PhysicsICSE Class 9 · Chapter 12

Current Electricity ICSE Class 9: Charge, Current, Potential Difference and Ohm's Law

Updated 15 October 2026 · Fact-checked

Electric current is the rate of flow of charge, I = Q/t, measured in amperes. A potential difference V = W/q (volts) drives charge through a conductor, which opposes it with resistance (ohms). Ohm's law: at constant temperature, V = IR. ICSE Class 9 also covers free electrons, conductors and insulators, charge quantisation, current direction and circuit symbols.

What Current Electricity is about

Electricity in wires is the movement of tiny charged particles. The CISCE 2028 Class IX syllabus covers:

free electrons and the types of conductors: good conductors, bad conductors (insulators), semiconductors and superconductors;
electric charge $Q$ , its SI unit (coulomb) and its quantum nature;
electric current as the rate of flow of charge; conventional and electronic directions;
potential difference $V = \frac{W}{q}$ (no derivation) and its SI unit; resistance and its SI unit;
circuit symbols; detecting current with a galvanometer or ammeter (their working is not included); a circuit with a cell, key and rheostat (qualitatively); a brief introduction to cells and accumulators as sources of direct current (construction and working excluded);
Ohm's law, $V = I R$ , with simple numericals.

Expect definitions with units, circuit diagrams with correct symbols, statements of Ohm's law, and 2 to 3-mark numericals on $Q = I t$ , $V = \frac{W}{q}$ , $V = I R$ and the number of electrons.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Current Electricity is in the 8. Electricity and Magnetism (i) part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

Free electrons and types of conductors

In a metal, the outermost electrons of each atom are loosely held and can move about inside the metal. These are free electrons. When a potential difference is applied, they drift through the metal, forming a current.

Good conductors: many free electrons; allow current easily (copper, silver, aluminium).
Bad conductors (insulators): almost no free electrons; do not allow current (rubber, plastic, glass, dry wood).
Semiconductors: conduct better than insulators but much less than metals; their conduction can be controlled (silicon, germanium) and they are used in electronic devices.
Superconductors: some materials, cooled below a very low critical temperature, have zero resistance, so current flows without loss.

Worked example

Classify: copper, silicon, rubber, mercury cooled to about −269 °C.

Try it first, then show the solutionSolution

Copper: good conductor. Rubber: insulator.
Silicon: semiconductor.
Mercury below its critical temperature: superconductor (zero resistance).

Answer: Good conductor; semiconductor; insulator; superconductor

Charge and its quantum nature

Electric charge is a property of matter: electrons carry negative charge, protons positive. SI unit: coulomb (C).

The smallest free charge is that of an electron (or proton), $e = 1.6 \times 1 0^{- 19}$ C. Every charge is a whole-number multiple of $e$ :

$Q = n e$ , where $n$ is an integer.

This is the quantum nature (quantisation) of charge. So 1 C is the charge of $\frac{1}{1.6 \times 1 0 ^{- 19}} = 6.25 \times 1 0^{18}$ electrons.

Worked example

How many electrons make up a charge of 1 C? Can a body carry a charge of $2.4 \times 1 0^{- 19}$ C?

Try it first, then show the solutionSolution

$n = \frac{1}{1.6 \times 1 0 ^{- 19}} = 6.25 \times 1 0^{18}$ .
$\frac{2.4 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 1.5$ , not a whole number.
So $2.4 \times 1 0^{- 19}$ C is not possible.

Answer: 6.25 × 10¹⁸; no

Electric current and its direction

Electric current is the rate of flow of charge: $I = \frac{Q}{t}$ . SI unit: ampere (A); $1 A = 1 C s^{- 1}$ . Small currents are in mA ( $1 0^{- 3}$ A) or μA ( $1 0^{- 6}$ A).

Direction:

Conventional current flows from the positive terminal of a cell, through the external circuit, to the negative terminal (chosen before electrons were discovered).
Electronic current (the actual flow of electrons) is in the opposite direction: from negative to positive.

Current is detected and measured with a galvanometer or an ammeter connected in series.

Worked example

A current of 0.5 A flows through a lamp for 10 minutes. Find the charge that passes and the number of electrons.

