Class 12 Physics · Chapter 3
⚡ Current Electricity — The Complete NCERT Master Guide
From Ohm's Law to Potentiometers — every concept, formula, question type, and exam strategy you need to score full marks.
📋 Table of Contents
01 Introduction & Chapter Overview
02 Key Concepts Explained
03 Complete Formula Table
04 Chapter Summary (Bullets)
05 NCERT Exercise Q&A
06 Learning Outcome Questions
07 Assertion–Reason Questions
08 Case-Based & HOTS Questions
09 MCQs with Answers
10 Previous-Year CBSE Questions
11 JEE / NEET / CUET Questions
12 Common Student Mistakes
13 Quick Revision & Formula Sheet
14 Chapter Mind Map
15 FAQs
16 Conclusion & Exam Tips
Introduction
Why Current Electricity Matters — And Why You'll Love It
Imagine a world without electric circuits — no mobile phone, no fan, no bulb, no hospital equipment. The invisible flow of electrons through wires powers nearly every device in modern life. Chapter 3: Current Electricity in Class 12 Physics (NCERT) is the chapter that explains this invisible magic with the precision of mathematics.
This chapter builds on the electrostatics concepts you studied in Chapters 1 and 2, but shifts focus from static charges to moving charges — and their spectacular real-world consequences. It is one of the most important chapters for CBSE Board exams (carrying approximately 7–8 marks in 3-mark and 5-mark questions) and is heavily tested in JEE, NEET, CUET, and all competitive exams.
Board Exam Weightage
~7–8 marks in CBSE Term exams. Questions from Kirchhoff's Laws, Wheatstone Bridge, and Potentiometer are asked almost every year.
JEE / NEET Importance
2–3 questions per paper, often involving multi-loop circuits, equivalent resistance, and internal resistance problems.
Links to Other Chapters
Connects directly to Electromagnetic Induction (Ch. 6), Alternating Current (Ch. 7), and Semiconductor Devices (Ch. 14).
Core Content
Key Concepts — Explained Simply
2.1 Electric Current
Electric current is the rate of flow of electric charge through a cross-section of a conductor. If a charge Q passes through a cross-section in time t:
Definition of Current
I = Q / t
I = Current (Ampere, A) | Q = Charge (Coulomb, C) | t = time (second, s)
Current is a scalar quantity (it does not follow vector addition laws at junctions — instead, it follows Kirchhoff's Current Law). The SI unit is the Ampere (A), named after André-Marie Ampère.
Conventional vs. Electron Current
By convention, current flows from positive to negative terminal (high to low potential), but electrons actually flow from negative to positive. Always use the conventional direction in circuit problems.
2.2 Drift Velocity and Its Relation to Current
Free electrons in a conductor move randomly at high speeds (~10⁶ m/s), but with no net displacement. When an electric field is applied, they acquire a small net velocity in a direction opposite to the field — this is called drift velocity (vd).
Current & Drift Velocity
I = n·e·A·vd
n = number density of free electrons (m⁻³) | e = 1.6 × 10⁻¹⁹ C | A = cross-sectional area | vd = drift speed
Drift Velocity from Electric Field
vd = eEτ / m = eV·τ / (m·L)
τ = relaxation time (average time between collisions) | m = mass of electron | E = electric field | V = potential difference | L = length
How slow is drift velocity?
Drift velocity is typically only ~10⁻⁴ m/s — much slower than walking! Yet a bulb lights up instantly because the electric field propagates at nearly the speed of light.
2.3 Ohm's Law and Its Limitations
Ohm's Law states that the current through a conductor is directly proportional to the potential difference across it, provided physical conditions (temperature, material, etc.) remain constant.
Ohm's Law
V = I · R
V = Potential difference (Volt) | I = Current (Ampere) | R = Resistance (Ohm, Ω)
Limitations of Ohm's Law: Not all materials obey Ohm's Law. Materials that do are called ohmic (e.g., metallic wires). Materials that don't are called non-ohmic:
- Diodes — current flows only in one direction (I–V curve is non-linear and asymmetric)
- Thyristors — resistance changes with conditions
- Electrolytic solutions — resistance changes with concentration
- Vacuum tubes — I–V curve follows different laws
2.4 Resistance and Resistivity
Resistance (R) is the opposition offered by a conductor to the flow of current. It depends on the material, length, and cross-sectional area.
