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State Board · Class 11 · Physics

Lesson 11: Electric Current Through Conductors

7 concepts · 56 practice questions with answers · Free to read

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Electric current and drift velocity

From daily life

Water flows in a pipe; electrons flow in a wire. Link to earlier classes : current, Ohm's law and resistance of classes 9-10 now extend to drift velocity and resistivity.

Simple definition

Current is the charge crossing per second.

Drift velocityIv_dE →I = n e A v_dv_d = e E τ ÷ m

In the figure : Under E electrons drift slowly in the opposite direction.

Rule 1: Electric current — I = Q ÷ t (A = C/s); direction : of positive charge (conventional). 1 A = 6.25 × 10¹⁸ electrons/s.

  • Easy : Q = 120 C, t = 1 min ⇒ I = 2 A.
  • Medium : I = 0.5 A, t = 10 s ⇒ Q = 5 C ⇒ n = 3.12 × 10¹⁹ electrons.
  • Exam (CET) : Direction of current? (1) from + to − (2) electron direction (3) both (4) none → answer 1)

Rule 2: Drift velocity — I = n e A v_d; copper n = 8.5 × 10²⁸ m⁻³. v_d is very small (≈ 10⁻⁴ m/s).

  • Easy : I = 1 A, A = 1 mm² ⇒ v_d = 7.4 × 10⁻⁵ m/s.
  • Medium : A doubled ⇒ v_d halves.
  • Exam (CET) : A bulb lights at once because? (1) the field travels at light speed (2) v_d is large (3) heat (4) magnet → answer 1)

Rule 3: Mobility — v_d = e E τ ÷ m; mobility μ = v_d ÷ E = e τ ÷ m (m² V⁻¹ s⁻¹).

  • Easy : E = 1 V/m, τ = 2.5 × 10⁻¹⁴ s ⇒ v_d = 4.4 × 10⁻³ m/s.
  • Medium : μ = 4.4 × 10⁻³ m²/Vs.
  • Exam (CET) : τ doubled ⇒ μ? (1) double (2) half (3) same (4) ¼ → answer 1)

Derivation : I = n e A v_d

  • Step 1 : A length v_d t of wire holds n A v_d t electrons.
  • Step 2 : Charge crossing in time t : Q = n e A v_d t.
  • Step 3 : I = Q / t.
  • Step 4 : I = n e A v_d.
Remember

I = Q/t; I = neAv_d; v_d = eEτ/m.

Common mistake / CET trap

v_d is small, the field is fast; electrons flow opposite to conventional I.

Shortcut

Memory clue : "I = n e A v".

Spot the difference

SpeedValueMeaning
Drift≈ 10⁻⁴ m/saverage
Thermal≈ 10⁵ m/srandom
Field spreads3 × 10⁸ m/sinstant bulb

Practice questions

Easy
  1. 1Define I.
  2. 2What is drift velocity?
  3. 3Unit of μ?
Medium
  1. 4Q = 300 C, t = 2 min : I?
  2. 5I = 2 A, A = 2 mm², n = 8.5 × 10²⁸ : v_d?
  3. 6τ = 10⁻¹⁴, e/m = 1.76 × 10¹¹ : μ?
Challenging
  1. 7Derive I = nAev_d.
  2. 8Why does the bulb light at once though v_d is small?
Show answers (try them yourself first)
  1. 1Q/t
  2. 2Average velocity of electrons due to the field
  3. 3m² V⁻¹ s⁻¹
  4. 42.5 A
  5. 57.4 × 10⁻⁵ m/s
  6. 61.76 × 10⁻³ m²/Vs
  7. 7Q = nAv_d t e
  8. 8The E field spreads at light speed

Ohm's law and resistance

From daily life

Water flows less through a narrow pipe; current flows less through a thin wire.

Simple definition

Resistance opposes current; R = V/I.

V–I graphohmicnon-ohmicVIslope = RV = I R ; R = ρ L ÷ A

In the figure : A straight graph means ohmic.

Rule 1: Ohm's law — At constant temperature V = I R; R constant (ohmic). Unit Ω = V/A; dimension [M L² T⁻³ A⁻²].

