ScienceCBSE Class 10 · Electricity

Electric Current, Potential Difference and Circuits Class 10

Updated 14 October 2026 · Fact-checked

Electric current is the rate of flow of charge through a conductor, measured in ampere. Potential difference is the work done to move unit charge between two points, measured in volt. The board asks for definitions and units, numericals on charge, electrons and work, and circuit diagrams with the ammeter in series and the voltmeter in parallel.

Notice the ammeter sits in the loop (series), the voltmeter branches across the bulb (parallel), and conventional current runs opposite to electrons.

Illustration drawn for QuestionBuddy.

Understand Electric Current, Potential Difference and Circuits

When electric charge flows through a conductor, such as a metal wire, we say an electric current flows. In metal wires the moving charges are electrons. Current tells you how much charge crosses a cross-section of the wire every second, so $I = \frac{Q}{t}$ . The SI unit of charge is the coulomb (C) and of current the ampere (A): 1 A means 1 C passes each second. One electron carries $1.6 \times 1 0^{- 19} C$ , so 1 C is the charge of about $6 \times 1 0^{18}$ electrons.

Current needs a circuit: a continuous, closed path. A switch or plug key completes the path. Break it anywhere and the current stops everywhere, which is why a torch goes dark the moment you switch it off.

Why is current drawn from the positive terminal to the negative terminal when electrons move the other way? Electricity was studied before electrons were discovered, so current was taken as a flow of positive charge. That convention stays: conventional current is opposite to electron flow.

What makes charge move? Think of water in a level pipe: it does not flow until one end is joined to a raised tank, creating a pressure difference. Electrons similarly need a difference of 'electric pressure', called potential difference. A cell produces it by chemical action, even when no current is drawn. To keep current flowing, the cell uses up its stored chemical energy.

Potential difference between two points is the work done to move a unit charge from one to the other: $V = \frac{W}{Q}$ . Its unit is the volt: $1 V = 1 J C^{- 1}$ . So a 6 V battery gives 6 J to every coulomb passing through it.

Current is measured with an ammeter, always in series, so the current being measured passes through it. Potential difference is measured with a voltmeter, always in parallel across the two points. Circuit diagrams use standard symbols for the cell, battery, plug key, resistor, rheostat, bulb, ammeter and voltmeter. These ideas feed straight into Ohm's law, which links V and I through resistance.

Electric Current, Potential Difference and Circuits: formulas and key facts

Electric current: $I = \frac{Q}{t}$; Q in coulomb, t in second, I in ampere. Rearrange as $Q = I t$ .
Unit of current: $1 A = \frac{1 C}{1 s}$; $1 m A = 1 0^{- 3} A$ , $1 μ A = 1 0^{- 6} A$ .
Number of electrons: $n = \frac{Q}{e}, e = 1.6 \times 1 0^{- 19} C$; 1 C is about $6 \times 1 0^{18}$ electrons (more precisely $6.25 \times 1 0^{18}$ ).
Potential difference: $V = \frac{W}{Q}$; Rearrange as $W = V Q$ . W in joule.
Unit of potential difference: $1 V = \frac{1 J}{1 C}$; 1 V: 1 J of work moves 1 C between the two points.
Direction of current: Conventional current flows from $+$ to $-$ terminal outside the cell, opposite to electron flow; Electrons move from the negative terminal round to the positive.
Meter connections: Ammeter in series; voltmeter in parallel; Asked in diagram and practical questions.

How to solve Electric Current, Potential Difference and Circuits questions

Use this routine for any numerical or diagram question on current and potential difference.

1Write down every given quantity with its symbol and unit.
2Convert to SI: minutes and hours to seconds, mA and μA to A, kJ to J.
3Pick the formula: $I = \frac{Q}{t}$ for charge and time, $V = \frac{W}{Q}$ for work and charge.
4Rearrange first, then substitute with units.
5For electrons, divide the charge by $1.6 \times 1 0^{- 19} C$ .
6Write the answer with its unit and check it makes sense.
7For diagrams, put the ammeter in series and the voltmeter across the part, and mark current from the + terminal.

Electric Current, Potential Difference and Circuits in real life

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.

