📌 Snapshot
- Defines steady electric current as net charge per unit time across an area,
I = q/t(and instantaneouslyI = lim ΔQ/Δt). - Builds the microscopic picture of conduction: free electrons drift through a lattice of fixed positive ions, giving rise to drift velocity
vd = −eEτ/m, mobilityμ = vd/E = eτ/m, and Ohm's lawV = IRwithR = ρl/A. - Classifies materials as conductors, semiconductors, and insulators by resistivity range, and gives the linear temperature law
ρT = ρ0[1 + α(T − T0)]for metals (α > 0), contrasting with semiconductors/insulators (ρ decreases with T). - Develops circuit tools used by CUET: electrical power
P = VI = I²R = V²/R, series/parallel resistors, EMF and internal resistanceV = ε − Ir, combinations of cells, Kirchhoff's junction and loop rules, and the Wheatstone bridge balanceR1/R2 = R3/R4. - High-weightage CUET unit because almost every concept yields a direct one-line MCQ (definitions, units, sign of α) and a 2-step numerical MCQ (drift velocity, terminal voltage, bridge balance).
📖 Detailed Notes
2.1 Core concepts
- Charges in motion constitute an electric current; in solid conductors the carriers are free electrons against a background of fixed positive ions (NCERT §3.1–3.3, p. 81–83).
- For steady current across an area in time t with net forward charge q,
I = q/t; more generallyI(t) = lim ΔQ/Δt as Δt→0. SI unit is the ampere; nerve currents are in µA while lightning carries tens of thousands of A (NCERT §3.2, p. 82). - Without an electric field, electrons move thermally with random directions, so the average velocity is zero and no net current flows (NCERT §3.3, p. 82–83).
- Ohm's law (1828, G.S. Ohm): for many conductors, V ∝ I, written
V = IR; R is the resistance of the conductor, SI unit ohm (Ω) (NCERT §3.4, p. 83). - Geometry of R: doubling length doubles R; halving cross-section doubles R; hence
R = ρl/A, where ρ (resistivity) depends only on material (NCERT §3.4, p. 83–84). - Current density
j = I/A(A m⁻²); in vector formE = ρjor equivalentlyj = σE, where σ = 1/ρ is conductivity (NCERT §3.4, p. 84–85). - Drift velocity: averaging electron motion under field E with average collision time τ (relaxation time) gives
vd = −(eE/m)τ, independent of time (NCERT §3.5, Eq. 3.17, p. 85–86). - This yields the microscopic form of Ohm's law:
j = (ne²τ/m) Eand henceσ = ne²τ/m,ρ = m/(ne²τ)(NCERT §3.5, Eqs. 3.21–3.23, p. 86). - Worked estimate (Example 3.1): in a Cu wire (A = 1.0 × 10⁻⁷ m², I = 1.5 A, n = 8.5 × 10²⁸ m⁻³), drift speed
vd ≈ 1.1 mm s⁻¹— about 10⁻⁵ times the thermal speed and 10⁻¹¹ times the speed of EM signal propagation (NCERT §3.5, Example 3.1, p. 86–87). - Mobility
μ = |vd|/E = eτ/m, SI unit m² V⁻¹ s⁻¹ (NCERT §3.5.1, Eqs. 3.24–3.25, p. 88–89). - Limitations of Ohm's law: (a) V not proportional to I, (b) V–I relation depends on sign of V (diode), (c) V–I relation not single-valued (GaAs) (NCERT §3.6, p. 89).
- Materials by resistivity: metals 10⁻⁸–10⁻⁶ Ω m; insulators ~10¹⁸ times that of metals; semiconductors in between, with ρ that decreases with rising temperature (NCERT §3.7, p. 89–90).
