arrows on circuit diagrams. In a metal wire the moving charges are electrons, and being negative they drift the opposite way โ€” from the negative terminal towards the positive. Both descriptions give the same current; just remember that electron flow and conventional current point in opposite directions.

Interactive 1 โ€” Counting Charge at a Gate

A battery drives current through a $3.0\ \Omega$ resistor. The clock runs in slow motion while the counter records the charge that crosses the dashed gate; each moving dot stands for $0.1\ \text{C}$ (about $6.25\times10^{17}$ electrons). Watch $Q/t$ settle on the ammeter reading. Change the number of cells, open the key, or switch between electron flow and conventional current. Play runs the clock; Reset starts the count again.

2
ammeter $I$โ€”
time $t$โ€”
charge counted $Q$โ€”
$Q/t$โ€”
electrons $n=Q/e$โ€”
โœ๏ธ Example 1 โ€” Current from charge and time
A charge of $30\ \text{C}$ flows through a bulb in $1.0$ minute. Find the current.
Units $Q=30\ \text{C}$, $t=1.0\ \text{min}=60\ \text{s}$ โ€” time must be in seconds.
Formula $I=\dfrac Qt=\dfrac{30\ \text{C}}{60\ \text{s}}=0.50\ \text{A}$.
Check Half a coulomb each second: in $60\ \text{s}$ that is $30\ \text{C}$. โœ“
โœ๏ธ Example 2 โ€” Counting electrons
A current of $0.40\ \text{A}$ flows through a fan for $5.0$ minutes. How much charge passes, and how many electrons is that?
Charge $t=5.0\times60=300\ \text{s}$, so $Q=It=0.40\times300=120\ \text{C}$.
Electrons $$n=\frac Qe=\frac{120}{1.6\times10^{-19}}=7.5\times10^{20}.$$
Meaning Seven hundred and fifty million million million electrons in five minutes โ€” which is why we count charge in coulombs, not in electrons.
โœ๏ธ Example 3 โ€” How long does it take?
How long does it take for $1.0\times10^{20}$ electrons to pass a point in a wire carrying a current of $0.80\ \text{A}$?
Charge $Q=ne=1.0\times10^{20}\times1.6\times10^{-19}=16\ \text{C}$.
Time $t=\dfrac QI=\dfrac{16\ \text{C}}{0.80\ \text{A}}=20\ \text{s}$.
Check $0.80\ \text{A}\times20\ \text{s}=16\ \text{C}$, and $16/(1.6\times10^{-19})=1.0\times10^{20}$. โœ“

Electric Circuits and Circuit Diagrams

A current flows only around a complete, unbroken loop โ€” an electric circuit. Open the switch (in the laboratory usually a plug key) and the loop is broken: the current stops everywhere at once, not just after the gap. Electricians and examiners do not draw realistic pictures of cells and bulbs; they use a small set of standard circuit symbols, joined by straight lines for the connecting wires. Learn these โ€” every circuit question uses them.

An electric cell (long line $+$, short thick line $-$)
A battery: cells joined $+$ to $-$
Plug key or switch, open (OFF)
Plug key or switch, closed (ON)
Wires joined (a junction dot)
Wires crossing without joining
An electric bulb
A resistor of resistance $R$
A variable resistor or rheostat
An ammeter
A voltmeter

A circuit diagram is a neat drawing of a circuit with these symbols and straight wires. Draw it as a rectangle where possible, put the cell or battery on one side, mark the direction of the conventional current with an arrow from the $+$ terminal, and label each component ($K$ for the key, $R$ for a resistor, โ’ถ and โ“‹ for the meters).

Electric Potential Difference

Water flows along a level pipe only if something makes the pressure at one end higher than at the other โ€” a pump, or a tank on the roof. Charge is the same: it flows along a wire only if there is a difference of electric potential between the ends. A cell creates and keeps up this difference by chemical reactions inside it. Each time a charge goes round the circuit, the cell does work on it, and the charge hands that energy over to the bulb, heater or motor it passes through. The work done per unit charge is the potential difference.

๐Ÿ“˜ Definition 3.3 โ€” Electric potential difference

The potential difference $V$ between two points of a circuit is the work $W$ done in moving a unit positive charge from one point to the other:

$$V=\frac WQ .$$

The SI unit is the volt (V): the potential difference between two points is $1\ \text{V}$ when $1\ \text{J}$ of work is done in moving $1\ \text{C}$ of charge between them, $1\ \text{V}=1\ \text{J/C}$. Potential difference is often called voltage, and it is measured with a voltmeter. Rearranged, the work done (energy transferred) is $W=VQ$.

