By the end of this chapter you'll be able to…

  • 1Relate current to drift velocity (I = neAv_d)
  • 2Apply Ohm's law and resistivity with temperature dependence
  • 3Combine resistors in series and parallel
  • 4Use Kirchhoff's junction and loop laws to solve circuits
  • 5Apply the Wheatstone bridge and potentiometer principles
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Why this chapter matters
Current electricity turns abstract charge into usable energy. Ohm's law, resistivity, Kirchhoff's laws, the Wheatstone bridge, and the potentiometer let us analyse circuits and measure resistance and emf accurately -- the engineering core of electronics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Current Electricity

'Current is the flow of charge — but the real story is about ENERGY transfer, not electron speed.'

1. Chapter Overview

This chapter moves from ELECTROSTATICS (charges at rest) to CURRENT ELECTRICITY (charges in motion). Topics include: electric CURRENT (I = dq/dt), OHM'S LAW (V = IR), RESISTIVITY and its temperature dependence, the DRIFT VELOCITY of electrons, KIRCHHOFF'S LAWS (junction and loop rules) for analysing complex circuits, the WHEATSTONE BRIDGE for precise resistance measurement, the POTENTIOMETER for comparing emfs, and ELECTRIC POWER.


2. Electric Current and Drift Velocity

Current

  • I = dq/dt. Direction: from + to − (conventional current). Unit: Ampere (A).
  • Current density: J = I/A = σE. J = n e v_d.

Drift Velocity

  • v_d = (eE/m)τ. 'The small net velocity of electrons under an applied field.'
  • I = n e A v_d — where n is number density of free electrons.

3. Ohm's Law and Resistivity

Ohm's Law

  • V = IR (at constant temperature). 'The current through a conductor is directly proportional to the potential difference across it.'
  • Ohmic materials: V-I graph is a STRAIGHT LINE through origin (metals). Non-ohmic: Diodes, semiconductors, electrolytes.

Resistivity and Conductivity

  • Resistivity ρ = RA/L. Unit: Ω·m. Intrinsic property of material.
  • Conductivity σ = 1/ρ. Unit: S/m.
  • Temperature dependence: ρ = ρ₀[1 + α(T − T₀)]. α = temperature coefficient of resistance.
MaterialResistivity ρ (Ω·m)Temperature Coefficient α
Copper1.7×10⁻⁸+0.0039/°C
Silicon2.3×10³ (intrinsic)−0.075/°C (NEGATIVE)
Glass10¹⁰ to 10¹⁴

4. Combination of Resistors

CombinationFormulaCharacteristic
SERIESR_eq = R₁ + R₂ + ...Same current, potential divides
PARALLEL1/R_eq = 1/R₁ + 1/R₂ + ...Same potential, current divides

5. Kirchhoff's Laws

Junction Law (KCL)

  • Σ I_in = Σ I_out. 'Charge is CONSERVED at a junction.'
  • 'What goes into a junction MUST come out.'

Loop Law (KVL)

  • Σ V = 0 around any closed loop. 'Energy is CONSERVED in a closed loop.'
  • 'The sum of potential rises equals the sum of potential drops.'

Worked Example 1

Problem: Find the current through the 2 Ω resistor in a circuit with E₁=6V (r₁=1Ω), E₂=4V (r₂=1Ω), and R=2Ω connected in a loop. Solution: Apply KVL: E₁ − Ir₁ − IR − Ir₂ − E₂ = 0 (assuming E₁ > E₂ direction). 6 − I(1) − I(2) − I(1) − 4 = 0 ⇒ 2 − 4I = 0 ⇒ I = 0.5 A.


6. Wheatstone Bridge

  • A circuit for measuring UNKNOWN resistance accurately.
  • Balanced condition: R₁/R₂ = R₃/R₄ (galvanometer shows ZERO current).
  • 'At balance, the ratio of resistances in one branch equals the ratio in the other.'
  • Applications: Metre bridge (practical form of Wheatstone bridge).

7. Potentiometer

  • A device for comparing emfs WITHOUT drawing current.
  • Principle: V = kl (potential drop proportional to length).
  • Comparing emfs: E₁/E₂ = l₁/l₂.
  • Advantage: It draws ZERO current from the source being measured — gives TRUE emf.