Try it first, then show the solutionSolution

$Q = I t = 0.5 \times 600 = 300$ C.
$n = \frac{300}{1.6 \times 1 0 ^{- 19}} = 1.875 \times 1 0^{21}$ .

Answer: 300 C; 1.875 × 10²¹ electrons

Potential difference

Charges flow only when there is a potential difference (p.d.) between two points, just as water flows from a higher level to a lower one.

The potential difference between two points is the work done per unit charge in moving charge from one point to the other: $V = \frac{W}{q}$ (no derivation).

SI unit: volt (V). 1 V is the p.d. when 1 J of work is done to move 1 C of charge: $1 V = 1 J C^{- 1}$ .

A cell maintains a p.d. between its terminals; it is measured with a voltmeter connected in parallel.

Worked example

12 J of work is done in moving 2 C of charge between two points. Find the potential difference, and the work needed to move 5 C between the same points.

Try it first, then show the solutionSolution

$V = \frac{12}{2} = 6$ V.
$W = V q = 6 \times 5 = 30$ J.

Answer: 6 V; 30 J

Resistance and circuit components

Resistance $R$ is the opposition a conductor offers to the flow of current. SI unit: ohm (Ω). $1 Ω = 1 V A^{- 1}$ .

A simple circuit has a source (cell or battery), a key (switch), connecting wires, and a load (bulb or resistance wire). It is closed when the key is on and current flows; open when the key is off.

A rheostat (variable resistance) changes the resistance in a circuit, and so changes the current. A resistance box gives chosen fixed values.

Sources of direct current: cells (e.g. a dry cell, which is used once) and accumulators (rechargeable storage cells, such as those in cars and inverters).

Circuit symbols to learn: cell (long line +, short line −), battery, key (open/closed), resistor, rheostat (resistor with an arrow), bulb, ammeter (A in a circle), voltmeter (V in a circle), galvanometer (G in a circle), connecting wire and joined wires (dot).

Worked example

In a circuit with a 6 V cell, the rheostat is adjusted from 10 Ω to 20 Ω (total resistance). How does the current change?

Try it first, then show the solutionSolution

At 10 Ω: $I = \frac{6}{10} = 0.6$ A.
At 20 Ω: $I = \frac{6}{20} = 0.3$ A.
Doubling the resistance halves the current.

Answer: From 0.6 A to 0.3 A

Ohm's law

Ohm's law: the current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature (and other physical conditions) remain the same.

$V \propto I$ , so $\frac{V}{I} = R$ (a constant), i.e. $V = I R$ .

The V–I graph of a conductor obeying Ohm's law is a straight line through the origin; its slope ( $\frac{V}{I}$ ) gives the resistance.

To verify it: connect a resistance wire, an ammeter (series), a voltmeter (across the wire), a rheostat, a key and a battery; record pairs of $V$ and $I$ for different rheostat settings and check that $\frac{V}{I}$ stays constant.

Worked example

A conductor has the readings $V$ = 1.0, 2.0, 3.0, 4.0 V for $I$ = 0.2, 0.4, 0.6, 0.8 A. Does it obey Ohm's law? Find its resistance.

Try it first, then show the solutionSolution

$\frac{V}{I} = 5$ Ω in every case.
The ratio is constant (straight-line graph through the origin), so it obeys Ohm's law.
$R = 5$ Ω.

Answer: Yes; 5 Ω

Physics in real life: Current Electricity

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Ohm's law in a lamp

Suppose a lamp in your room works on 220 V and draws a current of 0.5 A when glowing.

Find its resistance when glowing.

Try it first, then show the solutionSolution

$R = \frac{V}{I} = \frac{220}{0.5}$ .
$R = 440$ Ω.

Answer: 440 Ω.

Memory hook: R = V ÷ I.

Charge from a phone charger

Imagine a phone charger supplies a steady current of 2 A for 30 minutes.

How much charge flows, and how many electrons is that?

Try it first, then show the solutionSolution

$Q = 2 \times 1800 = 3600$ C.
$n = \frac{3600}{1.6 \times 1 0 ^{- 19}} = 2.25 \times 1 0^{22}$ .