- Temperature dependence for metals over a limited range:
ρT = ρ0 [1 + α(T − T0)]; α (temperature coefficient of resistivity, dimension K⁻¹) is positive for metals; alloys like nichrome, manganin, constantan have very weak α and are used in standard resistors (NCERT §3.8, Eq. 3.26, p. 90). - Microscopic explanation via
ρ = m/(ne²τ): in metals n ≈ const, τ falls with T so ρ rises; in semiconductors/insulators n rises strongly with T and dominates, so ρ falls (NCERT §3.8, p. 91). - Electrical energy and power: in time Δt charge IΔt drops through V, dissipating
ΔW = IVΔt, soP = IV = I²R = V²/R(ohmic loss) (NCERT §3.9, Eqs. 3.31–3.33, p. 92–93). - High-voltage transmission: with cable resistance Rc, wasted power
Pc = P²Rc/V², so raising V reduces transmission loss (NCERT §3.9, Eq. 3.35, p. 93). - A cell has emf ε = V₊ + V₋ — the open-circuit potential difference between terminals — and internal resistance r; when current I flows, terminal voltage
V = ε − Ir, andI = ε/(R + r); maximum currentImax = ε/r(NCERT §3.10, Eqs. 3.36–3.40, p. 93–95). - Cells in series (same orientation):
εeq = ε1 + ε2,req = r1 + r2; if a cell is reversed its emf enters with a negative sign (NCERT §3.11, Eqs. 3.45–3.47, p. 95–96). - Cells in parallel:
1/req = 1/r1 + 1/r2andεeq/req = ε1/r1 + ε2/r2; extends to n cells (NCERT §3.11, Eqs. 3.56–3.59, p. 96–97). - Kirchhoff's rules: (a) Junction rule — at any junction Σ I_in = Σ I_out (conservation of charge); (b) Loop rule — algebraic sum of changes of potential around any closed loop is zero (conservation of energy) (NCERT §3.12, p. 97–98).
- Wheatstone bridge: four resistors R1, R2, R3, R4 with battery across one diagonal and galvanometer across the other; balance (Ig = 0) gives
R2/R1 = R4/R3, equivalentlyR1/R2 = R3/R4; a practical realisation is the metre bridge, which uses this balance to find an unknown resistance (NCERT §3.13, Eqs. 3.62–3.64, p. 100–101).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Electric current I | Net charge crossing a cross-section per unit time; I = lim ΔQ/Δt. SI unit ampere (A). |
p. 82 |
| Current density j | Current per unit area normal to the flow; vector along E; SI unit A m⁻². | p. 84 |
| Drift velocity vd | Average velocity acquired by electrons under field E; vd = −eEτ/m. |
p. 86 |
| Relaxation time τ | Average time between successive collisions of an electron with ions. | p. 85 |
| Mobility μ | Magnitude of drift velocity per unit electric field; μ = vd/E = eτ/m; SI unit m² V⁻¹ s⁻¹. |
p. 88 |
| Resistance R | Ratio V/I for an ohmic conductor; SI unit ohm (Ω = 1 V A⁻¹). | p. 83 |
| Resistivity ρ | Material property in R = ρl/A; SI unit Ω m. |
p. 84 |
| Conductivity σ | Reciprocal of resistivity, σ = 1/ρ; SI unit S m⁻¹ (or Ω⁻¹ m⁻¹). | p. 85 |
| Temperature coefficient of resistivity α | Fractional change in ρ per unit rise in T; ρT = ρ0[1+α(T−T0)]; positive for metals. |
p. 90 |
| Electromotive force (emf) ε | Potential difference between the terminals of a source in open circuit; ε = V₊ + V₋. | p. 94 |
| Internal resistance r | Resistance of the electrolyte/source itself; gives V = ε − Ir when I flows. |
p. 94 |
| Junction rule (Kirchhoff I) | Σ currents entering a junction = Σ currents leaving (conservation of charge). | p. 97 |
| Loop rule (Kirchhoff II) | Algebraic sum of potential changes around any closed loop is zero. | p. 98 |
| Wheatstone bridge balance | Galvanometer null when R1/R2 = R3/R4 (equivalently R2/R1 = R4/R3). |
p. 101 |
2.3 Diagrams / processes to remember
- Fig. 3.1 (p. 83): metallic cylinder with +Q and −Q on end discs — electrons drift to neutralise charges, motivating the need for a cell to maintain steady current.
- Fig. 3.2 (p. 83–84): rectangular slab of length l and area A illustrating
R = ρl/A. - Fig. 3.3 (p. 85): zig-zag electron path A → B between collisions, with a slight drift B → B′ opposite to E.
- Fig. 3.4 (p. 86): cylinder of unit area and length vd containing charge carriers — used to derive
I = neAvd. - Fig. 3.5 (p. 89): V vs I for a good conductor — solid curve deviates from dashed Ohm's-law line at large V.
