Class 12 Physics · Chapter 3

Current Electricity

Master electric current, Ohm's Law, drift velocity, Kirchhoff's rules and Wheatstone Bridge — high-weightage topics for the AAI ATC exam.

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8 Subtopics
Full NCERT Coverage
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24+ MCQs
Numerical Practice
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All Key Formulas
Quick Reference
3.1
Electric Current Definition · Steady vs instantaneous current · SI unit: Ampere

What is Electric Current?

Electric current is the net charge flowing across a given cross-section per unit time. For steady currents: I = q/t. More generally, the instantaneous current is defined as the limit of ΔQ/Δt as Δt → 0.

Both positive charges moving forward and negative charges moving backward contribute to net current. The SI unit of current is the Ampere (A). Domestic appliances use currents of order 1 A; lightning carries tens of thousands of amperes; nerve signals are in microamperes.

📐 Key Formulas
I = q/t    — steady current (charge per unit time)
I(t) = limΔt→0 ΔQ/Δt    — instantaneous current
1 A = 1 C/s    — SI unit definition
AAI ATC Tip: Current is a scalar quantity even though drawn with arrows. Currents do NOT obey vector addition law. I = j · ΔS (scalar product of current density vector and area vector).
Practice MCQs — Section 3.1
Q1 A charge of 180 C flows through a conductor in 1 minute. The current in the conductor is:
180 A
3 A
0.5 A
10800 A
I = q/t = 180/60 = 3 A
Q2 In 10 s, 6.25 × 10¹⁸ electrons cross a cross-section. The current is (e = 1.6 × 10⁻¹⁹ C):
0.1 A
1.0 A
10 A
0.01 A
q = 6.25×10¹⁸ × 1.6×10⁻¹⁹ = 1 C; I = q/t = 1/10 = 0.1 A
Q3 The current in a nerve signal is 2 μA. Charge flowing in 5 s is:
10 C
2.5 × 10⁻⁶ C
10 × 10⁻⁶ C
4 × 10⁻⁷ C
q = I×t = 2×10⁻⁶ × 5 = 10 × 10⁻⁶ C = 10 μC
3.2
Ohm's Law and Resistivity V = IR · R = ρl/A · Conductivity · Current density

Ohm's Law

Discovered by G.S. Ohm in 1828, Ohm's law states that the current I through a conductor is proportional to the potential difference V across its ends: V = IR, where R is the resistance (unit: ohm, Ω). Resistance depends on material and geometry, not just material alone.

Resistivity and Conductivity

Resistance depends on length l and area A: R = ρl/A, where ρ (resistivity) is a material property. Its reciprocal σ = 1/ρ is called conductivity. The vector form of Ohm's law: j = σE, where j is current density (A/m²).

Metals: ρ ~ 10⁻⁸ to 10⁻⁶ Ω·m  |  Insulators: ~10¹⁸ times higher  |  Semiconductors: intermediate

📐 Key Formulas
V = IR    — Ohm's law (R = resistance in Ω)
R = ρl/A    — resistance from geometry and resistivity
j = I/A    — current density (A/m²)
E = jρ    or    j = σE    — vector form of Ohm's law
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Insert Diagram: Slab of conductor showing R = ρl/A — doubling length doubles R; halving area doubles R (Fig 3.2 NCERT)

Practice MCQs — Section 3.2
Q4 A wire of length 15 m and cross-section 6×10⁻⁷ m² has resistance 5 Ω. Its resistivity is:
2×10⁻⁷ Ω·m
2×10⁻⁷ Ω·m
1.25×10⁷ Ω·m
5×10⁻⁷ Ω·m
ρ = RA/l = 5 × 6×10⁻⁷ / 15 = 30×10⁻⁷/15 = 2×10⁻⁷ Ω·m
Q5 A conductor has resistance 10 Ω. If its length is doubled and area halved, the new resistance is:
10 Ω
20 Ω
40 Ω
5 Ω
R' = ρ(2l)/(A/2) = 4ρl/A = 4R = 4×10 = 40 Ω
Q6 Current density in a wire is 5×10⁶ A/m² and E = 250 V/m. The conductivity of the material is:
5×10⁻⁵ S/m
2×10⁴ S/m
1.25×10⁹ S/m
4×10⁻⁵ S/m
σ = j/E = 5×10⁶/250 = 2×10⁴ S/m
3.3
Drift of Electrons & Mobility Drift velocity · Relaxation time τ · I = neAv_d · Mobility μ

Origin of Drift Velocity

In the absence of an electric field, electrons move randomly with zero average velocity. When an electric field E is applied, electrons experience acceleration a = eE/m (opposite to E). Due to random collisions with ions occurring every τ seconds (relaxation time), electrons acquire a steady average drift velocity:

vd = eEτ/m (magnitude)

This drift is extremely slow (~mm/s) compared to thermal speeds (~10² m/s), yet large currents flow because electron number density n is enormous (~10²⁸ m⁻³).