Answer: 3600 C; 2.25 × 10²² electrons.

Memory hook: Charge = current × time.

Rheostat as a dimmer

Suppose a torch circuit with a 3 V battery has a variable resistor so that the total resistance can be 5 Ω or 15 Ω.

Find the current in each case and say when the bulb is brighter.

Try it first, then show the solutionSolution

5 Ω: $I = 0.6$ A. 15 Ω: $I = 0.2$ A.
More current gives a brighter bulb: brighter at 5 Ω.

Answer: 0.6 A (brighter) and 0.2 A.

Memory hook: Less resistance, more current.

Topics in this chapter

Current Electricity formula sheet

Current: $I = \frac{Q}{t}$; Ampere = coulomb per second.
Quantisation: $Q = n e, e = 1.6 \times 1 0^{- 19} C$; n a whole number.
Potential difference: $V = \frac{W}{q}$; Volt = joule per coulomb; no derivation.
Ohm's law: $V = I R$; At constant temperature.
Units: $1 Ω = 1 V A^{- 1}$; Resistance.
Electrons per coulomb: $\frac{1}{e} = 6.25 \times 1 0^{18}$; Electrons in 1 C.

Current Electricity: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Circuit to verify Ohm's law

Draw a circuit diagram to verify Ohm's law; 3 marks.

Draw

A battery and a key in series.
A rheostat and an ammeter in series with them.
A resistance wire (resistor) R in the same series loop.
A voltmeter connected in parallel across R only.
Mark + and − on the meters and the battery.

Label

Battery
Key K
Rheostat Rh
Ammeter A (series)
Resistor R
Voltmeter V (parallel across R)
direction of conventional current

Watch out: Use standard symbols; the ammeter goes in series, the voltmeter in parallel. Draw with a ruler and right-angled corners.

How the school exam asks Current Electricity

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Definitions and units	Define current, potential difference, resistance; give SI units; define 1 ampere, 1 volt, 1 ohm.	Write the relation (I = Q/t, V = W/q, R = V/I) and the unit in words.
Conductors	Classify materials; explain free electrons; what is a semiconductor or superconductor.	Number of free electrons; zero resistance below a critical temperature.
Charge numericals	Charge from current and time; number of electrons; whether a charge is possible.	Q = It; n = Q/e must be a whole number.
Current direction	Distinguish conventional and electronic current; mark directions on a circuit.	Conventional: + to − outside the cell; electrons: opposite.
Circuit symbols and diagrams	Draw symbols; draw a circuit with cell, key, rheostat, ammeter and voltmeter.	Ammeter series, voltmeter parallel; standard symbols.
Ohm's law	State the law; interpret a V–I graph; numericals with V = IR.	Include 'at constant temperature'; slope of V–I = R.

Tricks to remember Current Electricity

Short memory aids. Say each one aloud once, then use it on a question the same day.

VIR triangle
Cover the one you want: V = IR, I = V/R, R = V/I.
Use it when: Ohm's law numericals.
Conventional is positive-minded
Conventional current leaves the + terminal; electrons leave the − terminal.
Use it when: Current direction.
Ammeter in line, voltmeter alongside
Ammeter in series (in line), voltmeter in parallel (alongside).
Use it when: Circuit diagrams.
Six-point-two-five
1 coulomb = 6.25 × 10¹⁸ electrons.
Use it when: Electron-count questions.

Current Electricity: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Unit of p.d. · 1 mark · MCQExam level

One volt is equal to:
(a) 1 J C⁻¹
(b) 1 C J⁻¹
(c) 1 A Ω⁻¹
(d) 1 C s⁻¹

Try it first, then show the solutionSolution

$V = \frac{W}{q}$ : joule per coulomb.

Answer: (a)

Example 2: Current from charge · 2 marks · Short answerExam level

A charge of 2 C flows through a wire in 4 s. Find the current, and the time for 30 C to flow at the same current.

Try it first, then show the solutionSolution

$I = \frac{2}{4} = 0.5$ A.
$t = \frac{30}{0.5} = 60$ s.

