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Physics · Class 12 · Chapter 3

Current Electricity

This chapter links the microscopic picture of drifting electrons to the circuit laws you use every day. Numerical questions on cells, Kirchhoff's rules and bridges are very common, so practise drawing and labelling circuits neatly.

In this chapter: electric current and current density, Ohm's law, drift velocity, mobility and the origin of resistivity, limitations of Ohm's law, resistivity of materials and its temperature dependence, electrical energy and power, emf and internal resistance, cells in series and parallel, Kirchhoff's rules, the Wheatstone bridge.

Electric current

Current is the rate of flow of charge through a cross-section, I = dq/dt. Its SI unit is the ampere (A), a base unit. By convention the direction of current is the direction in which positive charge would move, so in a metal it is opposite to the motion of electrons. Current has a direction but it is a scalar, since currents add algebraically and do not follow the laws of vector addition.

Current density j is current per unit area normal to the flow, a vector with unit A m−2.

Ohm's law and resistivity

For a metallic conductor at constant temperature, V = IR. The resistance depends on the dimensions and the material:

R = ρlAE = ρj  or  j = σEρ is resistivity (Ω m), σ = 1/ρ is conductivity (S m−1)

Resistivity depends on the material and on temperature, not on the length or thickness of the wire. When a wire is stretched, its volume stays the same, so if the length becomes n times the area becomes 1/n times and the resistance becomes n2 times.

Drift of electrons and the origin of resistivity

Free electrons in a metal move randomly at high speeds, but with no field their average velocity is zero. When a field E is applied, each electron accelerates at a = −eE/m between collisions. Averaged over many electrons, this gives a small steady drift velocity opposite to E:

vd = eEτmI = neAvdρ = mne2τn = number density of free electrons, τ = average relaxation time between collisions

This derivation shows why Ohm's law holds: vd is proportional to E, so j is proportional to E. The mobility μ is the drift speed per unit field, μ = vd/E = eτ/m, with unit m2 V−1 s−1.

Drift speeds are tiny, around 1 mm s−1, yet a bulb lights the moment the switch is closed. The field is set up along the whole wire almost at the speed of light, and electrons everywhere in the wire start drifting at once.

Worked example: A copper wire of cross-section 1.0 × 10−7 m2 carries 1.5 A. Taking n = 8.5 × 1028 m−3, find the drift speed.
Solution: vd = I/(neA) = 1.5/(8.5 × 1028 × 1.6 × 10−19 × 1.0 × 10−7) = 1.5/1360 ≈ 1.1 × 10−3 m s−1, about 1.1 mm per second.

Limitations of Ohm's law

Ohm's law is an empirical rule, not a basic law. It fails when V is not proportional to I (for example in a filament lamp as it heats up), when the relation depends on the sign of V (a diode), and when the same current can occur at more than one voltage (GaAs). Such devices are called non-ohmic.

Resistivity of materials and temperature

Metals have low resistivity (10−8 to 10−6 Ω m), insulators very high (up to 1018 times more), and semiconductors lie in between. For metals, over a limited range,

ρT = ρ0[1 + α(T − T0)]α = temperature coefficient of resistivity, unit K−1 or °C−1
  • Metals: α is positive. As T rises, ions vibrate more, collisions become more frequent, τ falls and ρ rises.
  • Nichrome, manganin, constantan: high resistivity and very weak temperature dependence, which is why they are used for standard resistors and heating elements.
  • Semiconductors: ρ decreases as T rises (α negative), because n increases with temperature faster than τ falls.

Electrical energy and power

When charge flows through a potential difference V, energy is converted at the rate

P = VI = I2R = V2R

This power appears as heat (Joule heating). It explains why power is transmitted at high voltage: for the same power the current is smaller, and the loss I2Rc in the transmission cables drops sharply.

Cells, emf and internal resistance

The emf ε of a cell is the potential difference between its terminals when no current is drawn. The electrolyte has an internal resistance r, so when a current I flows through an external resistor R:

I = εR + rV = ε − Irwhile a cell is being charged, current flows into its positive terminal and V = ε + Ir

The terminal voltage is always less than ε while the cell supplies current. The maximum current (short circuit, R = 0) is ε/r.

Cells in series and parallel

ConnectionEquivalent emfEquivalent internal resistance
Series (+ of one to − of next)ε1 + ε2 (subtract if one is reversed)r1 + r2
Parallel (like terminals joined)(ε1r2 + ε2r1)/(r1 + r2)r1r2/(r1 + r2)

For n identical cells in series, the emf is nε and the internal resistance nr; for n in parallel, the emf stays ε and the internal resistance is r/n.