So a $1.5\ \text{V}$ cell gives $1.5\ \text{J}$ of energy to every coulomb that passes through it, and a $12\ \text{V}$ car battery gives $12\ \text{J}$ per coulomb. Joining cells $+$ to $-$ in a battery adds their voltages: two $1.5\ \text{V}$ cells in a torch give $3.0\ \text{V}$.

โœ๏ธ Example 4 โ€” Potential difference from work
$24\ \text{J}$ of work is done in moving a charge of $2.0\ \text{C}$ from one end of a heater coil to the other. Find the potential difference across the coil.
Formula $V=\dfrac WQ=\dfrac{24\ \text{J}}{2.0\ \text{C}}=12\ \text{V}$.
Meaning Every coulomb that passes through the coil delivers $12\ \text{J}$ of energy to it.
โœ๏ธ Example 5 โ€” Energy from a battery
A $12\ \text{V}$ battery supplies a current of $0.50\ \text{A}$ to a lamp for $2.0$ minutes. How much charge flows, and how much energy does the battery give to the lamp?
Charge $t=2.0\times60=120\ \text{s}$; $Q=It=0.50\times120=60\ \text{C}$.
Energy $W=VQ=12\ \text{V}\times60\ \text{C}=720\ \text{J}$.
Check Together, $W=VIt=12\times0.50\times120=720\ \text{J}$. โœ“ (Section 3.4 turns $VI$ into the power.)

Measuring Current and Potential Difference

An ammeter counts the charge flowing through it, so the current you want to measure must pass through the meter: it goes in series, in the same loop. A voltmeter compares the potential at two points, so its two terminals are connected across the component โ€” in parallel with it. To disturb the circuit as little as possible, an ammeter is built with a very low resistance and a voltmeter with a very high resistance (resistance is defined properly in Section 3.2).

๐Ÿงญ Connecting and reading meters
  1. Ammeter in series with the component whose current you want โ€” never across it.
  2. Voltmeter in parallel across the component whose potential difference you want โ€” never in the main line.
  3. Polarity: the $+$ terminal of each meter goes towards the $+$ terminal of the cell, or the needle kicks backwards.
  4. Range: choose a range larger than the value you expect; start with the largest if unsure.
  5. Least count $=\dfrac{\text{value of one main interval}}{\text{number of small divisions in it}}$. Reading $=$ (number of divisions) $\times$ (least count). Look straight down at the needle to avoid parallax.
โœ๏ธ Example 6 โ€” Reading the meters
An ammeter reads from $0$ to $1\ \text{A}$ with $20$ equal divisions on its scale, and its needle rests on the $13$th division. A voltmeter reads from $0$ to $6\ \text{V}$ with $30$ equal divisions, and its needle rests on the $17$th division. Find the least count and the reading of each meter.
Ammeter Least count $=\dfrac{1\ \text{A}}{20}=0.05\ \text{A}$; reading $=13\times0.05=0.65\ \text{A}$.
Voltmeter Least count $=\dfrac{6\ \text{V}}{30}=0.2\ \text{V}$; reading $=17\times0.2=3.4\ \text{V}$.
Tip Write a reading to the least count: $0.65\ \text{A}$, not $0.6\ \text{A}$ or $0.650\ \text{A}$. Check your answers on the dials of Interactive 2.
Interactive 2 โ€” The Meter Connection Lab

A battery drives current through a $10\ \Omega$ resistor $R$. Put the ammeter โ’ถ and the voltmeter โ“‹ in series or in parallel with $R$ and read the dials (ammeter $0$โ€“$1\ \text{A}$, $20$ divisions; voltmeter $0$โ€“$6\ \text{V}$, $30$ divisions). The ammeter has almost no resistance ($0.01\ \Omega$), the voltmeter a huge one ($1\ \text{M}\Omega$). Try the wrong connections to see why the rules exist. Play runs the electrons and steps through the cells.

3
batteryโ€”
ammeter readsโ€”
voltmeter readsโ€”
current through $R$โ€”
โš ๏ธ Common mistakes

1. Minutes and milliamperes. $I=\dfrac Qt$ needs $Q$ in coulombs, $t$ in seconds and gives $I$ in amperes. Convert $250\ \text{mA}=0.25\ \text{A}$ and $2\ \text{min}=120\ \text{s}$ first.