- Fig. 3.6 (p. 89): characteristic curve of a diode (asymmetric forward/reverse) — Ohm's-law failure type (b).
- Fig. 3.7 (p. 89): V vs I for GaAs showing non-unique V for the same I — Ohm's-law failure type (c).
- Fig. 3.8 (p. 90): ρ of copper vs T — linear in the usual range, deviates at very low T.
- Fig. 3.9 (p. 90): ρ of nichrome — very weak dependence on T.
- Fig. 3.10 (p. 90): ρ of a typical semiconductor — decreases with T.
- Fig. 3.11 (p. 93): cell driving a resistor R; heat dissipated in R comes from chemical energy of the electrolyte.
- Fig. 3.12 (p. 94): electrolytic cell with positive (P) and negative (N) terminals; emf is V₊ + V₋.
- Fig. 3.13 (p. 95): two cells in series — yields
εeq = ε1 + ε2,req = r1 + r2. - Fig. 3.14 (p. 96): two cells in parallel — yields
1/req = 1/r1 + 1/r2. - Fig. 3.15 (p. 98): junction a where I3 = I1 + I2; loop rules for closed loops
ahdcbaandahdefga. - Fig. 3.16 (p. 98) and Fig. 3.17 (p. 99): cubical network of 12 equal resistors (Example 3.5, Req = 5R/6) and a non-symmetric mesh (Example 3.6) — classic Kirchhoff applications.
- Fig. 3.18 (p. 101): Wheatstone bridge — battery on AC, galvanometer on BD, balance
R2/R1 = R4/R3.
2.4 Common confusions / NTA trap points
- Current is a scalar, not a vector — even though we draw arrows; currents do not add by the parallelogram law. The arrow only indicates direction along the wire (Points to Ponder 1, p. 104).
- "V = IR is Ohm's law" is wrong as stated: V = IR is the definition of R and applies to any conducting device. Ohm's law is the stronger claim that R is independent of V (i.e., the V–I plot is a straight line through the origin) (Points to Ponder 2, p. 105).
- Drift speed vs signal speed vs thermal speed: drift speed in a metal is ~mm s⁻¹, thermal speed of ions ~10² m s⁻¹, while the electric field itself propagates near the speed of light. Distractors swap these in MCQs (NCERT Example 3.1 & 3.2, p. 87–88).
- Sign of α: positive for metals (ρ increases with T) but negative for semiconductors and insulators (ρ decreases with T). NTA likes to flip this.
- Terminal voltage: when the cell is discharging through R, V = ε − Ir (V < ε); when it is being charged from outside, the formula reverses sign (V = ε + Ir for the terminal voltage of the battery being charged). Many students always write ε − Ir.
- Wheatstone bridge balance condition is sometimes mis-written. The NCERT writes it as
R1/R2 = R3/R4(Summary, p. 104) and equivalentlyR2/R1 = R4/R3from the loop derivation (Eq. 3.64a, p. 101) — both forms are correct provided the arms are paired consistently (adjacent arms on the same side of the galvanometer).
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. Steady electric current I across a cross-section in time t, carrying net charge q, is defined as
▸ Show answer & explanation
Answer: B
For a steady current, the charge passing per unit time defines the current; (A) and (C) are dimensionally wrong, (D) is meaningless.
Q2. The SI unit of mobility μ of a charge carrier is
▸ Show answer & explanation
Answer: B
Since μ = vd/E with vd in m s⁻¹ and E in V m⁻¹, μ has units m² V⁻¹ s⁻¹. (A) is the unit of drift velocity, (C) of electric field.
Q3. According to the NCERT, Ohm's law is best stated as
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Answer: B
NCERT explicitly notes V = IR merely defines R; Ohm's law is the stronger statement that R is independent of V, i.e., I–V is linear. (A) is incorrect because diodes also satisfy V = IR as a definition but disobey Ohm's law.
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Q4. For a metallic conductor the resistivity ρT varies with temperature as ρT = ρ0[1 + α(T − T0)]. The temperature coefficient α for a typical metal is
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Answer: B
NCERT states the dimension of α is (Temperature)⁻¹ and that α is positive for metals. (C) describes semiconductors/insulators.