  • Easy : V = 12 V, I = 3 A ⇒ R = 4 Ω.
  • Medium : R = 20 Ω, I = 0.5 A ⇒ V = 10 V.
  • Exam (CET) : A straight V–I line ⇒ ? (1) ohmic (2) non-ohmic (3) semiconductor (4) bulb → answer 1)

Rule 2: Resistance and shape — R = ρ L ÷ A. L doubled ⇒ R doubles; A doubled ⇒ R halves; radius doubled ⇒ R ¼.

  • Easy : A 10 Ω wire stretched to three times its length (same volume) ⇒ R × 9 = 90 Ω.
  • Medium : Copper ρ = 1.7 × 10⁻⁸, L = 10 m, A = 1 mm² ⇒ R = 0.17 Ω.
  • Exam (CET) : Wire stretched n times ⇒ R? (1) n² times (2) n times (3) 1/n (4) same → answer 1) Trick : L × n, A ÷ n ⇒ n².

Rule 3: Microscopic form — J = σ E (J = I/A, σ = 1/ρ). ρ = m ÷ (n e² τ).

  • Easy : J = 2 × 10⁶ A/m², σ = 5.9 × 10⁷ ⇒ E = 0.034 V/m.
  • Medium : In a 1 mm² wire I = 3 A ⇒ J = 3 × 10⁶ A/m².
  • Exam (CET) : Unit of J? (1) A/m² (2) A (3) V/m (4) Ω → answer 1)

Derivation : ρ = m ÷ (n e² τ)

  • Step 1 : v_d = e E τ / m.
  • Step 2 : J = n e v_d = n e² τ E / m.
  • Step 3 : J = σ E ⇒ σ = n e² τ / m.
  • Step 4 : ρ = 1/σ = m ÷ (n e² τ).
Remember

V = IR; R = ρL/A; J = σE.

Common mistake / CET trap

When stretching, volume is constant : R ∝ L²; Ohm's law is not universal.

Shortcut

Memory clue : "V = I R".

Spot the difference

GraphSlopeExample
Straightconstant Rcopper wire
Curvedvarying Rbulb, diode

Practice questions

Easy
  1. 1State Ohm's law.
  2. 2Ω in terms of V and A?
  3. 3What is J?
Medium
  1. 4V = 24, I = 6 : R?
  2. 5L = 5 m, A = 0.5 mm², ρ = 1.7 × 10⁻⁸ : R?
  3. 6Wire radius doubled : R?
Challenging
  1. 7Wire stretched to twice the length : R?
  2. 8Derive ρ.
Show answers (try them yourself first)
  1. 1V ∝ I (constant temperature)
  2. 2V/A
  3. 3I/A
  4. 44 Ω
  5. 50.17 Ω
  6. 6R/4
  7. 74R
  8. 8From σ = ne²τ/m

Resistivity and its temperature dependence

From daily life

A cold bulb draws more current at switch-on than when hot.

Simple definition

Resistivity ρ is the resistance of a unit cube of the material.

Temperature and resistancemetalsemiconductoralloyRTR_t = R₀ (1 + α ΔT)

In the figure : Metals rise, semiconductors fall.

Rule 1: Temperature coefficient — R_t = R₀ (1 + α ΔT); α = (R_t − R₀) ÷ (R₀ ΔT) (K⁻¹). Metals : α positive (≈ 4 × 10⁻³).

  • Easy : R₀ = 100 Ω, α = 4 × 10⁻³, ΔT = 50 ⇒ R = 120 Ω.
  • Medium : R₁ (20 °C) = 10 Ω ⇒ R₂ (100 °C) = 10 × (1 + 0.004 × 100)/(1 + 0.004 × 20) = 12.96 Ω.
  • Exam (CET) : Metal R with rising temperature? (1) increases (2) decreases (3) same (4) zero → answer 1)

Rule 2: By material — Metals : T ↑ ⇒ τ ↓ ⇒ ρ ↑. Semiconductors : T ↑ ⇒ n ↑ ⇒ ρ ↓ (α negative). Alloys (manganin, nichrome) : tiny α, large ρ.

