Definition
Electric current
Rate of flow of charge: I = dq/dt. SI unit: ampere (A) = C/s. Conventional current is in the direction of positive charge flow (opposite to electron flow).
-- NCERT Class 12 Physics, Ch. 3, p. 82Electric Current
8 MCQs1 revision card9-step worked example
Source: NCERT Current ElectricityOfficial key: NTA-verifiedLast updated: 21 Sep 2026
Electric Current, explained for NEET
The common confusion: "Electrons race through the wire, so the bulb lights at once." They do not race. In NCERT Class 12 Physics Part I, Chapter 3, Example 3.1 (page 87), a copper wire of cross-section 1.0×10⁻⁷ m² carrying 1.5 A has a drift speed of only about 1.1×10⁻³ m/s (1.1 mm/s). The bulb still lights almost instantly because the electric field is set up throughout the circuit almost at once — it propagates at about 3×10⁸ m/s — so electrons everywhere in the wire start drifting together (Example 3.2, page 88).
Electric current is the net charge flowing across a cross-section per unit time. For a steady current, I = q/t, so q = It. The SI unit is the ampere (1 A = 1 C/s).
Linking current to drift. In an electric field E, free electrons acquire an average drift velocity v_d = −(eE/m)τ (Equation 3.17). The minus sign means electrons drift opposite to E, so conventional current points opposite to electron motion. The current in a conductor of cross-section A is
I = neA|v_d| (Equation 3.18, page 86)
where n is the number density of free electrons (free electrons per m³), e = 1.6×10⁻¹⁹ C, and A is in m².
NEET bridge. Questions ask you to (a) find v_d from I, n and A; (b) count electrons crossing per second (I/e); or (c) compare drift speeds in wires of different area. For a fixed material and current, v_d ∝ 1/A.
Watch out:
- n is a density, not the total number of electrons in the wire.
- 1 mm² = 10⁻⁶ m², not 10⁻³ m². A wrong conversion shifts v_d by three powers of ten.
- In a steady current, charge is conserved, so the current is the same at every cross-section of a single wire. Where the wire is thinner, v_d is larger — the current is not.
Can you answer these Electric Current MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the relation I = neA|v_d| for current in a metallic conductor, the symbol n stands for:
Show answer and why every option is right or wrong
Answer: D. n is the number density of free electrons, in m⁻³, which is why neA|v_d| has units of C/s (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86).
Why A is wrong: A is wrong because a total count has no volume in its unit; using it would make neA|v_d| come out in C·m³/s instead of amperes. n is a density, not a total.
Why B is wrong: B is wrong because moles measure amount of substance, not free-electron concentration; the formula needs free electrons per cubic metre.
Why C is wrong: C is wrong because the number of electrons crossing per second is I/e — that is a result you compute from the current, not the n in the formula.
MCQ 2Easy RecallPractice
In the NCERT worked example for a copper wire of cross-section 1.0×10⁻⁷ m² carrying 1.5 A, the order of magnitude of the electron drift speed is:
Show answer and why every option is right or wrong
Answer: A. The drift speed works out to about 1.1×10⁻³ m/s, i.e. 1.1 mm/s (NCERT Class 12 Physics Part I, Chapter 3, Example 3.1, page 87).
Why B is wrong: B is wrong because it assumes electrons must move fast for a bulb to light at once; the actual drift speed is about a millimetre per second.
Why C is wrong: C is wrong because about 3×10⁸ m/s is the speed at which the electric field propagates along the circuit, not the speed of the electrons themselves.
Why D is wrong: D is wrong because a non-zero current needs a non-zero drift: I = neA|v_d| is zero only if v_d is zero.
MCQ 3Easy RecallPractice
Electron drift speed in a wire is only about a millimetre per second, yet a bulb lights almost immediately when the switch is closed. This is because:
Show answer and why every option is right or wrong
Answer: C. The field is established throughout the circuit almost at once, so the current starts almost instantly even though each electron gains only a small average drift speed (NCERT Class 12 Physics Part I, Chapter 3, Example 3.2, page 88).
Why A is wrong: A is wrong because it confuses the propagation speed of the field (about 3×10⁸ m/s) with the motion of the electrons, which only drift at about 1 mm/s.
Why B is wrong: B is wrong because there is no brief high-speed burst; electrons acquire only an average drift speed of the order of millimetres per second.