Electric current

Imagine a power cut at home in the evening. You use a torch whose bulb draws a current of 300 mA from a 3 V battery, and you keep it on for 20 minutes while finishing homework.

Find (a) the charge that flows through the bulb, (b) the number of electrons that pass, and (c) the energy the battery supplies, in kJ.

Try it first, then show the solutionSolution

Convert: I = 300 mA = 0.3 A, t = 20 min = 1200 s.
(a) $Q = I t = 0.3 A \times 1200 s = 360 C$ .
(b) $n = \frac{Q}{e} = \frac{360}{1.6 \times 1 0 ^{- 19}} = 2.25 \times 1 0^{21}$ electrons.
(c) $W = V Q = 3 V \times 360 C = 1080 J = 1.08 k J$ .

Answer: 360 C; $2.25 \times 1 0^{21}$ electrons; 1.08 kJ

Memory hook: Torch on longer: more coulombs ( $Q = I t$ ), more joules ( $W = V Q$ ).

Electric potential difference

Suppose you and your cousin each have a battery-run toy car. Your car runs on 9 V and draws 0.5 A; your cousin's runs on 3 V and draws 1 A. Both cars are run for 4 minutes.

Which car moves more charge through its circuit, and which takes more energy from its battery? Explain why the answers differ.

Try it first, then show the solutionSolution

t = 4 min = 240 s.
Your car: $Q = 0.5 A \times 240 s = 120 C$ ; $W = 9 V \times 120 C = 1080 J$ .
Cousin's car: $Q = 1 A \times 240 s = 240 C$ ; $W = 3 V \times 240 C = 720 J$ .
Cousin's car moves more charge; your car takes more energy.
Reason: energy is $W = V Q$ ; your battery gives 9 J to each coulomb against 3 J, which outweighs the smaller charge.

Answer: Cousin's car moves more charge (240 C vs 120 C); your car takes more energy (1080 J vs 720 J).

Memory hook: Volts tell you joules per coulomb, so energy needs both V and Q.

Tricks to remember Electric Current, Potential Difference and Circuits

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

Ammeter in a line, voltmeter across
Ammeter = in a line with the part (series). Voltmeter = across the part (parallel).
Use it when: Drawing circuits and practical questions.
QIT
Read 'Q = It' as 'Queue In Time': charge is current multiplied by time.
Use it when: Finding charge from current and time.
WVQ
Work = Volts × Queue of charge: $W = V Q$ . Volt is joule per coulomb.
Use it when: Work and energy questions.
Electrons rebel
Electrons always go against the arrow of conventional current.
Use it when: Direction questions and assertion-reason items.

Electric Current, Potential Difference and Circuits: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Charge from current and time · 2 marks · Very short answerBoard level

A current of 0.4 A flows through a bulb for 5 minutes. Find the charge that flows through the bulb.

Try it first, then show the solutionSolution

I = 0.4 A, t = 5 min = 300 s; formula $Q = I t$ (1 mark).
$Q = 0.4 A \times 300 s = 120 C$ (1 mark).

Answer: 120 C

Example 2: Mixed units: charge, electrons and p.d. · 3 marks · Short answerHarder board level

A cell drives a current of 250 mA through a resistor for 8 minutes. During this time the cell supplies 1.44 kJ of energy to the charge moving through the resistor. Find (a) the charge that flows, (b) the number of electrons that pass, (c) the potential difference across the resistor. ( $e = 1.6 \times 1 0^{- 19} C$ )

Try it first, then show the solutionSolution

(a) Convert: I = 250 mA = 0.25 A, t = 8 min = 480 s; $Q = I t = 0.25 A \times 480 s = 120 C$ (1 mark).
(b) $n = \frac{Q}{e} = \frac{120}{1.6 \times 1 0 ^{- 19}} = 7.5 \times 1 0^{20}$ electrons (1 mark).
(c) W = 1.44 kJ = 1440 J; $V = \frac{W}{Q} = \frac{1440 J}{120 C} = 12 V$ (1 mark).