  • Easy : A thermistor (semiconductor) measures temperature.
  • Medium : Manganin resistors serve as standard resistances.
  • Exam (CET) : Heater wire? (1) nichrome (2) copper (3) silver (4) aluminium → answer 1)

Rule 3: Resistivity values — Copper 1.7 × 10⁻⁸, nichrome 1.1 × 10⁻⁶ Ω m; insulators ≈ 10¹⁰–10¹⁴.

  • Easy : Nichrome L = 2 m, A = 0.2 mm² ⇒ R = 11 Ω.
  • Medium : Unit of ρ is Ω m.
  • Exam (CET) : Unit of ρ? (1) Ω m (2) Ω (3) Ω/m (4) S → answer 1)

Derivation : R_t = R₀ (1 + α ΔT)

  • Step 1 : Experiment : change in R ∝ R₀ and ΔT.
  • Step 2 : R_t − R₀ = α R₀ ΔT.
  • Step 3 : R_t = R₀ + α R₀ ΔT.
  • Step 4 : R_t = R₀ (1 + α ΔT).
Remember

R_t = R₀(1 + αΔT); ρ = m/(ne²τ).

Common mistake / CET trap

ρ depends on the material, not the shape; α is negative for semiconductors.

Shortcut

Memory clue : "Hot metal = more resistance".

Spot the difference

MaterialαR with temperature
Metal+rises
Semiconductor−falls
Alloy≈ 0almost constant

Practice questions

Easy
  1. 1Define α.
  2. 2Sign of α for a semiconductor?
  3. 3Heater wire?
Medium
  1. 4R₀ = 50, α = 5 × 10⁻³, ΔT = 80 : R?
  2. 5R = 8 → 10.4, ΔT = 100 : α?
  3. 6Nichrome L = 1, A = 0.1 mm² : R?
Challenging
  1. 7Why does ρ fall in a semiconductor?
  2. 8Wire R at 0 °C and 100 °C : 10, 14 : α?
Show answers (try them yourself first)
  1. 1(R_t − R₀)/(R₀ΔT)
  2. 2Negative
  3. 3Nichrome
  4. 470 Ω
  5. 53 × 10⁻³ K⁻¹
  6. 611 Ω
  7. 7The number of carriers n grows
  8. 84×10⁻³ K⁻¹

Electrical energy, power and heating effect

From daily life

Irons and heaters warm up from I² R heating in the wire.

Simple definition

Power is the electrical energy used per second.

Electric powerP = V I= I² R = V² ÷ RH = I² R t1 kWh = 3.6 × 10⁶ J220 V, 100 W

In the figure : P = VI = I²R = V²/R.

Rule 1: Power — P = V I = I² R = V² ÷ R (W); energy W = P t; 1 kWh = 3.6 × 10⁶ J.

  • Easy : A 100 W bulb for 5 h ⇒ 0.5 kWh.
  • Medium : 2 kW heater, 220 V ⇒ I = 9.09 A, R = 24.2 Ω.
  • Exam (CET) : A 220 V/100 W bulb on 110 V ⇒ P? (1) 25 W (2) 50 W (3) 100 W (4) 200 W → answer 1) Trick : P ∝ V².

Rule 2: Joule's law — H = I² R t (J); I doubled ⇒ H four times. H = I² R t ÷ 4.2 in calories.

  • Easy : I = 2 A, R = 5 Ω, t = 10 s ⇒ H = 200 J.
  • Medium : A 1 kW heater warms 2 kg water by 30 K ⇒ t = 252 s.
  • Exam (CET) : H ∝ ? (1) I² (2) I (3) 1/I (4) √I → answer 1)

Rule 3: Appliances — Fuse : low melting point (melts at high I). Thick wire ⇒ small R ⇒ less heat. Electricity bill = kWh.