Q5. A copper wire of cross-section 1.0 × 10⁻⁷ m² carries a current of 1.5 A. Taking n = 8.5 × 10²⁸ m⁻³, the drift speed of conduction electrons is approximately
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Answer: A
Substituting in `vd = I/(neA)` gives ≈1.1 mm s⁻¹. (D) is the speed of the EM signal that travels through the conductor, not the drift speed.
Q6. Two identical wires of resistance R are first connected in series and then in parallel across the same battery. The ratio of the equivalent resistance in series to that in parallel is
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Answer: C
Rs/Rp = 2R / (R/2) = 4. Hence 4 : 1. Students who confuse the parallel formula may pick (B).
Q7. A battery of emf ε and internal resistance r is connected to an external resistance R. The current in the circuit is
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Answer: C
Total resistance in the loop is R + r, so I = ε/(R + r). The terminal voltage is then V = IR = ε − Ir.
Q8. The maximum current that can be drawn from a cell of emf ε and internal resistance r occurs when the external resistance R equals
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Answer: D
From I = ε/(R + r), I is largest when R = 0, giving Imax = ε/r. (A) confuses maximum-power transfer with maximum current.
Q9. Two cells of emfs ε₁ = 1.5 V and ε₂ = 2.0 V and internal resistances r₁ = 0.2 Ω and r₂ = 0.3 Ω are connected in series in the same orientation (positive of one to negative of the other). The equivalent emf and internal resistance are
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Answer: A
Series-aiding cells add their emfs and internal resistances: εeq = 1.5 + 2.0 = 3.5 V; req = 0.2 + 0.3 = 0.5 Ω. (C) uses the parallel formula by mistake.
Q10. Two identical cells, each of emf ε and internal resistance r, are connected in parallel and the combination is connected across an external resistor. The equivalent internal resistance of the combination is
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Answer: C
For identical cells in parallel, 1/req = 1/r + 1/r = 2/r, hence req = r/2 and εeq = ε. Choice (A) is the series result.
Q11. Kirchhoff's junction rule (Σ I_in = Σ I_out at every junction) is a direct consequence of
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Answer: C
Steady current means no charge piles up at a junction, so total in = total out — that is conservation of charge. The loop rule, in contrast, expresses energy conservation.
Q12. In a Wheatstone bridge the four arms have resistances P, Q, R, S in order, with the galvanometer connected between the junctions of P–Q and R–S. The balance condition (null deflection in the galvanometer) is
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Answer: B
The balance condition equates the ratio of resistances in adjacent arms of the bridge: P/Q = R/S. Choice (C) is the wrong (diagonal) pairing.
Q13. In a Wheatstone bridge P = 10 Ω in the first arm, Q = 20 Ω in the second arm, and R = 15 Ω in the third arm. The value of the unknown resistance S in the fourth arm at balance is
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Answer: C
Using P/Q = R/S, S = R · Q/P = 15 × 20/10 = 30 Ω. Picking (A) reverses the ratio.
Q14. Assertion (A): The resistivity of a typical semiconductor decreases with rise in temperature. Reason (R): In a semiconductor the number density n of charge carriers increases sharply with temperature, more than compensating any decrease in the relaxation time τ.
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Answer: A
NCERT gives exactly this reasoning using ρ = m/(ne²τ): n rising with T dominates over τ falling, so ρ falls. The Reason is the textbook explanation of the Assertion.
Q15. Match the quantity in List I with the SI unit / value-range in List II. List I (i) Resistivity of a metal (ii) Mobility of a charge carrier (iii) Conductivity (iv) Drift speed in a copper wire carrying a few amperes List II (P) m² V⁻¹ s⁻¹ (Q) ~10⁻³ m s⁻¹ (a few mm s⁻¹) (R) 10⁻⁸ Ω m to 10⁻⁶ Ω m (S) S m⁻¹ (= Ω⁻¹ m⁻¹)
▸ Show answer & explanation
Answer: A
Metallic resistivity lies in 10⁻⁸–10⁻⁶ Ω m, mobility's SI unit is m² V⁻¹ s⁻¹, conductivity's SI unit is S m⁻¹, and drift speeds in everyday wires are of the order of mm s⁻¹.
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