Answer: (a) 120 C; (b) $7.5 \times 1 0^{20}$ electrons; (c) 12 V

Example 3: Definition plus a linked calculation · 5 marks · Long answerHarder board level

(a) Define potential difference between two points and state its SI unit. (1)
(b) A 12 V battery drives a current of 0.5 A through a lamp for 2 minutes. Find the charge that flows, the number of electrons that pass and the energy supplied. ( $e = 1.6 \times 1 0^{- 19}$ C) (3)
(c) A 6 V battery is to supply the same energy to a lamp in the same time. What current must flow? Justify. (1)

Try it first, then show the solutionSolution

(a) The potential difference between two points is the work done to move a unit charge from one point to the other, $V = \frac{W}{Q}$ ; SI unit volt (V). (1 mark)
(b) t = 2 min = 120 s; $Q = I t = 0.5 A \times 120 s = 60 C$ . (1 mark)
$n = \frac{Q}{e} = \frac{60}{1.6 \times 1 0 ^{- 19}} = 3.75 \times 1 0^{20}$ electrons. (1 mark)
$W = V Q = 12 V \times 60 C = 720 J$ . (1 mark)
(c) With 6 V, each coulomb gets only 6 J, so twice the charge is needed: $Q = \frac{720}{6} = 120 C$ in 120 s, so $I = 1 A$ . (1 mark)

Answer: (b) 60 C, $3.75 \times 1 0^{20}$ electrons, 720 J. (c) 1 A, because halving the p.d. halves the energy per coulomb, so the charge (and current) must double.

Example 4: Draw a circuit with meters · 3 marks · Short answerHarder board level

Draw a schematic circuit diagram of a battery of three 2 V cells, a resistor, a plug key, an ammeter to measure the current through the resistor and a voltmeter to measure the potential difference across it. Mark the direction of current. What is the voltage of the battery?

Try it first, then show the solutionSolution

Draw and label: three cell symbols joined + to −, labelled 'battery 6 V' (long line +, short line −) (½ mark).
Draw the plug key, resistor R and ammeter (circle with A, + side towards battery +) all in one loop in series (1 mark).
Draw the voltmeter (circle with V) on a separate pair of wires joined across the two ends of R only (1 mark).
Arrow from + terminal through key, ammeter, resistor back to −; battery = 3 × 2 V = 6 V (½ mark).
Common slip: putting the voltmeter in the main loop or the ammeter across R.

Answer: Battery 6 V; ammeter in series, voltmeter across R, current from + to −.

Example 5: Case: cells in a torch · 4 marks · Case-basedHarder board level

Read and answer: A student tests two torches. Torch X uses a 3 V battery and draws 0.5 A. Torch Y uses a 4.5 V battery and draws 0.2 A. Both are on for 10 minutes.
(a) Find the charge through each torch bulb. (1)
(b) Find the energy supplied by each battery. (2)
(c) The student says 'Y's battery has a higher voltage, so it must supply more energy.' Is the student right? Justify. (1)

Try it first, then show the solutionSolution

(a) t = 600 s; $Q_{X} = 0.5 \times 600 = 300 C$ , $Q_{Y} = 0.2 \times 600 = 120 C$ (1 mark).
(b) $W_{X} = 3 \times 300 = 900 J$ (1 mark).
$W_{Y} = 4.5 \times 120 = 540 J$ (1 mark).
(c) No: energy is $W = V Q$ , so it depends on charge too; X moves more charge and supplies more energy (1 mark).

Answer: 300 C, 120 C; 900 J, 540 J; student is wrong, X supplies more.