Answer: 0.5 A; 60 s

Example 3: Ohm's law basics · 2 marks · Short answerExam level

A resistance of 30 Ω carries a current of 0.2 A. Find the p.d. across it, and the current if the p.d. is raised to 9 V (temperature constant).

Try it first, then show the solutionSolution

$V = 0.2 \times 30 = 6$ V.
$I = \frac{9}{30} = 0.3$ A.

Answer: 6 V; 0.3 A

Example 4: Conventional vs electronic · 2 marks · Short answerHarder exam level

Distinguish between conventional current and electronic current, and state the direction of each in the external circuit of a cell.

Try it first, then show the solutionSolution

Conventional current is the direction in which positive charge would flow: from the + terminal through the circuit to the − terminal.
Electronic current is the actual flow of electrons: from the − terminal through the circuit to the + terminal.

Answer: Opposite directions: + to − (conventional), − to + (electrons).

Example 5: Electrons and work · 3 marks · Short answerHarder exam level

$5 \times 1 0^{19}$ electrons pass through a bulb in 10 s across a p.d. of 3 V. Find the current and the work done.

Try it first, then show the solutionSolution

$Q = 5 \times 1 0^{19} \times 1.6 \times 1 0^{- 19} = 8$ C.
$I = \frac{8}{10} = 0.8$ A.
$W = V Q = 3 \times 8 = 24$ J.

Answer: 0.8 A; 24 J

Example 6: Reading a V–I table · 3 marks · Short answerHarder exam level

For a wire, V (V): 0.5, 1.0, 1.5, 2.0 and I (A): 0.1, 0.2, 0.3, 0.4. Find R, state the shape of the V–I graph, and predict I at 3.5 V.

Try it first, then show the solutionSolution

$R = \frac{V}{I} = 5$ Ω for every pair.
V–I graph: a straight line through the origin (Ohm's law obeyed).
$I = \frac{3.5}{5} = 0.7$ A.

Answer: 5 Ω; straight line through the origin; 0.7 A

Example 7: Is this charge possible? · 3 marks · Short answerHOTS / competency

An oil drop is found to carry a charge of $4.0 \times 1 0^{- 19}$ C, and another $4.8 \times 1 0^{- 19}$ C. Which value is possible? Explain using the quantum nature of charge, and find the number of extra electrons on the possible drop.

Try it first, then show the solutionSolution

$\frac{4.0 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 2.5$ : not a whole number, so impossible.
$\frac{4.8 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 3$ : possible.
The second drop carries 3 extra electrons (if negatively charged).

Answer: Only 4.8 × 10⁻¹⁹ C; 3 electrons

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 9-mark question split into 2 + 3 + 4 marks.

Testing a heater element

In a lab, students connect a heater element in a circuit with a battery, key, rheostat, ammeter and voltmeter. Their readings: V = 2, 4, 6, 8 V; I = 0.5, 1.0, 1.5, 2.0 A. Later, the element is used at 12 V for 5 minutes.

(i) · 2 marks
Draw (describe) the circuit, stating how the ammeter and voltmeter are connected.
Answer
Answer: Ammeter in series; voltmeter in parallel
Battery, key, rheostat, ammeter and element in series; voltmeter connected across the element only.
(ii) · 3 marks
Show that the element obeys Ohm's law and find its resistance.
Answer
Answer: 4 Ω
$\frac{V}{I} = 4$ Ω for every reading; constant ratio, so V ∝ I (straight-line V–I graph through the origin). $R = 4$ Ω.
(iii) · 4 marks
At 12 V, find the current, the charge passing in 5 minutes and the number of electrons (assuming R is unchanged).
Answer
Answer: 3 A; 900 C; 5.625 × 10²¹
$I = \frac{12}{4} = 3$ A. $Q = 3 \times 300 = 900$ C. $n = \frac{900}{1.6 \times 1 0 ^{- 19}} = 5.625 \times 1 0^{21}$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Define electric current and state its SI unit. (2 marks)

Step	Marks
Electric current is the rate of flow of electric charge: $I = \frac{Q}{t}$	1
SI unit: ampere (1 A = 1 C s⁻¹)	1

State Ohm's law and draw the V–I graph for a conductor obeying it. (3 marks)