Worked example: A battery of emf 12 V and internal resistance 0.4 Ω is connected to a 3.6 Ω resistor. Find the current, the terminal voltage and the power dissipated in the resistor.
Solution: I = 12/(3.6 + 0.4) = 3 A. Terminal voltage V = 12 − 3 × 0.4 = 10.8 V (check: 3 × 3.6 = 10.8 V). Power in the resistor = I2R = 9 × 3.6 = 32.4 W.

Kirchhoff's rules

  • Junction rule: at any junction, the sum of currents entering equals the sum leaving. It follows from conservation of charge.
  • Loop rule: the algebraic sum of changes in potential around any closed loop is zero. It follows from the fact that electrostatic force is conservative.

Sign convention for the loop rule: going through a resistor in the direction of the assumed current, the potential drops by IR (write −IR); going through a cell from − to +, the potential rises by ε (write +ε). Mark a current in each branch before you start. If a current comes out negative, it simply flows the other way; do not redraw the circuit.

Wheatstone bridge

Four resistors are arranged as a diamond with a cell across one diagonal (A to C) and a galvanometer across the other (B to D). Applying the loop rule to ABDA and BCDB with zero galvanometer current gives the balance condition:

R2R1 = R3R4
Wheatstone bridgewww.iitmedicoguide.comG+−εABCDR₁R₂R₃R₄IgI₁I₂IBalanced bridgeIg = 0VB = VDR₂/R₁ = R₃/R₄(equivalently R₁/R₂ = R₄/R₃)www.iitmedicoguide.com
In a balanced Wheatstone bridge B and D are at the same potential, so no current flows through the galvanometer. Three known resistances then give the fourth.

At balance, the galvanometer arm carries no current and can be removed or shorted without changing any other current, which is a quick way to simplify symmetric networks. The metre bridge is the practical form of this circuit; it and the potentiometer are no longer in the rationalised NCERT text, but the metre bridge is still worth knowing from the lab.

Common mistakes: (1) Thinking a stretched wire keeps the same area; volume is constant, so R goes as l2. (2) Confusing emf with terminal voltage when current is being drawn. (3) Getting loop-rule signs wrong halfway round the loop; write the rule for each element before adding. (4) Assuming a 100 W bulb always glows brighter than a 60 W bulb; in series at the same supply, the 60 W bulb has the higher resistance and glows brighter. (5) Using ρT = ρ0(1 + αΔT) for semiconductors.

JEE and NEET focus

  • Drift velocity, relaxation time, mobility and I = neAvd.
  • Resistance changes on stretching a wire or changing temperature.
  • Terminal voltage, maximum power transfer (R = r) and combinations of cells.
  • Kirchhoff's rules on two-loop circuits.
  • Balance condition of the Wheatstone bridge and its use to simplify networks.

Practice questions

A wire of resistance R is stretched uniformly to twice its length. Its new resistance is:

  1. R/2
  2. 2R
  3. 4R
  4. R/4
Show answer
C. l doubles and A halves, so R = ρl/A becomes 4 times.

The potential difference across a given metal wire is doubled at constant temperature. The drift velocity of electrons:

  1. is halved
  2. stays the same
  3. doubles
  4. becomes four times
Show answer
C. vd = eEτ/m and E = V/l doubles.

The resistivity of a pure semiconductor, when its temperature rises:

  1. increases
  2. decreases
  3. stays the same
  4. first rises then falls
Show answer
B. The number of charge carriers increases with temperature.

A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 3.5 Ω resistor. The terminal voltage is:

  1. 2.0 V
  2. 1.75 V
  3. 1.5 V
  4. 0.25 V
Show answer
B. I = 2/4 = 0.5 A; V = 2 − 0.5 × 0.5 = 1.75 V.

In a Wheatstone bridge (labelled as in the diagram), R1 = 10 Ω, R2 = 20 Ω and R4 = 15 Ω. For balance, R3 must be:

  1. 7.5 Ω
  2. 30 Ω
  3. 15 Ω
  4. 40 Ω
Show answer
B. R3 = R2R4/R1 = 20 × 15/10 = 30 Ω.

A 100 W and a 60 W bulb, both rated for 220 V, are connected in series across 220 V. Then:

  1. the 100 W bulb glows brighter
  2. the 60 W bulb glows brighter
  3. both glow equally
  4. neither glows
Show answer
B. The 60 W bulb has larger R = V2/P; with the same current, I2R is larger for it.

The SI unit of mobility is:

  1. m s−1
  2. m2 V−1 s−1
  3. V m−1 s−1
  4. m V−1
Show answer
B. μ = vd/E has unit (m s−1)/(V m−1).

Four identical cells, each of emf ε and internal resistance r, are connected in parallel. The equivalent emf and internal resistance are:

  1. 4ε, 4r
  2. ε, r/4
  3. ε/4, r/4
  4. 4ε, r/4
Show answer
B. Identical cells in parallel keep the emf and divide r by the number of cells.
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