2. "The current is used up in the bulb." It is not: the same current flows into and out of a bulb. What the bulb takes from the charge is energy โ€” that is what the potential difference measures.

3. Electron flow versus current. Arrows on circuit diagrams show conventional current, from $+$ to $-$ outside the cell. Electrons in the wire drift the other way.

4. Swapping the meters. An ammeter across a component short-circuits it and can be damaged; a voltmeter in series stops the current. Ammeter in series, voltmeter in parallel โ€” every time.

5. Charge that is not a multiple of $e$. No body can carry a charge such as $2.4\times10^{-19}\ \text{C}$, which would be $1.5$ electrons. Charge always comes in whole numbers of $e$.

๐Ÿš€ Beyond the board syllabus (JEE Main ยท NEET) โ€” optional: how fast do the electrons move?

In JEE Main and NEET 2026; not in the CBSE Class 10 syllabus or JEE Advanced.

If a wire of cross-sectional area $A$ contains $N$ free electrons per cubic metre, drifting with average speed $v_d$, then in time $t$ the electrons in a length $v_dt$ cross any section: $Q=NAv_dt\,e$, so $I=NeAv_d$. Copper has about $N=8.5\times10^{28}$ free electrons per $\text{m}^3$. For $I=1.0\ \text{A}$ in a wire of area $1.0\ \text{mm}^2=1.0\times10^{-6}\ \text{m}^2$:

$$v_d=\frac{I}{NeA}=\frac{1.0}{8.5\times10^{28}\times1.6\times10^{-19}\times1.0\times10^{-6}}\approx7.4\times10^{-5}\ \text{m/s},$$

less than a tenth of a millimetre per second โ€” slower than a snail. The lamp still lights at once because the push travels round the whole circuit almost instantly and every free electron starts moving together. You will meet drift speed properly in Class 12 Physics (Current Electricity).