Current from Drift Velocity

Consider a conductor of cross-section A with n free electrons per unit volume. In time Δt, all electrons in volume A·|vd|·Δt cross the area. Since each carries charge e:

I = neAvd

Mobility μ = |vd|/E = eτ/m  (unit: m²/V·s)

📐 Key Formulas
vd = eEτ/m    — drift velocity (τ = relaxation time)
I = neAvd    — current from drift (n = electron density)
σ = ne²τ/m    — conductivity from microscopic model
μ = vd/E = eτ/m    — mobility (m²/V·s)
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Insert Diagram: Electron drift — random zigzag path without field (A to B), slight net drift opposite to E with field (A to B') (Fig 3.3 NCERT)

Practice MCQs — Section 3.3
Q7 A copper wire (A = 1×10⁻⁷ m², I = 1.5 A, n = 8.5×10²⁸ m⁻³, e = 1.6×10⁻¹⁹ C). The drift velocity is approximately:
1.1 m/s
1.1 × 10⁻³ m/s
1.1 × 10⁻⁶ m/s
1.1 × 10⁶ m/s
vd = I/(neA) = 1.5/(8.5×10²⁸ × 1.6×10⁻¹⁹ × 1×10⁻⁷) = 1.5/(1.36×10³) ≈ 1.1×10⁻³ m/s = 1.1 mm/s
Q8 An electron has relaxation time τ = 2.5×10⁻¹⁴ s, mass m = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C. Its mobility is approximately:
2.5×10⁻³ m²/V·s
4.4×10⁻³ m²/V·s
6.0×10⁻³ m²/V·s
1.1×10⁻³ m²/V·s
μ = eτ/m = 1.6×10⁻¹⁹ × 2.5×10⁻¹⁴ / 9.1×10⁻³¹ = 4×10⁻³³/9.1×10⁻³¹ ≈ 4.4×10⁻³ m²/V·s
Q9 A 3 m long copper wire (A = 2×10⁻⁶ m², I = 3 A, n = 8.5×10²⁸ m⁻³). Time for electron to drift end-to-end is:
1.6 × 10³ s
2.7 × 10⁴ s
3.0 × 10² s
1.0 × 10⁵ s
vd = I/(neA) = 3/(8.5×10²⁸ × 1.6×10⁻¹⁹ × 2×10⁻⁶) = 3/(2.72×10⁴) ≈ 1.1×10⁻⁴ m/s; t = L/vd = 3/1.1×10⁻⁴ ≈ 2.7×10⁴ s
3.4
Temperature Dependence & Electrical Power ρ_T = ρ₀[1+α(T−T₀)] · P = IV = I²R = V²/R

Temperature Dependence of Resistivity

For metals, resistivity increases with temperature (α > 0) since τ decreases as collision frequency increases. For semiconductors and insulators, resistivity decreases with temperature because n (electron density) increases more than τ decreases.

Materials like Nichrome, Manganin, Constantan have very low α and are used in standard resistors precisely for this stability.

Electrical Power Dissipation

When current I flows through a conductor with potential difference V, energy is dissipated as heat (Joule heating). Power P = IV = I²R = V²/R. In power transmission, power loss in cables Pc = P²Rc/V² — minimised by using high voltage V.