Step	Marks
Current through a conductor is directly proportional to the p.d. across its ends	1
provided the temperature and other physical conditions remain constant	1
Graph: straight line through the origin, axes labelled V and I	1

A current of 0.4 A flows through a resistor when a p.d. of 10 V is applied. Find (a) the resistance (b) the charge passing in 2 minutes (c) the work done in moving this charge (d) the current if the p.d. is reduced to 6 V. (4 marks)

Step	Marks
(a) $R = \frac{10}{0.4} = 25$ Ω	1
(b) $Q = 0.4 \times 120 = 48$ C	1
(c) $W = V Q = 10 \times 48 = 480$ J	1
(d) $I = \frac{6}{25} = 0.24$ A	1

Where students lose marks in Current Electricity

The slips examiners see most often, and what to write instead.

Mistake: Stating Ohm's law without 'constant temperature'
CISCE examiners' comments, ICSE Physics 2024 (Class X): some candidates did not mention constant temperature, gave only the mathematical relation instead of a statement, or wrote 'I is directly proportional to R'.
Wrong working
'V = IR' alone, or 'I ∝ R'
Right working
'Current is directly proportional to the potential difference across the conductor, provided the temperature remains constant'
Do this: Learn the statement in words, with the condition.
Usually costs: 1 mark.
Mistake: Connecting the voltmeter in series
A voltmeter measures p.d. between two points, so it must be connected across (in parallel with) the component.
Do this: Ammeter in series, voltmeter in parallel.
Usually costs: Diagram marks.
Mistake: Forgetting to convert minutes to seconds
Q = It needs t in seconds for coulombs.
Wrong working
$Q = 0.5 \times 10 = 5$ C for 10 min
Right working
$Q = 0.5 \times 600 = 300$ C
Do this: Write t in seconds first.
Usually costs: The answer.
Mistake: Confusing current direction
Electrons flow from − to + in the external circuit; conventional current from + to −.
Do this: Label both arrows on circuit sketches when asked.
Usually costs: 1 mark.
Mistake: Describing the working of a galvanometer or cell
Their construction and working are excluded from the Class IX syllabus.
Do this: Only state their use (detecting current; providing a steady p.d.).
Usually costs: Time.