Where students lose marks in Electric Current, Potential Difference and Circuits

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

Mistake: Not converting time to seconds
Formulas need SI units.
Wrong working
Q = 0.5 A × 10 min = 5 C
Right working
Q = 0.5 A × 600 s = 300 C
Do this: Convert minutes and hours to seconds before substituting.
Usually costs: 1 mark
Mistake: Leaving mA as it is
Ampere is the SI unit.
Wrong working
Q = 250 × 4 = 1000 C for 250 mA, 4 s
Right working
Q = 0.25 A × 4 s = 1 C
Do this: Divide mA by 1000.
Usually costs: 1 mark
Mistake: Swapping meter connections in diagrams
Ammeter must carry the full current; voltmeter compares two points.
Wrong working
Voltmeter drawn in the main loop, ammeter across resistor
Right working
Ammeter in series, voltmeter across resistor
Do this: Draw the main loop first, then add the voltmeter on its own two wires.
Usually costs: 1-2 marks
Mistake: Saying current flows in the direction of electrons
Convention is opposite.
Wrong working
Current flows from − to + terminal through the bulb.
Right working
Conventional current flows from + to − through the external circuit.
Do this: Mark arrows starting at the long line of the cell.
Usually costs: 1 mark

Exam tips

Write the formula and SI unit with a definition to make it complete.
In 'what is meant by 1 V' answers, mention both 1 J and 1 C.
Show the formula, substitution with units and final unit in every numerical.
In diagrams, use standard symbols, label every part, and show the plug key.
For case-based questions, pull out V, I and t from the passage before calculating.

Practice questions on Electric Current, Potential Difference and Circuits

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 · Very short answerBoard level
How much work is done in moving a charge of 5 C across two points with a potential difference of 9 V?
Hint
Use $W = V Q$ .
Answer
$W = 9 \times 5 = 45 J$ .
Q2 · 1 mark · MCQBoard level
A charge of 60 C flows through a wire in 2 minutes. The current in the wire is
(a) 30 A
(b) 0.5 A
(c) 120 A
(d) 2 A
Hint
Convert minutes to seconds before using $I = Q / t$ .
Answer
(b) 0.5 A, because t = 120 s and I = 60 C / 120 s = 0.5 A.
Q3 · 3 marks · Short answerHarder board level
Battery X drives a current of 300 mA for 1 hour at 1.5 V. Battery Y drives a current of 1.2 A for 10 minutes at 9 V. (a) Which battery moves more charge? (b) Which battery supplies more energy? Show your working and explain why the answers to (a) and (b) are different.
Hint
Convert to A and s, find Q = It for each, then W = VQ.
Answer
(a) X: charge = 0.3 A × 3600 s = 1080 C; Y: charge = 1.2 A × 600 s = 720 C. X moves more charge (1 mark).
(b) X: W = 1.5 V × 1080 C = 1620 J; Y: W = 9 V × 720 C = 6480 J. Y supplies more energy (1 mark).
Energy depends on both charge and p.d. ( $W = V Q$ ); Y gives 9 J to each coulomb against only 1.5 J for X, which outweighs X's larger charge (1 mark).
Q4 · 3 marks · Short answerHarder board level
A wall charger supplies a current of 500 mA for 1.5 hours at 5 V. A power bank supplies 2 A for 20 minutes at 5 V. (a) Find the charge each delivers and the number of electrons each moves. (b) Find the energy each supplies in kJ and the ratio of the energies. (c) Which delivers charge at the faster rate, and does that one also supply more energy? Justify. ( $e = 1.6 \times 1 0^{- 19}$ C)
Hint
Convert mA to A and hours or minutes to seconds before using $Q = I t$ .
Answer
(a) Charger: $Q = 0.5 \times 5400 = 2700$ C, $n = \frac{2700}{1.6 \times 1 0 ^{- 19}} \approx 1.69 \times 1 0^{22}$ . Power bank: $Q = 2 \times 1200 = 2400$ C, $n = 1.5 \times 1 0^{22}$ (1 mark).
(b) Charger $W = 5 \times 2700 = 13500$ J $= 13.5$ kJ; power bank $W = 5 \times 2400 = 12000$ J $= 12$ kJ; ratio $9 : 8$ (1 mark).
(c) The power bank has the larger current (2 A), so it delivers charge faster, but it runs for a much shorter time, so it moves less charge and, at the same 5 V, supplies less energy (1 mark).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A cell has a potential difference across its terminals even when no current is drawn from it.
Reason (R): The potential difference across a cell is produced by electrons flowing through the external circuit.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Ask what creates the p.d. of a cell, and whether it needs current to exist.