  • Easy : A 1.5 kW geyser for 2 h ⇒ 3 kWh.
  • Medium : A 5 A fuse : an 1100 W appliance on 220 V ⇒ I = 5 A.
  • Exam (CET) : Fuse wire has? (1) low melting point (2) high (3) thick (4) magnetic → answer 1)

Derivation : P = V I = I² R

  • Step 1 : Work done on charge q across V : W = q V.
  • Step 2 : Power P = W/t = V (q/t).
  • Step 3 : q/t = I ⇒ P = V I.
  • Step 4 : V = I R ⇒ P = I² R = V²/R.
Remember

P = VI = I²R = V²/R; H = I²Rt; 1 kWh = 3.6 × 10⁶ J.

Common mistake / CET trap

Use V²/R in parallel and I²R in series; kWh is energy, not power.

Shortcut

Memory clue : "Series = I squared R".

Spot the difference

FormulaWhenExample
I² Rseries (same I)larger R ⇒ more P
V²/Rparallel (same V)smaller R ⇒ more P
V Igeneral—

Practice questions

Easy
  1. 1Formulas for P?
  2. 21 kWh = ? J
  3. 3State Joule's law.
Medium
  1. 4R of a 60 W, 220 V bulb?
  2. 5I = 3 A, R = 10, t = 1 min : H?
  3. 62 kW for 3 h : units?
Challenging
  1. 7220 V/100 W bulb at 110 V : P?
  2. 8Why is a fuse used?
Show answers (try them yourself first)
  1. 1VI, I²R, V²/R
  2. 23.6 × 10⁶
  3. 3H = I²Rt
  4. 4807 Ω
  5. 55400 J
  6. 66 kWh
  7. 725 W
  8. 8Melts at high I and breaks the circuit

Series and parallel combinations of resistors

From daily life

House lamps are in parallel — one failing does not stop the others.

Simple definition

Series = one after another; parallel = the same V across all.

Series and parallelR₁R₂seriesR = R₁ + R₂R₁R₂parallel1/R = 1/R₁ + 1/R₂R = R₁R₂/(R₁+R₂)same IV divides

In the figure : Series above, parallel below.

Rule 1: Series — R = R₁ + R₂ + …; same I, V divides (V ∝ R); R exceeds the largest.

  • Easy : 2, 3, 5 Ω in series ⇒ 10 Ω.
  • Medium : V = 12, R = 2 + 4 ⇒ I = 2 A; V = 4 V and 8 V.
  • Exam (CET) : What is equal in series? (1) I (2) V (3) R (4) P → answer 1)

Rule 2: Parallel — 1/R = 1/R₁ + 1/R₂ + …; same V, I divides (I ∝ 1/R); R is below the smallest.

  • Easy : 6 and 3 Ω ⇒ R = 2 Ω; with 12 V ⇒ I = 2 A and 4 A.
  • Medium : Three 4 Ω in parallel ⇒ 1.33 Ω.
  • Exam (CET) : n equal R in parallel ⇒ ? (1) R/n (2) nR (3) R (4) n/R → answer 1) Trick : two equal ⇒ half.

Rule 3: Power distribution — In series P ∝ R (I² R); in parallel P ∝ 1/R (V²/R). In series the larger R glows brighter.

  • Easy : 3 Ω and 6 Ω in series, 9 V ⇒ I = 1 A; P = 3 W and 6 W.
  • Medium : In parallel, 9 V ⇒ P = 27 W and 13.5 W.
  • Exam (CET) : In parallel more P in? (1) smaller R (2) larger R (3) same (4) zero → answer 1)

Derivation : Parallel resistors

  • Step 1 : Same V across both; I = I₁ + I₂.
  • Step 2 : V/R = V/R₁ + V/R₂.
  • Step 3 : Divide by V.
  • Step 4 : 1/R = 1/R₁ + 1/R₂.
Remember

Series R = ΣR; parallel 1/R = Σ1/R.

Common mistake / CET trap

R decreases in parallel; two equal ⇒ half.

Shortcut

Memory clue : "Series add, parallel add reciprocals".