Practice questions on Current Electricity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The SI unit of electric charge is:
(a) ampere
(b) volt
(c) coulomb
(d) ohm
Hint
Named after Coulomb.
Answer
(c)
Q2 · 1 mark · MCQExam level
Which of these is a semiconductor?
(a) copper
(b) germanium
(c) glass
(d) aluminium
Hint
Used in chips.
Answer
(b)
Q3 · 1 mark · Short answerExam level
Find the resistance of a bulb that draws 0.25 A from a 6 V battery.
Hint
R = V/I.
Answer
24 Ω.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): A charge of $3.2 \times 1 0^{- 19}$ C can exist on a body.
Reason (R): Charge on a body is always an integral multiple of $1.6 \times 1 0^{- 19}$ C.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
3.2 ÷ 1.6.
Answer
(c): it equals 2e.
Q5 · 2 marks · Short answerHarder exam level
How much work is done in moving a charge of 15 C across a p.d. of 220 V, and how many electrons make up this charge?
Hint
W = Vq.
Answer
$W = 220 \times 15 = 3300$ J; $n = \frac{15}{1.6 \times 1 0 ^{- 19}} \approx 9.4 \times 1 0^{19}$ electrons.
Q6 · 3 marks · Short answerHarder exam level
A current of 200 mA flows for 5 minutes. Find the charge and the number of electrons that pass.
Hint
200 mA = 0.2 A.
Answer
$Q = 0.2 \times 300 = 60$ C; $n = \frac{60}{1.6 \times 1 0 ^{- 19}} = 3.75 \times 1 0^{20}$ .
Q7 · 3 marks · Short answerExam level
Draw the circuit symbols for a cell, a battery of two cells, an open key, a rheostat, an ammeter and a voltmeter.
Hint
Standard symbols.
Answer
Cell: long thin line (+) and short thick line (−). Battery: two such pairs joined. Open key: a gap with a small circle or a raised lever. Rheostat: zig-zag (or rectangle) with an arrow across it. Ammeter: A in a circle. Voltmeter: V in a circle.
Q8 · 2 marks · Short answerHarder exam level
The p.d. across a conductor is doubled and its resistance is halved (by changing the conductor). What happens to the current?
Hint
I = V/R.
Answer
$I^{'} = \frac{2 V}{R /2} = 4 I$ : the current becomes four times.
Q9 · 3 marks · Short answerHarder exam level
In a circuit, the ammeter reads 1.5 A and the voltmeter across a resistance coil reads 4.5 V. What is the resistance? If the rheostat is changed so the voltmeter reads 6 V, what will the ammeter read?
Hint
Find R first.
Answer
$R = 3$ Ω; $I = \frac{6}{3} = 2$ A.
Q10 · 3 marks · Short answerHarder exam level
Explain why metals are good conductors while rubber is an insulator, and why superconductors are useful.
Hint
Free electrons; zero resistance.
Answer
Metals have many free electrons that can drift when a p.d. is applied; rubber has almost none. Superconductors have zero resistance below a critical temperature, so current flows without energy loss (useful for powerful electromagnets and lossless transmission).
Q11 · 4 marks · Long answerHOTS / competency
A V–I graph for a wire is a straight line through the origin with $V = 9$ V at $I = 1.5$ A. Find its resistance. When the wire becomes hot, the graph bends towards the V-axis. What does this show about its resistance and about Ohm's law?
Hint
Slope = R.
Answer
$R = \frac{9}{1.5} = 6$ Ω. Bending towards the V-axis means V/I increases: the resistance rises with temperature. Ohm's law holds only while the temperature stays constant.
Q12 · 3 marks · Short answerHOTS / competency
A student says: 'A current of 1 A means one electron passes each second.' Correct the statement, and find the time for one coulomb to pass if the current is 0.25 A.
Hint
1 C = 6.25 × 10¹⁸ electrons.
Answer
1 A means 1 C, i.e. $6.25 \times 1 0^{18}$ electrons, passes per second. At 0.25 A, $t = \frac{1}{0.25} = 4$ s.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.SI unit of charge?Show answer
Coulomb (C).
2.Charge on an electron?Show answer
1.6 × 10⁻¹⁹ C (negative).
3.Define 1 ampere.Show answer
A current of 1 C of charge flowing per second.
4.Define potential difference.Show answer
Work done per unit charge in moving charge between two points, V = W/q.
5.SI unit of resistance?Show answer
Ohm (Ω).
6.Name a semiconductor.Show answer
Silicon (or germanium).
7.What is a superconductor?Show answer
A material with zero resistance below a critical temperature.
8.How is an ammeter connected?Show answer
In series.

Best order to study it

Learn free electrons and types of conductors.
Learn charge, its quantisation and current with directions.
Learn potential difference and resistance with units.
Learn circuit symbols and the circuit with cell, key, rheostat and meters.
Learn Ohm's law and practise numericals.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain free electrons and classify conductors?
Can you show a charge is or is not a multiple of e?
Can you define 1 A, 1 V and 1 Ω?
Can you mark conventional and electronic current?
Can you draw the standard symbols?
Can you state Ohm's law with its condition?
Can you use Q = It, V = W/q and V = IR?

Current Electricity: frequently asked questions

Are series and parallel combinations in Class IX?

No. Combinations of resistors are studied in Class X. Class IX covers simple circuits, Ohm's law and simple numericals.

Do I need to know how a galvanometer works?

No. The syllabus says the functioning of the galvanometer and ammeter is not to be introduced; only their use to detect and measure current.

What is the difference between a cell and an accumulator?

Both supply direct current. A simple dry cell is used once; an accumulator is a storage cell that can be recharged. Their construction is not in the syllabus.

Why is conventional current opposite to electron flow?

The direction was chosen before electrons were discovered, assuming positive charges moved. The convention is still used in circuit diagrams.

Does Ohm's law apply to all conductors?

No. It holds for metallic conductors at constant temperature; for some devices the V–I graph is not a straight line.

Based on: CISCE ICSE 2028 Science (52) Physics syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028