Answer
(c) A is true but R is false. The p.d. is produced by chemical action inside the cell, so it exists even with no current; it is the p.d. that sets electrons moving, not the other way round.
Q6 · 3 marks · Short answerBoard level
A student connects an ammeter across a bulb and a voltmeter in series in the main line. Identify the errors and redraw the correct arrangement in words. What will each meter then measure?
Hint
Recall which meter goes in series.
Answer
Ammeter must be in series in the main loop (measures current through the bulb); voltmeter must be across the bulb's two ends (measures p.d. across it). Draw: cell, key, ammeter, bulb in one loop; voltmeter on wires joined to both ends of the bulb.
Q7 · 3 marks · Short answerHOTS / competency
A student measures the charge that has passed through a resistor at different times while a steady current flows. Her readings are:
t = 10 s, Q = 4 C
t = 20 s, Q = 8 C
t = 30 s, Q = 13 C
t = 40 s, Q = 16 C
(a) Find the steady current. (b) Which reading is wrong, and what should it be? (c) If the p.d. across the resistor is 6 V, find the energy supplied in 40 s.
Hint
For a steady current, Q/t is the same for every reading.
Answer
(a) Q/t = 4/10 = 8/20 = 16/40 = 0.4 A (1 mark).
(b) The 30 s reading is wrong; it should be Q = 0.4 A × 30 s = 12 C, not 13 C (1 mark).
(c) W = VQ = 6 V × 16 C = 96 J (1 mark).
Q8 · 4 marks · Case-basedHOTS / competency
Case study: A student sets up a circuit with a 6 V battery, a plug key, a resistor, an ammeter and a voltmeter. She runs three trials and records:
Trial 1: current 0.2 A for 5 minutes
Trial 2: current 0.3 A for 4 minutes
Trial 3: current 0.25 A for 8 minutes
(i) Find the total charge that flowed through the resistor in the three trials. (1 mark)
(ii) In her first set-up she had joined the voltmeter in the main line in series with the resistor. Explain why that position cannot give the p.d. across the resistor, and say where it should go. (1 mark)
(iii) With the voltmeter correctly placed it reads 6 V. Find the energy supplied to the resistor in Trial 3 and the number of electrons that passed in Trial 2. ( $e = 1.6 \times 1 0^{- 19}$ C) (2 marks)
Hint
Convert each time to seconds. A voltmeter compares two points.
Answer
(i) $Q_{1} = 0.2 \times 300 = 60$ C, $Q_{2} = 0.3 \times 240 = 72$ C, $Q_{3} = 0.25 \times 480 = 120$ C; total 252 C. (1)
(ii) The p.d. across the resistor is measured between its two ends, so the voltmeter must be joined across those two points, in parallel with the resistor. In the main line it is not connected to both ends of the resistor, so it cannot read the p.d. across it. (1)
(iii) Trial 3: $W = V Q = 6 \times 120 = 720$ J. Trial 2: $n = \frac{72}{1.6 \times 1 0 ^{- 19}} = 4.5 \times 1 0^{20}$ electrons. (2)

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).
2.1 ampere equals?Show answer
1 coulomb per second.
3.Charge on an electron?Show answer
$1.6 \times 1 0^{- 19} C$ (negative).
4.1 volt equals?Show answer
1 joule per coulomb.
5.Device that maintains p.d.?Show answer
A cell or battery.
6.How is an ammeter connected?Show answer
In series.

Electric Current, Potential Difference and Circuits: frequently asked questions

What is the difference between current and potential difference?

Current is how much charge flows per second. Potential difference is the work done per unit charge that pushes the charge. Without a potential difference, no current flows.

Why does a cell have potential difference even when switched off?

Chemical action inside the cell keeps a potential difference across its terminals whether or not current is drawn. Current flows only when the circuit is closed.

Do I need to learn circuit symbols?

Yes. You must draw cells, battery, plug key, resistor, rheostat, bulb, ammeter, voltmeter and wire joints correctly in diagram questions.

Why must an ammeter be connected in series?

An ammeter measures the current through a part of the circuit, so that same current has to pass through the ammeter. That happens only when it is joined in the same line (series) with that part. A voltmeter is different: it compares two points, so it is joined across them (in parallel).