Resistance in terms of Resistivity
R = ρ · L / A
ρ = resistivity (Ω·m) | L = length of conductor (m) | A = cross-sectional area (m²)
Microscopic form of Ohm's Law:
J = σE where J = current density (A/m²), σ = conductivity = 1/ρ (S/m), E = electric field (V/m)
| Material | Resistivity ρ (Ω·m) at 20°C | Category |
|---|---|---|
| Silver | 1.6 × 10⁻⁸ | Best conductor |
| Copper | 1.7 × 10⁻⁸ | Common conductor |
| Aluminium | 2.7 × 10⁻⁸ | Conductor |
| Nichrome | 1.0 × 10⁻⁶ | Alloy (heating elements) |
| Silicon | 6.4 × 10² | Semiconductor |
| Glass | 10¹⁰ – 10¹⁴ | Insulator |
2.5 Temperature Dependence of Resistance
For most metallic conductors, resistance increases with temperature (positive temperature coefficient). For semiconductors and insulators, resistance decreases with temperature.
Temperature Dependence
RT = R₀ [1 + α(T − T₀)]
R₀ = resistance at reference temperature T₀ | α = temperature coefficient of resistance (per °C or per K) | T = final temperature
| Material | Temperature Coefficient α (°C⁻¹) |
|---|---|
| Copper | +3.9 × 10⁻³ |
| Nichrome | +1.7 × 10⁻⁴ (very small — used in heaters) |
| Manganin | ≈ 0 (used in standard resistors) |
| Semiconductors | Negative (resistance decreases with T) |
2.6 Combination of Resistors — Series and Parallel
Series Combination
Equivalent Resistance — Series
Req = R₁ + R₂ + R₃ + ...
Same current flows through each resistor. Voltage divides. Req > any individual R.
Parallel Combination
Equivalent Resistance — Parallel
1/Req = 1/R₁ + 1/R₂ + 1/R₃ + ...
Same voltage across each resistor. Current divides. Req < smallest individual R.
For two resistors in parallel:
Req = (R₁ × R₂) / (R₁ + R₂) — the "product over sum" formula. Very handy in exams!
2.7 Cells, EMF, and Internal Resistance
A cell converts chemical energy into electrical energy. The Electromotive Force (EMF, ε) is the maximum potential difference across the terminals when no current flows (open circuit). Internal resistance (r) is the resistance offered by the electrolyte inside the cell.
Terminal Voltage During Discharge
V = ε − I·r
ε = EMF | I = current drawn | r = internal resistance | V = terminal voltage (V < ε during discharge)
Current in External Circuit
I = ε / (R + r)
R = external load resistance | r = internal resistance | ε = EMF
| Condition | V = ? | Meaning |
|---|---|---|
| Open circuit (I = 0) | V = ε | Terminal voltage equals EMF |
| Discharging (current drawn) | V = ε − Ir | V < ε; energy lost in internal resistance |
| Charging (current forced in) | V = ε + Ir | V > ε; external source does extra work |
Cells in Series and Parallel
n identical cells in Series
εnet = nε, rnet = nr
Use series when external resistance R >> r (to maximize current)
n identical cells in Parallel
εnet = ε, rnet = r/n
Use parallel when R << r (to reduce effective internal resistance)
2.8 Kirchhoff's Laws
Kirchhoff's Laws (KCL and KVL) are the backbone of circuit analysis. They are based on the fundamental laws of conservation of charge and energy.
Kirchhoff's Current Law (KCL) — Junction Rule
ΣIin = ΣIout or ΣI = 0 at a junction
Based on Conservation of Charge. The algebraic sum of currents at any junction is zero.
Kirchhoff's Voltage Law (KVL) — Loop Rule
ΣV = 0 in any closed loop
Based on Conservation of Energy. The algebraic sum of all potential differences in any closed loop is zero.
Sign Convention for KVL (VERY IMPORTANT for exams!):
- Traversing a resistor in the direction of current → −IR (potential drops)
- Traversing a resistor against current → +IR (potential rises)
- Traversing a cell from − to + → +ε (going from low to high potential)
- Traversing a cell from + to − → −ε (going from high to low potential)
2.9 Wheatstone Bridge and Meter Bridge
A Wheatstone Bridge is a circuit of four resistors arranged in a diamond shape, used for precise measurement of unknown resistance. It was devised by Charles Wheatstone.