📐 Key Formulas
ρT = ρ₀[1 + α(T − T₀)]    — temperature dependence
P = IV = I²R = V²/R    — electrical power (watts)
Pc = P²Rc/V²    — cable power loss (minimise by high V)
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Insert Diagram: ρ vs T graphs for copper (linear increase), nichrome (flat), and semiconductor (decreasing) (Figs 3.8, 3.9, 3.10 NCERT)

Practice MCQs — Section 3.4
Q10 A nichrome heater (R₁ = 75.3 Ω at 27°C) draws I = 2.68 A from 230 V supply. Steady-state temperature is (α = 1.70×10⁻⁴ °C⁻¹):
500 °C
847 °C
1000 °C
347 °C
R₂ = 230/2.68 = 85.8 Ω; T₂−T₁ = (85.8−75.3)/(75.3×1.70×10⁻⁴) = 10.5/0.01281 ≈ 820°C; T₂ = 27+820 = 847°C
Q11 A 100 W bulb is connected to 220 V supply. Resistance of bulb filament is:
22 Ω
220 Ω
484 Ω
2200 Ω
R = V²/P = (220)²/100 = 48400/100 = 484 Ω
Q12 A device consumes 1000 W at 250 V via cables (Rc = 1 Ω). Power wasted in cables is:
4 W
10 W
16 W
100 W
I = P/V = 1000/250 = 4 A; Pc = I²Rc = 16×1 = 16 W
3.5
Cells, EMF and Internal Resistance ε = V + Ir · Terminal voltage · I = ε/(R+r)

EMF and Internal Resistance of a Cell

A cell maintains a potential difference by chemical energy. The EMF (ε) is the potential difference between terminals when no current flows (open circuit). The electrolyte has finite internal resistance r.

When current I flows through external resistance R:
Terminal voltage V = ε − Ir  (during discharge)

I = ε/(R+r); Maximum current Imax = ε/r (when R = 0)

📐 Key Formulas
ε = V + Ir    — EMF = terminal voltage + voltage drop across r
I = ε/(R+r)    — current with internal resistance
V = ε − Ir    — terminal voltage (less than EMF during use)
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Insert Diagram: Cell with EMF ε, internal resistance r connected to external R; current flow and terminal voltage (Fig 3.12 NCERT)

Practice MCQs — Section 3.5
Q13 A battery (ε = 10 V, r = 3 Ω) is connected to a resistor. Current in circuit = 0.5 A. The external resistance R is:
3 Ω
23 Ω
17 Ω
20 Ω
R = ε/I − r = 10/0.5 − 3 = 20 − 3 = 17 Ω. Terminal V = ε − Ir = 10 − 0.5×3 = 8.5 V
Q14 A car battery (ε = 12 V, r = 0.4 Ω). Maximum current drawable without damage:
12 A
4.8 A
30 A
0.4 A
Imax = ε/r = 12/0.4 = 30 A (theoretical max at R=0)
Q15 A battery is being charged by 120 V supply via series resistor 15.5 Ω. Battery: ε = 8 V, r = 0.5 Ω. Terminal voltage during charging is:
8 V
11.5 V
120 V
4.5 V
I = (120−8)/(15.5+0.5) = 112/16 = 7 A; Vterminal = ε + Ir = 8 + 7×0.5 = 11.5 V (adds during charging)
3.6
Cells in Series and Parallel Equivalent EMF · Equivalent internal resistance

Combination Rules

Like resistors, cells can be combined into equivalent single cells:

CombinationEquivalent EMF (εeq)Equivalent r (req)
Series (n cells)ε1 + ε2 + … + εnr1 + r2 + … + rn
Parallel (2 cells)(ε1r2 + ε2r1)/(r1+r2)r1r2/(r1+r2)
AAI ATC Tip: Series combination increases both EMF and internal resistance. Parallel combination gives lower internal resistance — useful for large current requirements. If any cell in series is reversed, its EMF is subtracted: εeq = ε1 − ε2.
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Insert Diagram: Two cells in series (Fig 3.13) and two cells in parallel (Fig 3.14) NCERT — showing equivalent single cell

Practice MCQs — Section 3.6
Q16 Three cells, each ε = 2 V and r = 0.5 Ω, are connected in series across external resistance R = 4.5 Ω. Current in circuit is:
0.5 A
1.0 A
2.0 A
4.0 A
εeq = 3×2 = 6 V; req = 3×0.5 = 1.5 Ω; I = 6/(4.5+1.5) = 6/6 = 1.0 A
Q17 Two cells (ε₁ = 6 V, r₁ = 1 Ω) and (ε₂ = 4 V, r₂ = 1 Ω) are in parallel. The equivalent EMF is:
10 V
2 V
5 V
6 V
εeq = (ε₁r₂ + ε₂r₁)/(r₁+r₂) = (6×1 + 4×1)/(1+1) = 10/2 = 5 V
Q18 Two identical cells (ε = 3 V, r = 0.5 Ω each) in parallel feed R = 2 Ω. Current through R is:
0.5 A
2.5 A
1.33 A
3 A
req = 0.5×0.5/(0.5+0.5) = 0.25 Ω; εeq = 3 V; I = 3/(2+0.25) = 3/2.25 ≈ 1.33 A
3.7
Kirchhoff's Rules Junction Rule (KCL) · Loop Rule (KVL) · Circuit analysis