Spot the difference

ConnectionRSameDivides
SeriesΣRIV
Parallel1/R = Σ1/RVI

Practice questions

Easy
  1. 1R in series?
  2. 2R in parallel?
  3. 3Which connection is used at home?
Medium
  1. 45 and 20 Ω in parallel : R?
  2. 5Three 12 Ω in series : R?
  3. 610 V, 4 and 6 Ω in series : I?
Challenging
  1. 7A 10 Ω wire cut into 5 equal pieces, joined in parallel : R?
  2. 8If one bulb fuses in a series string?
Show answers (try them yourself first)
  1. 1R₁ + R₂ + …
  2. 21/R = Σ1/R
  3. 3Parallel
  4. 44 Ω
  5. 536 Ω
  6. 61 A
  7. 70.4 Ω
  8. 8All go off (the circuit breaks)

Cell emf and internal resistance

From daily life

A torch dims with an old cell — its internal resistance has grown.

Simple definition

emf (ε) is the cell's V in an open circuit; internal resistance r lies inside the cell.

Cell : emf and r+rRεr (internal)I = ε ÷ (R + r) ; V = ε − I r

In the figure : A cell with r, external R.

Live figure

Cell emf and internal resistance

In the app you can move this figure with your finger.

Open in the app

Rule 1: emf and internal resistance — ε = V + I r; I = ε ÷ (R + r); terminal V = ε − I r. In an open circuit V = ε.

  • Easy : ε = 12 V, r = 1 Ω, R = 5 Ω ⇒ I = 2 A; V = 10 V.
  • Medium : Short circuit : I = ε/r = 12 A.
  • Exam (CET) : V in an open circuit? (1) ε (2) 0 (3) ε/2 (4) Ir → answer 1)

Rule 2: Finding r — r = R (ε ÷ V − 1); while charging V = ε + I r.

  • Easy : ε = 1.5, R = 7 Ω, V = 1.4 ⇒ r = 0.5 Ω.
  • Medium : Charging : ε = 12, I = 2, r = 0.5 ⇒ V = 13 V.
  • Exam (CET) : Terminal V while charging? (1) ε + Ir (2) ε − Ir (3) ε (4) 0 → answer 1)

Rule 3: Maximum power — Power in the load P = ε² R ÷ (R + r)²; maximum at R = r : P_max = ε² ÷ 4r; efficiency η = R/(R + r).

  • Easy : ε = 12, r = 1 ⇒ P_max = 36 W (R = 1 Ω).
  • Medium : R = r ⇒ η = 50 %.
  • Exam (CET) : P is maximum when? (1) R = r (2) R = 0 (3) R = ∞ (4) R = 2r → answer 1)

Derivation : P_max = ε² / 4r

  • Step 1 : I = ε/(R + r); P = I² R = ε² R / (R + r)².
  • Step 2 : dP/dR = 0 ⇒ (R + r)² − 2R (R + r) = 0.
  • Step 3 : R + r = 2R ⇒ R = r.
  • Step 4 : P = ε² r / (4 r²) = ε² ÷ 4r.
Remember

V = ε − Ir; I = ε/(R + r); P_max = ε²/4r.

Common mistake / CET trap

emf ≠ terminal V; do not forget r.

Shortcut

Memory clue : "ε = V + Ir".

Spot the difference

StateVI
Openε0
Normalε − Irε/(R + r)
Short0ε/r

Practice questions

Easy
  1. 1What is emf?
  2. 2Formula for terminal V?
  3. 3Short-circuit current?
Medium
  1. 4ε = 9, r = 0.5, R = 4.5 : I, V?
  2. 5ε = 3, V = 2.7, R = 9 : r?
  3. 6ε = 24, r = 2 : P_max?
Challenging
  1. 7Prove P_max = ε²/4r.
  2. 8Why is V higher while charging?
Show answers (try them yourself first)
  1. 1Energy per unit charge from the cell (V when open)
  2. 2ε − Ir
  3. 3ε/r
  4. 4I = 1.8 A; V = 8.1 V
  5. 5r = 1 Ω
  6. 672 W
  7. 7dP/dR = 0
  8. 8The external source must also supply Ir

Cells in series and parallel, types of cells

From daily life

A torch has two or three cells in a row — to raise the voltage.

Simple definition

A battery is a group of cells; connection changes ε and r.

Arrays of cellsseriesnε ; nrparallel rowsnε ; nr/mmax I : R = nr/mI = nε ÷ (R + nr/m)

In the figure : Series above, parallel rows beside.