8. Comparison Table: Voltmeter vs Potentiometer

AspectVoltmeterPotentiometer
What it measuresPotential differenceEmf or potential difference
Current drawnSmall but non-zeroZERO (null-point method)
AccuracyLimited by internal resistanceVERY HIGH
True emf?No (draws some current)YES (no current drawn)

9. Electric Power and Energy

  • Power: P = VI = I²R = V²/R. Unit: Watt (W).
  • Joule heating: H = I²Rt. 'The heat produced by current in a resistor.'
  • Kilowatt-hour (kWh) : Commercial unit of energy. 1 kWh = 3.6×10⁶ J.

10. Common Mistakes

  1. Conventional current vs electron flow: Conventional current flows from + to −. Electrons flow from − to +. For circuit analysis, use CONVENTIONAL current.
  2. Kirchhoff's sign convention: 'Potential RISE from − to + (battery), Potential DROP from + to − (battery).' Be consistent in your loop direction.
  3. Resistors in parallel: 1/R_eq is the SUM of RECIPROCALS. Students often add resistances directly (that's series).
  4. Voltmeter connection: Always in PARALLEL. Ammeter in SERIES. Connecting them wrong can damage the instrument.

11. CBSE Exam Focus

  1. Ohm's law and V-I characteristics — ohmic vs non-ohmic materials
  2. Drift velocity — derivation of I = n e A v_d
  3. Resistivity — temperature dependence, α
  4. Kirchhoff's laws — solving complex circuits (junction and loop equations)
  5. Wheatstone bridge — balanced condition, metre bridge problems
  6. Potentiometer — comparing emfs, finding internal resistance of a cell

12. Self-Test

Q1: A copper wire of length 2 m and area 0.5 mm² has resistivity 1.7×10⁻⁸ Ω·m. Find its resistance. A1: R = ρL/A = (1.7×10⁻⁸)(2)/(0.5×10⁻⁶) = (3.4×10⁻⁸)/(5×10⁻⁷) = 0.068 Ω.

Q2: Find the current through a 10 Ω resistor when connected to a 12 V battery with internal resistance 2 Ω. A2: I = E/(R+r) = 12/(10+2) = 12/12 = 1 A.

Q3: In a Wheatstone bridge, R₁=2Ω, R₂=4Ω, R₃=3Ω. Find R₄ for balance. A3: R₁/R₂ = R₃/R₄ ⇒ 2/4 = 3/R₄ ⇒ R₄ = 6 Ω.

Q4: A potentiometer wire has length 10 m and resistance 20 Ω. A 2V driver cell is connected. Find the potential gradient. A4: Total current I = 2/20 = 0.1 A. Potential gradient k = IR/L = (0.1×20)/10 = 0.2 V/m.

Q5: Compare the emfs of two cells using a potentiometer. Balance lengths are 80 cm and 60 cm. A5: E₁/E₂ = l₁/l₂ = 80/60 = 4/3. E₁:E₂ = 4:3.


13. Conclusion

Current electricity is the ENGINEERING side of electrostatics:

  • OHM'S LAW: 'The simplest relation between V and I — linear, predictable, fundamental.'
  • KIRCHHOFF: 'The two rules that let you analyse ANY circuit — no matter how complex.'
  • WHEATSTONE BRIDGE: 'Precision resistance measurement — balanced when the galvanometer reads zero.'
  • POTENTIOMETER: 'The instrument that measures TRUE emf — no current, no error.'