Two Fundamental Rules

Junction Rule (KCL): At any junction, the sum of currents entering = sum of currents leaving. This follows from conservation of charge — no charge accumulates at junctions.

Loop Rule (KVL): The algebraic sum of all potential changes (across resistors and cells) around any closed loop is zero. This follows from the fact that electric potential is a state function — the net change around a closed path is zero.

Sign Convention for Loop Rule: Traversing a resistor in the direction of assumed current → potential drops (−IR). Traversing a cell from −ve to +ve terminal → potential rises (+ε). Opposite directions reverse signs.
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Insert Diagram: Kirchhoff's rules applied to a network — junction rule at node 'a', loop rule for two loops (Fig 3.15 NCERT)

Practice MCQs — Section 3.7
Q19 At a junction, currents of 3 A, 4 A enter and currents of 2 A, 1 A leave. The unaccounted current leaving the junction is:
0 A
3 A
4 A
10 A
Entering = 3+4 = 7 A; Leaving (known) = 2+1 = 3 A; Remaining = 7−3 = 4 A
Q20 In a loop: EMF 10 V, resistors 2 Ω, 3 Ω in series, internal resistance 0 Ω. The current by loop rule is:
10 A
3.33 A
2 A
5 A
Loop rule: 10 − 2I − 3I = 0; 5I = 10; I = 2 A
Q21 A cubical network of 12 resistors each 1 Ω, connected diagonally to 10 V battery. The equivalent resistance is:
1 Ω
5/6 Ω
12 Ω
3/2 Ω
Using Kirchhoff's rules with symmetry: −IR − (1/2)IR − IR + ε = 0 → ε = (5/2)IR; Req = ε/3I = 5R/6 = 5/6 Ω
3.8
Wheatstone Bridge Balance condition · Unknown resistance · Meter Bridge

Wheatstone Bridge Principle

Four resistors R₁, R₂, R₃, R₄ arranged in a bridge. A galvanometer G connects the mid-points (B and D). A battery connects A to C. When the bridge is balanced (Ig = 0), applying Kirchhoff's laws to loops ADBA and CBDC gives the balance condition:

R₁/R₂ = R₃/R₄

This allows determination of an unknown resistance R₄ by varying R₃ until null deflection. The practical device is called the Meter Bridge.

📐 Balance Condition
R₁/R₂ = R₃/R₄    — Wheatstone bridge null condition
R₄ = R₃ × (R₂/R₁)    — unknown resistance from balance
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Insert Diagram: Wheatstone bridge circuit with R₁, R₂, R₃, R₄, galvanometer G and battery ε (Fig 3.18 NCERT)

Practice MCQs — Section 3.8
Q22 In a Wheatstone bridge: R₁ = 100 Ω, R₂ = 10 Ω, R₃ = 5 Ω. For balance, R₄ must be:
50 Ω
1 Ω
0.5 Ω
5 Ω
R₄ = R₃ × R₂/R₁ = 5 × 10/100 = 0.5 Ω
Q23 Wheatstone bridge with R₁ = 200 Ω, R₂ = 20 Ω is balanced when R₃ = 50 Ω. The unknown R₄ is:
50 Ω
500 Ω
5 Ω
100 Ω
R₄ = R₃ × R₂/R₁ = 50 × 20/200 = 5 Ω
Q24 Wheatstone bridge: AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, DA = 60 Ω, galvanometer G = 15 Ω across BD, 10 V across AC. Ig is approximately:
10 mA
0 mA (balanced)
4.87 mA
2.5 mA
Check balance: AB/DA = 100/60 ≠ BC/CD = 10/5 — bridge NOT balanced. Applying KVL to three meshes gives Ig = 4.87 mA (NCERT Example 3.7)

📋 Chapter Summary — Quick Revision

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