Current Electricity

Current Electricity

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Current electricity studies the steady, directional drift of charge carriers through a conductor. The microscopic link between the macroscopic current I and the slow drift velocity v_d of electrons is given by

I = n A q v_d,

where n is the free-electron number density, A the cross-section, and q = e. A typical drift velocity in a copper wire is ~10⁻⁴ m/s, so electric signals travel near light-speed but electrons barely creep.

Ohm’s law states V = IR, valid only at constant temperature. Resistance depends on geometry through R = ρL/A, where ρ (resistivity, Ω·m) is a material property. For series resistors, R’s add; for parallel, the reciprocals add.

JEE Main quick pointers: (1) Memorise V = ε − Ir for cells with internal resistance r. (2) Remember the maximum-power condition R_load = r, giving P_max = ε²/(4r). (3) Wheatstone bridge balance: P/Q = R/S.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Ohm’s Law and Resistivity

A conductor obeys V = IR when temperature, strain, and other physical conditions are held constant; the ratio R = V/I defines its resistance in ohms. Unlike resistance, resistivity ρ (Ω·m) is intrinsic — copper is ~1.7 × 10⁻⁸ Ω·m, nichrome ~1.1 × 10⁻⁶ Ω·m. Because R = ρL/A, doubling a wire’s length doubles its resistance, while doubling its cross-section halves it. Conductance G = 1/R has SI unit siemens (S).

Combinations

• Series: the same current passes through every resistor, so voltages add: V = I(R₁ + R₂ + …), and R_eq = ΣRᵢ.
• Parallel: the same potential appears across each branch, so currents add: I = V(1/R₁ + 1/R₂ + …), and 1/R_eq = Σ(1/Rᵢ). Parallel combination always lowers R_eq below the smallest branch.

Kirchhoff’s Rules

KCL (junction rule): ΣI_in = ΣI_out. KVL (loop rule): traversing any closed loop, Σε = ΣIR, with sign conventions for emf and IR drops. These rules convert any network into a solvable linear system.

Cells and Internal Resistance

A real cell of emf ε and internal resistance r delivers terminal voltage V = ε − Ir. The current through an external load R is I = ε/(R + r). Maximum power transfer to the load occurs when R = r, giving P_max = ε²/(4r) — a common JEE assertion-reason trap.

Wheatstone Bridge and Potentiometer

A Wheatstone bridge is balanced when P/Q = R/S; the galvanometer shows null deflection, and an unknown resistance is found from the ratio of known arms. A potentiometer compares EMFs by sliding a jockey along a uniform wire until the galvanometer reads zero: ε₁/ε₂ = ℓ₁/ℓ₂. Because it draws no current from the source, it gives a true EMF comparison.

Joule Heating

H = I²Rt joules of heat are dissipated in a resistor in time t. Power forms: P = VI = I²R = V²/R.

Quantity	Formula	Typical JEE form
Drift velocity	v_d = I/(nAq)	numerical with n given
Resistance vs temperature	ρ_T = ρ₀[1 + α(T − T₀)]	α in /°C, sign matters
Cells in series (same ε, r)	ε_eq = nε, r_eq = nr	mix with parallel cells carefully

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Temperature Dependence and Superconductivity

Resistivities of metals rise linearly with temperature because lattice vibrations scatter conduction electrons more frequently. Semiconductors (Si, Ge) behave oppositely: more charge carriers are thermally liberated as T rises, so α < 0. Alloys like constantan or manganin have α ≈ 0, making them ideal for standard resistors. Below a material-specific critical temperature T_c, a superconductor’s resistivity collapses to zero — a true zero, not merely a very small value; persistent currents in superconducting loops have been observed to flow for years.

Drift Velocity vs Signal Speed

A common misconception is that electrons “rush” through a wire. In reality, the drift velocity (∼mm/s in household wiring) is orders of magnitude slower than the propagation of the electric field, which sets up at nearly the speed of light. JEE problems exploit this by giving n and A and asking for the time for one electron to traverse a length L: use t = L/v_d = nAqL/I.

Maximum Power Transfer — Why Half is Lost

At R = r, only half the generated power is delivered to the load; the other half is dissipated inside the source. Power transfer is not efficient at the matched condition — it is maximum, useful for communication circuits, not for power lines (which use R ≫ r).

Common JEE Traps

• Forgetting to convert ε and r when mixing series and parallel cells.
• Treating “potential difference across a resistor” and “terminal voltage of a cell” as the same — they differ by the IR drop across r.
• Confusing resistance (geometry-dependent) with resistivity (material property).
• Sign errors in KVL loops when a battery is traversed from − to + terminal.

Worked Micro-Example

A 12 V battery with internal resistance 1 Ω drives a 5 Ω resistor. I = 12/(5 + 1) = 2 A, V_load = 2 × 5 = 10 V, and P_load = 10 × 2 = 20 W (theoretical max would be ε²/4r = 36 W at R = 1 Ω).

Practice Prompts

1. Two cells (ε₁ = 6 V, r₁ = 1 Ω) and (ε₂ = 3 V, r₂ = 2 Ω) oppose each other across an 8 Ω load. Find the current and the potential difference across each cell.
2. A potentiometer wire is 100 cm long with a 2 V driver cell. A standard cell of 1.018 V balances at 50.9 cm. An unknown EMF balances at 75 cm. Determine the unknown EMF and comment on why no current flows in the standard cell branch.

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