Why D is wrong: D is wrong because a bulb does not store charge to light itself; the current is carried by free electrons already present throughout the wire and filament.
MCQ 4Direct ApplicationPractice
A steady current of 3.0 A flows through a wire for 4.0 s. How much charge passes through any cross-section of the wire?
Show answer and why every option is right or wrong
Answer: D. For a steady current, q = It = 3.0 A × 4.0 s = 12 C (NCERT Class 12 Physics Part I, Chapter 3, definition of electric current).
Why A is wrong: A is wrong because 0.75 = 3.0/4.0 — it divides current by time instead of multiplying (q = It, not I/t).
Why B is wrong: B is wrong because 7.0 = 3.0 + 4.0 — it adds current and time, which have different units and cannot be added.
Why C is wrong: C is wrong because 1.3 ≈ 4.0/3.0 — it divides time by current, inverting the relation I = q/t.
MCQ 5Direct ApplicationPractice
A metal wire of cross-sectional area 2.5 mm² has a free-electron density of 5.0×10²⁸ m⁻³. The electrons drift with speed 2.0×10⁻⁴ m/s. Taking e = 1.6×10⁻¹⁹ C, the current in the wire is:
Show answer and why every option is right or wrong
Answer: B. A = 2.5×10⁻⁶ m². I = neA|v_d| = 5.0×10²⁸ × 1.6×10⁻¹⁹ × 2.5×10⁻⁶ × 2.0×10⁻⁴; coefficients 5.0 × 1.6 × 2.5 × 2.0 = 40, exponent 28 − 19 − 6 − 4 = −1, so I = 40×10⁻¹ = 4.0 A (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86).
Why A is wrong: A is wrong because it converts 2.5 mm² as 2.5×10⁻³ m²; the correct conversion is 1 mm² = 10⁻⁶ m², so the area and the current come out a thousand times too large.
Why C is wrong: C is wrong because it divides by the drift speed instead of multiplying: neA = 2.0×10⁴ C/m, and 2.0×10⁴ ÷ 2.0×10⁻⁴ = 1.0×10⁸. Current is proportional to v_d, I = neA|v_d|.
Why D is wrong: D is wrong because it leaves out e: nAv_d = 5.0×10²⁸ × 2.5×10⁻⁶ × 2.0×10⁻⁴ = 2.5×10¹⁹, which is the number of electrons crossing per second, not the current in amperes.
MCQ 6Direct ApplicationPractice
A wire carries a steady current of 3.2 A. Taking e = 1.6×10⁻¹⁹ C, the number of electrons crossing a cross-section of the wire per second is:
Show answer and why every option is right or wrong
Answer: B. Charge per second is 3.2 C, so the number of electrons per second is I/e = 3.2/(1.6×10⁻¹⁹) = 2.0×10¹⁹ (NCERT Class 12 Physics Part I, Chapter 3, definition of electric current).
Why A is wrong: A is wrong because 5.1×10⁻¹⁹ ≈ 3.2 × 1.6×10⁻¹⁹ — it multiplies current by e instead of dividing by e.
Why C is wrong: C is wrong because it slips a power of ten: 1/10⁻¹⁹ = 10¹⁹, so 3.2/1.6 × 10¹⁹ = 2.0×10¹⁹, not 2.0×10¹⁸.
Why D is wrong: D is wrong because 5.0×10⁻²⁰ = 1.6×10⁻¹⁹/3.2 — it divides e by the current, inverting the ratio.
MCQ 7CalculationPractice
A metal wire carrying a steady current has cross-sectional area 2.0×10⁻⁶ m² at section X and 5.0×10⁻⁷ m² at section Y. Which statement is correct?
Show answer and why every option is right or wrong
Answer: A. Step 1: in a steady state charge is conserved, so the same current crosses every section. Step 2: with I, n and e fixed, |v_d| = I/(neA) ∝ 1/A, so v_Y/v_X = A_X/A_Y = (2.0×10⁻⁶)/(5.0×10⁻⁷) = 4 (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86).
Why B is wrong: B is wrong because it treats drift speed as a fixed property of the material; with the same current, v_d must rise where the area falls.
Why C is wrong: C is wrong because it lets the current change along a single wire, which would mean charge piling up or vanishing between X and Y; in a steady current it is the drift speed that changes.