Exercises

I
Charge, current and electrons. Take $e=1.6\times10^{-19}\ \text{C}$.
  • 1 A charge of $15\ \text{C}$ passes a point in $5.0\ \text{s}$. Find the current.
  • 2 A current of $0.25\ \text{A}$ flows for $2.0$ minutes. How much charge flows?
  • 3 How long does a current of $2.0\ \text{A}$ take to carry $500\ \text{C}$?
  • 4 Find the total charge of $5.0\times10^{19}$ electrons.
  • 5 How many electrons make up a charge of $1.0\ \text{C}$?
  • 6 How many electrons pass a point each second in a wire carrying $3.2\ \text{mA}$?
  • 7 A current of $200\ \text{mA}$ flows for $10$ minutes. Find the charge and the number of electrons.
  • 8 Can a body carry a charge of $2.4\times10^{-19}\ \text{C}$? Explain.
  • 9 In a horizontal wire, electrons drift from left to right. In which direction is the conventional current?
Check: enter any two of $Q$, $I$, $t$ (leave the third blank) and compare with your answer
$Q$โ€”
$I$โ€”
$t$โ€”
electrons $n=Q/e$โ€”
1 $I=\tfrac Qt=\tfrac{15}{5.0}=3.0\ \text{A}$.
2 $Q=It=0.25\times120=30\ \text{C}$.
3 $t=\tfrac QI=\tfrac{500}{2.0}=250\ \text{s}$ (about $4.2$ minutes).
4 $Q=ne=5.0\times10^{19}\times1.6\times10^{-19}=8.0\ \text{C}$.
5 $n=\tfrac{1.0}{1.6\times10^{-19}}=6.25\times10^{18}$ electrons.
6 In $1\ \text{s}$, $Q=3.2\times10^{-3}\ \text{C}$, so $n=\tfrac{3.2\times10^{-3}}{1.6\times10^{-19}}=2.0\times10^{16}$ electrons per second.
7 $Q=0.200\times600=120\ \text{C}$; $n=\tfrac{120}{1.6\times10^{-19}}=7.5\times10^{20}$.
8 No: $\tfrac{2.4\times10^{-19}}{1.6\times10^{-19}}=1.5$, and a body cannot carry half an electron's charge. Any charge must be a whole-number multiple of $e$.
9 From right to left โ€” conventional current is opposite to the motion of electrons.
II
Circuits and meters.
  • 1 Draw the symbols for a cell, a battery of three cells, an open plug key and a closed plug key.
  • 2 In the symbol for a cell, which line is the positive terminal?
  • 3 What is the difference between a cell and a battery?
  • 4 A bulb, a cell and a key form a circuit. Why does the bulb go out at once when the key is opened, even though the key is far from the bulb?
  • 5 Draw a circuit diagram with a battery of two cells, a plug key, a resistor and an ammeter in series, and a voltmeter measuring the potential difference across the resistor. Mark the direction of the current.
  • 6 Why should an ammeter have a very low resistance?
  • 7 Why should a voltmeter have a very high resistance?
  • 8 A student connects a voltmeter in series with a bulb and a battery. What will she observe?
1 Cell: a long thin line and a short thick line. Battery: three such pairs joined long-to-short in a row. Open key: two terminals with a gap (lever raised, or plug removed). Closed key: the gap bridged. (See the symbol chart above.)
2 The longer, thinner line is the positive terminal; the short thick line is negative.
3 A cell is a single source of potential difference; a battery is two or more cells joined together ($+$ of one to $-$ of the next), giving a larger potential difference.
4 Opening the key breaks the only loop, so charge stops flowing everywhere in the circuit together; the free electrons in the bulb's filament stop as soon as those at the key do.
5 A rectangle: battery on the left ($+$ at the top), then along the loop the key $K$, the ammeter โ’ถ ($+$ terminal towards the battery's $+$) and the resistor $R$; the voltmeter โ“‹ is connected by two wires to the two ends of $R$, in parallel. The current arrow goes from the battery's $+$ terminal round to its $-$ terminal.
6 It is placed in series, so all the current passes through it; a low resistance means it hardly reduces the current it is measuring.
7 It is connected in parallel; a very high resistance means it draws almost no current, so it does not change the potential difference it is measuring.
8 The bulb does not glow โ€” the voltmeter's huge resistance allows almost no current โ€” and the voltmeter reads almost the full battery voltage.
III
Potential difference and energy.
  • 1 $20\ \text{J}$ of work is done in moving $4.0\ \text{C}$ between two points. Find the potential difference between them.
  • 2 How much work is done in moving $3.0\ \text{C}$ across a potential difference of $6.0\ \text{V}$?
  • 3 How much energy does a $1.5\ \text{V}$ cell give to $10\ \text{C}$ of charge?
  • 4 A $9.0\ \text{V}$ battery does $270\ \text{J}$ of work. How much charge has it moved?
  • 5 State in words what is meant by "the potential difference between two points is $1\ \text{V}$".
  • 6 How much energy, in joules, does one electron gain in moving through a potential difference of $1.0\ \text{V}$?
  • 7 A $6.0\ \text{V}$ battery drives a current of $0.50\ \text{A}$ for $1.0$ minute. Find the charge moved and the energy supplied.
  • 8 The potential difference across a torch bulb is $3.0\ \text{V}$ and the current is $0.20\ \text{A}$. How much energy does the bulb receive in $50\ \text{s}$?
1 $V=\tfrac WQ=\tfrac{20}{4.0}=5.0\ \text{V}$.
2 $W=VQ=6.0\times3.0=18\ \text{J}$.
3 $W=VQ=1.5\times10=15\ \text{J}$.
4 $Q=\tfrac WV=\tfrac{270}{9.0}=30\ \text{C}$.
5 One joule of work is done in moving one coulomb of charge from one point to the other.
6 $W=Ve=1.0\times1.6\times10^{-19}=1.6\times10^{-19}\ \text{J}$.
7 $Q=0.50\times60=30\ \text{C}$; $W=VQ=6.0\times30=180\ \text{J}$.
8 $Q=It=0.20\times50=10\ \text{C}$; $W=VQ=3.0\times10=30\ \text{J}$.
IV
Reading meters and applications.
  • 1 An ammeter reads $0$โ€“$3\ \text{A}$ with $30$ equal divisions. Find its least count, and its reading when the needle is on the $17$th division.
  • 2 A voltmeter reads $0$โ€“$10\ \text{V}$ with $50$ equal divisions. Find its least count, and its reading when the needle is on the $23$rd division.
  • 3 Phone battery. A phone battery is rated $4000\ \text{mAh}$ (it can supply $4000\ \text{mA}$ for one hour). How much charge, in coulombs, can it deliver?
  • 4 The phone in part 3 draws an average current of $0.20\ \text{A}$. How long does a full charge last?
  • 5 Lightning. A lightning flash carries $15\ \text{C}$ of charge in $0.50\ \text{ms}$ across a potential difference of $1.0\times10^{8}\ \text{V}$. Find the average current and the energy released.
  • 6 Car starter. A $12\ \text{V}$ car battery supplies $150\ \text{A}$ to the starter motor for $2.0\ \text{s}$. Find the charge and the energy delivered.
  • 7 Challenge. An electron moving from point A to point B gains $8.0\times10^{-19}\ \text{J}$ of energy. Find the potential difference between A and B. Which point is at the higher potential?
1 Least count $=\tfrac{3}{30}=0.1\ \text{A}$; reading $=17\times0.1=1.7\ \text{A}$.
2 Least count $=\tfrac{10}{50}=0.2\ \text{V}$; reading $=23\times0.2=4.6\ \text{V}$.
3 $Q=It=4.0\ \text{A}\times3600\ \text{s}=1.44\times10^{4}\ \text{C}$ ($14\,400\ \text{C}$).
4 $t=\tfrac QI=\tfrac{14\,400}{0.20}=72\,000\ \text{s}=20\ \text{h}$.
5 $I=\tfrac{15}{0.50\times10^{-3}}=3.0\times10^{4}\ \text{A}$; $W=VQ=1.0\times10^{8}\times15=1.5\times10^{9}\ \text{J}$.
6 $Q=150\times2.0=300\ \text{C}$; $W=VQ=12\times300=3600\ \text{J}$ ($3.6\ \text{kJ}$).
7 $V=\tfrac We=\tfrac{8.0\times10^{-19}}{1.6\times10^{-19}}=5.0\ \text{V}$. A negative charge gains energy moving towards higher potential, so B is at the higher potential.
๐ŸŽฏ Exam corner โ€” board and entrance-exam practice BoardNEETJEE Main