'Current electricity transforms abstract charge into usable energy — the flow of electrons powers our world.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Ohm's law and drift
V = IR; I = neAv_d
Resistance R depends on material and geometry.
Resistivity
rho = RA/L; rho = rho0[1 + alpha(T - T0)]
Conductivity sigma = 1/rho.
Kirchhoff's laws
Sum I at junction = 0; Sum V around loop = 0
Charge and energy conservation.
Wheatstone bridge balance
R1/R2 = R3/R4
Galvanometer reads zero at balance.
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Adding parallel resistors directly
In parallel use 1/R_eq = sum of 1/R; direct addition is for series.
WATCH OUT
Confusing conventional current with electron flow
Conventional current flows from + to -; electrons flow the opposite way. Use conventional current in analysis.
WATCH OUT
Connecting meters incorrectly
Ammeter goes in series (low resistance), voltmeter in parallel (high resistance).
WATCH OUT
Inconsistent signs in Kirchhoff's loop law
Fix a loop direction and apply consistent sign conventions for rises and drops.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Resistivity
A 2 m copper wire of area 0.5 mm^2 has resistivity 1.7e-8 ohm m. Find its resistance.
Show solution
R = rho L/A = 1.7e-8 x 2 / 0.5e-6 = 0.068 ohm.
Q2EASY· Circuit
Find the current through a 10 ohm resistor connected to a 12 V battery of internal resistance 2 ohm.
Show solution
I = E/(R + r) = 12/(10 + 2) = 1 A.
Q3EASY· Wheatstone
In a Wheatstone bridge R1 = 2, R2 = 4, R3 = 3 ohm. Find R4 for balance.
Show solution
R1/R2 = R3/R4 gives 2/4 = 3/R4, so R4 = 6 ohm.
Q4MEDIUM· Potentiometer
A potentiometer wire of length 10 m and resistance 20 ohm is driven by a 2 V cell. Find the potential gradient.
Show solution
Current I = 2/20 = 0.1 A. Potential gradient = IR/L = (0.1 x 20)/10 = 0.2 V/m.
Q5MEDIUM· Kirchhoff
Two cells E1 = 6 V (r1 = 1) and E2 = 4 V (r2 = 1) oppose across R = 2 ohm in a loop. Find the current.
Show solution
KVL: 6 - I(1) - I(2) - I(1) - 4 = 0, so 2 - 4I = 0 and I = 0.5 A.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Current I = dq/dt; I = neAv_d; conventional current flows + to -.
  • Ohm's law V = IR; ohmic materials give a straight V-I line.
  • Resistivity rho = RA/L; metals increase rho with temperature.
  • Series R_eq = sum R; parallel 1/R_eq = sum 1/R.
  • Kirchhoff: junction (charge) and loop (energy) laws.
  • Wheatstone bridge balance: R1/R2 = R3/R4.
  • Potentiometer draws no current and gives true emf; E1/E2 = l1/l2.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 7-9 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Kirchhoff / circuits3-51Multi-loop circuit analysis
Potentiometer / bridge31Comparing emf and measuring resistance
Ohm's law / resistivity2-31Resistance and drift velocity
Prep strategy
  • Master series and parallel resistor formulas
  • Set consistent signs in Kirchhoff loops
  • Learn the balance conditions for bridge and potentiometer
  • Practise drift velocity and resistivity numericals

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Electrical engineering

Circuit analysis with Ohm's and Kirchhoff's laws underlies all electrical design.

Measurement instruments

Wheatstone bridges and potentiometers give precise measurements of resistance and emf.

Power distribution

Resistance and Joule heating govern transmission losses and the design of heaters and fuses.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Reduce networks using series/parallel rules first
  2. Apply Kirchhoff with a fixed loop direction
  3. State balance conditions for bridges and potentiometers
  4. Track units in resistivity and power calculations

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Analyse balanced and unbalanced Wheatstone networks and symmetry shortcuts.
  • Derive the temperature dependence of resistivity from the relaxation-time model.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Physics examHigh
JEE Main and Advanced (Current Electricity)High
NEET PhysicsHigh

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

A voltmeter, although high in resistance, still draws a small current from the cell, so the voltage it reads is slightly less than the true emf because of the internal-resistance drop. A potentiometer works by the null method: the balance point is found when no current flows through the cell being measured. With zero current there is no internal-resistance drop, so the potential measured equals the true emf, making the potentiometer far more accurate.

Free electrons in a conductor move randomly at high speeds, but their average (drift) velocity under an applied field is very small -- often less than a millimetre per second. Current is large nonetheless because of the enormous number density n of free electrons: I = neAv_d. Moreover, the electric field is established almost instantly along the whole wire, so all electrons start drifting together and the current appears immediately, even though individual electrons drift slowly.
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Last reviewed on 30 May 2026. Written and reviewed by subject-matter experts — read about our process.
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