Rule 1: Series — ε_eq = n ε, r_eq = n r; I = n ε ÷ (R + n r). R ≫ r ⇒ series is better.

  • Easy : 4 cells ε = 1.5, r = 0.5, R = 4 ⇒ I = 6 ÷ (4 + 2) = 1 A.
  • Medium : 6 cells of 2 V ⇒ 12 V (lead-acid battery).
  • Exam (CET) : Emf in series? (1) nε (2) ε (3) ε/n (4) 0 → answer 1)

Rule 2: Parallel — ε_eq = ε, r_eq = r ÷ n; I = ε ÷ (R + r/n). R ≪ r ⇒ parallel is better (lasts longer).

  • Easy : 4 cells in parallel, R = 4 ⇒ I = 1.5 ÷ (4 + 0.125) = 0.364 A.
  • Medium : Emf does not rise in parallel.
  • Exam (CET) : r_eq in parallel? (1) r/n (2) nr (3) r (4) 0 → answer 1)

Rule 3: Mixed grouping and types of cells — m rows of n cells : ε = nε, r = nr/m; maximum I : R = nr/m. Primary (Leclanché 1.5 V), secondary (lead-acid 2 V, rechargeable), fuel cells.

  • Easy : 12 cells, 3 rows × 4 : ε = 6 V, r = 0.667 Ω (r = 0.5 each).
  • Medium : R = nr/m = 0.667 Ω ⇒ maximum I = 4.5 A.
  • Exam (CET) : Rechargeable? (1) lead-acid (2) Leclanché (3) Daniell (4) dry → answer 1)

Derivation : r of a parallel group

  • Step 1 : n cells in parallel : each supplies I/n.
  • Step 2 : V = ε − (I/n) r.
  • Step 3 : Compare with V = ε_eq − I r_eq.
  • Step 4 : ε_eq = ε, r_eq = r/n.
Remember

Series nε, nr; parallel ε, r/n; mixed nε, nr/m.

Common mistake / CET trap

Parallel does not raise emf; maximum I when R = nr/m.

Shortcut

Memory clue : "Series = voltage, parallel = endurance".

Spot the difference

Groupε_eqr_eq
Series (n)nεnr
Parallel (n)εr/n
Mixed (m×n)nεnr/m

Practice questions

Easy
  1. 1ε and r of a series group?
  2. 2When is a parallel group preferred?
  3. 3What is a secondary cell?
Medium
  1. 45 cells of 1.5 V, r = 0.2, R = 2 : I?
  2. 53 parallel cells of 1.5 V, r = 0.6 : r_eq?
  3. 62 rows × 3 cells (ε = 2, r = 1) : ε, r?
Challenging
  1. 7R for maximum I in a mixed group?
  2. 8Prove r_eq for a parallel group.
Show answers (try them yourself first)
  1. 1nε; nr
  2. 2When R is small
  3. 3A rechargeable cell
  4. 42.50 A
  5. 50.2 Ω
  6. 6ε = 6 V; r = 1.5 Ω
  7. 7R = nr/m
  8. 8V = ε − (I/n)r

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
CurrentI = Q/t = nAev_dA; A
OhmV = IR; R = ρL/AΩ; M L² T⁻³ A⁻²
ρm/(ne²τ)Ω m
TemperatureR_t = R₀(1 + αΔT)K⁻¹
PowerP = VI = I²R = V²/RW; M L² T⁻³
Series/parallelΣR; 1/R = Σ1/RΩ
CellV = ε − IrV; M L² T⁻³ A⁻¹
Max powerR = r; ε²/4rW

🎯 Most important points for the exam

  • I = nAev_d; v_d is small, the field is fast.
  • V = IR; a stretched wire has R ∝ L².
  • Metals α +, semiconductors α −.
  • P = VI = I²R = V²/R; 1 kWh = 3.6 × 10⁶ J.
  • Series ΣR (same I); parallel 1/R = Σ1/R (same V).
  • V = ε − Ir; P is maximum at R = r.
  • Mixed group : nε, nr/m.
Study mantra

I → V = IR → ρ → P → series/parallel → ε, r → cell groups.