Why D is wrong: D is wrong because it takes v_d ∝ A instead of v_d ∝ 1/A; the narrower section needs the faster drift.
MCQ 8CalculationPractice
Two copper wires P and Q have the same free-electron density. P has diameter 1.0 mm and carries 2.0 A; Q has diameter 2.0 mm and carries 3.0 A. The ratio of drift speeds v_P : v_Q is:
Show answer and why every option is right or wrong
Answer: C. |v_d| = I/(neA) with A ∝ d². So v ∝ I/d². For P: 2.0/1.0² = 2.0; for Q: 3.0/2.0² = 0.75. Ratio 2.0 : 0.75 = 8 : 3 (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86).
Why A is wrong: A is wrong because it compares only the currents and ignores that Q has four times the cross-sectional area of P.
Why B is wrong: B is wrong because it uses A ∝ d instead of A ∝ d²: (2.0/1.0) : (3.0/2.0) = 4 : 3.
Why D is wrong: D is wrong because it multiplies by the area instead of dividing: (2.0×1.0²) : (3.0×2.0²) = 2 : 12 = 1 : 6.
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Electric Current: quick recall before you leave
How do you solve a Electric Current question? A worked example
- 1
Given
A metal wire has free-electron density n = 6.0×10²⁸ m⁻³ and cross-sectional area A = 2.0×10⁻⁶ m². Its conduction electrons drift with speed |v_d| = 2.5×10⁻⁴ m/s. Electron charge e = 1.6×10⁻¹⁹ C.
- 2
Required
The current I in the wire, i.e. the rate at which charge crosses a cross-section.
- 3
Concept
Current is the charge crossing a cross-section per second. In one second the drifting electrons sweep through a volume A|v_d|; that volume holds nA|v_d| free electrons, each carrying charge e. So the charge per second is neA|v_d|.
- 4
Formula
I = neA|v_d| (NCERT Class 12 Physics Part I, Chapter 3, Equation 3.18, page 86)
- 5
Substitution
I = 6.0×10²⁸ × 1.6×10⁻¹⁹ × 2.0×10⁻⁶ × 2.5×10⁻⁴
- 6
Calculation
Coefficients: 6.0 × 1.6 × 2.0 × 2.5 = 48
Powers of ten: 28 − 19 − 6 − 4 = −1
I = 48 × 10⁻¹ = 4.8 A
No exact counting numbers or mathematical constants appear here; e = 1.6×10⁻¹⁹ C is a rounded measured constant, and the answer is kept to two significant figures to match the data (each given to two significant figures). - 7
Final answer
I = 4.8 A (4.8 C of charge crosses any cross-section each second)
- 8
Common trap
Dropping e and reporting nA|v_d| = 3.0×10¹⁹ as the current — that is the number of electrons crossing per second, not a current in amperes. A second trap is dividing by v_d instead of multiplying; current rises in proportion to drift speed. If the area is given in mm², convert with 1 mm² = 10⁻⁶ m² before substituting.
- 9
Similar NEET-style question
"A metal wire has free-electron density 8.0×10²⁸ m⁻³ and cross-sectional area 1.5×10⁻⁶ m². The electrons drift at 5.0×10⁻⁵ m/s. Find the current. Take e = 1.6×10⁻¹⁹ C."
Strategy: I = neA|v_d| = 8.0×10²⁸ × 1.6×10⁻¹⁹ × 1.5×10⁻⁶ × 5.0×10⁻⁵. Coefficients 8.0 × 1.6 × 1.5 × 5.0 = 96; powers of ten 28 − 19 − 6 − 5 = −2. I = 96×10⁻² = 0.96 A (9.6×10⁻¹ A).
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What to remember before solving Electric Current questions
Which Electric Current formulas do you need for NEET?
1 formula — click to collapse
Current from drift velocity
Macroscopic current expressed in terms of microscopic drift velocity. n = electron density.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | current | A |
| n | electron density | 1/m^3 |
| e | electron charge | C |
| A | cross-section | m^2 |
| v_d | drift velocity | m/s |
Valid when
- Steady current
- Single-carrier conductor
More in Current Electricity: 3 exam traps and mistakes · 6 formulas · 6 question patterns from its other lessons.
Electric Current questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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