E1. Board ยท 2 marks Define one ampere. How many electrons flow through a cross-section of a wire each second when the current is $1\ \text{A}$?

Definition The current is $1\ \text{A}$ when $1\ \text{C}$ of charge flows through a cross-section of the conductor in $1\ \text{s}$.
Working $n=\dfrac Qe=\dfrac{1}{1.6\times10^{-19}}=6.25\times10^{18}$.
Answer $6.25\times10^{18}$ electrons per second. Exam tip: write $I=\frac Qt$ beside the definition โ€” the formula line earns its own mark.

E2. Board ยท 3 marks Draw a circuit diagram of a battery of three cells, a plug key, a resistor, an ammeter and a voltmeter, arranged to measure the current through the resistor and the potential difference across it. State how each meter is connected and why.

Diagram One closed loop: battery (three cells, long line $+$) โ†’ key $K$ โ†’ ammeter โ’ถ โ†’ resistor $R$ โ†’ back to the battery's $-$ terminal. The voltmeter โ“‹ is connected across the two ends of $R$. Arrow for the current leaving the $+$ terminal.
Ammeter In series, so that the whole current through $R$ also passes through it; its resistance is very low, so it hardly changes that current.
Voltmeter In parallel with $R$, so that it compares the potential at the two ends of $R$; its resistance is very high, so it takes almost no current. Exam tip: mark the $+$ terminals of both meters on the side nearer the battery's $+$ terminal.

E3. Board ยท Assertionโ€“Reason Assertion (A): An ammeter is always connected in series with the component whose current is to be measured. Reason (R): An ammeter has a very high resistance. (A) Both A and R are true and R explains A. (B) Both are true but R does not explain A. (C) A is true but R is false. (D) A is false but R is true.

Idea A is true. R is false: an ammeter has a very low resistance, so that putting it in series does not reduce the current.
Answer (C). Exam tip: "high resistance" belongs to the voltmeter โ€” swapping the two is the usual trap.

E4. Board ยท Case study A power bank is labelled "$10\,000\ \text{mAh}$, $3.7\ \text{V}$". (i) How much charge, in coulombs, can it deliver? [2 marks] (ii) How much energy does it store? [1 mark] (iii) For how long can it supply a steady $2.0\ \text{A}$? [1 mark]

(i) $10\,000\ \text{mAh}=10\ \text{A}$ for $1\ \text{h}$: $Q=It=10\times3600=3.6\times10^{4}\ \text{C}$.
(ii) $W=VQ=3.7\times3.6\times10^{4}=1.3\times10^{5}\ \text{J}$ (about $133\ \text{kJ}$).
(iii) $t=\dfrac QI=\dfrac{3.6\times10^{4}}{2.0}=1.8\times10^{4}\ \text{s}=5.0\ \text{h}$. Exam tip: mAh is a unit of charge: $1\ \text{mAh}=3.6\ \text{C}$.

E5. NEET A wire carries a steady current of $4.8\ \text{A}$. The number of electrons crossing a cross-section of the wire each second is (A) $3.0\times10^{19}$ (B) $3.0\times10^{18}$ (C) $7.7\times10^{-19}$ (D) $4.8\times10^{19}$

Working $n=\dfrac{It}{e}=\dfrac{4.8\times1}{1.6\times10^{-19}}=3.0\times10^{19}$.
Answer (A). Exam tip: (C) is what you get by multiplying by $e$ instead of dividing โ€” a quick sense check: there must be a huge number of electrons.

E6. JEE Main An electron moves through a potential difference of $200\ \text{V}$. The energy it gains is (A) $3.2\times10^{-17}\ \text{J}$ (B) $1.6\times10^{-17}\ \text{J}$ (C) $200\ \text{J}$ (D) $3.2\times10^{-21}\ \text{J}$

Working $W=Ve=200\times1.6\times10^{-19}=3.2\times10^{-17}\ \text{J}$.
Answer (A). Exam tip: $W=QV$ works for any charge โ€” for one electron, $Q=e$.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 45 minutes

Research question. How much charge does a phone really draw while you use it โ€” and does the "mAh" printed on the battery tell you honestly how long it will last?

Why it matters. Billions of lithium-ion batteries are made every year for phones, scooters and electric cars, and more and more end up as electronic waste. Knowing what a battery rating means, and how screen brightness or video streaming changes the current, helps people buy sensibly and makes batteries last longer. The same arithmetic โ€” charge $=$ current $\times$ time โ€” is what engineers use to size the battery packs of electric buses in Indian and European cities.

What to do.

  1. Plug an inexpensive USB power meter (it shows volts and amperes) between a charger and a phone. Record the charging current every 5 minutes for half an hour.
  2. Calculate the charge delivered in each 5-minute interval, $Q=It$, and add them up. Convert your total to mAh ($1\ \text{mAh}=3.6\ \text{C}$).
  3. With the phone fully charged, note the battery percentage before and after 30 minutes of (a) reading with low brightness, (b) streaming video at full brightness. Use the battery's rated mAh to estimate the average current in each case.
  4. Predict how many hours each activity would take to empty a full battery, and compare with what actually happens.

Tools: USB power meter, phone and charger, stopwatch, spreadsheet.

Hand in: A table of current against time, your total charge in coulombs and mAh, the two estimated average currents, and a paragraph on whether the printed rating is a fair guide.

Going further. The voltmeter shows about $5\ \text{V}$ at the USB plug but the battery itself is about $3.7\ \text{V}$. Use $W=VQ$ to compare the energy taken from the charger with the energy stored, and suggest where the difference goes.

Instructor note: a $5000\ \text{mAh}$ battery holds $5\times3600=18\,000\ \text{C}$; at an average $0.50\ \text{A}$ it lasts $10\ \text{h}$. Typical phone currents are $0.1$โ€“$0.3\ \text{A}$ for light use and $0.5$โ€“$1\ \text{A}$ for gaming or video. Charging currents fall in the last 20% of charge, so students who measure only the first half-hour will over-estimate the total; that is a useful discussion point about sampling. For "going further", $5\ \text{V}\times Q$ versus $3.7\ \text{V}\times Q$ suggests roughly a quarter of the energy is lost as heat in the charging circuitry.

๐Ÿงช Self-check quiz

0/5 correct

Q1. One ampere is the same as

Q2. The number of electrons in a charge of $1\ \text{C}$ is

Q3. To measure the potential difference across a resistor, a voltmeter is connected

Q4. $2.0\ \text{J}$ of work is done in moving $0.50\ \text{C}$ of charge between two points. The potential difference between them is

Q5. A current of $0.50\ \text{A}$ flows for $1$ minute. The charge that flows is

โ† Chapter 3 โ€” Electricity3.2 Ohm's Law and the Factors Affecting Resistance โ†’