Wheatstone Bridge Balanced Condition
P/Q = R/S (when galvanometer reads zero)
P, Q = ratio arms | R = known resistance | S = unknown resistance | No current through galvanometer = balanced bridge
The Meter Bridge is a practical form of the Wheatstone Bridge. It uses a 1-metre resistance wire (usually manganin or constantan) along which a jockey slides to find the balance point (null point).
Meter Bridge Formula
S = R × (100 − ℓ) / ℓ
ℓ = balance length from left end (cm) | R = known resistance | S = unknown resistance
Why is a balanced Wheatstone Bridge more accurate?
When balanced, no current passes through the galvanometer, eliminating the galvanometer's resistance from the calculation. This gives a highly accurate null method measurement.
2.10 Potentiometer and Its Applications
A potentiometer is a device consisting of a long, uniform resistance wire (typically 10 m) connected to a battery (driver cell). It is used to measure potential differences and EMFs without drawing any current from the circuit (null method).
Potential Gradient
k = V / L (V·m⁻¹)
k = potential gradient | V = voltage across wire | L = total length of wire. At null point: ε = k × ℓ
Application 1: Comparison of EMFs of Two Cells
EMF Comparison
ε₁ / ε₂ = ℓ₁ / ℓ₂
ℓ₁ and ℓ₂ are the null-point lengths for cells 1 and 2 respectively
Application 2: Measurement of Internal Resistance
Internal Resistance
r = R × (ℓ₁ − ℓ₂) / ℓ₂
ℓ₁ = null length (open circuit) | ℓ₂ = null length (with external R connected) | r = internal resistance of cell
Potentiometer vs. Voltmeter — Why Potentiometer is Better
A voltmeter draws some current from the circuit, altering the very quantity it measures. A potentiometer draws zero current at the null point, giving a perfect, non-disturbing measurement of EMF.
Quick Reference
Important Formulas — Complete List
⚡ Current Electricity — Master Formula Sheet
Electric Current
I = Q / t = nevdA
A (Ampere)
Drift Velocity
vd = eEτ/m
m/s
Ohm's Law
V = IR
V = Volt
Current Density
J = I/A = σE = nevd
A/m²
Resistivity
ρ = m / (ne²τ) = 1/σ
Ω·m
Resistance
R = ρL/A
Ω (Ohm)
Temp. dependence of R
RT = R₀[1 + α(T−T₀)]
—
Series combination
Rs = R₁ + R₂ + R₃ + ...
Ω
Parallel combination
1/Rp = 1/R₁ + 1/R₂ + ...
Ω
Terminal Voltage
V = ε − Ir (discharge)
V
Current in circuit
I = ε / (R + r)
A
Power dissipated
P = VI = I²R = V²/R
W (Watt)
Wheatstone Bridge
P/Q = R/S (balanced)
—
Meter Bridge
S = R(100−ℓ)/ℓ
Ω
Potential Gradient
k = V/L; ε = kℓ
V/m
EMF comparison (Potentiometer)
ε₁/ε₂ = ℓ₁/ℓ₂
—
Internal resistance (Potentiometer)
r = R(ℓ₁−ℓ₂)/ℓ₂
Ω
Chapter Summary
Lesson Summary — Key Points at a Glance
Electric current = rate of flow of charge. I = Q/t. Unit: Ampere.
Electrons flow opposite to conventional current direction.
Drift velocity vd is small (~10⁻⁴ m/s) but creates macroscopic current.
I = nevdA connects drift velocity to current.
Ohm's Law: V = IR, valid only for ohmic conductors at constant temperature.
Resistance R = ρL/A. Resistivity ρ is a material property, not shape-dependent.
For metals: R increases with T (positive α). For semiconductors: R decreases with T.
Series: Req = ΣR. Parallel: 1/Req = Σ(1/R).
EMF (ε) ≠ Terminal Voltage (V). V = ε − Ir during discharge.
KCL: ΣI = 0 at junction (conservation of charge).
KVL: ΣV = 0 in a closed loop (conservation of energy).
Wheatstone Bridge balanced when P/Q = R/S and IG = 0.
Meter Bridge: S = R(100−ℓ)/ℓ. Balance point depends only on resistance ratio.
Potentiometer: null method, draws no current — more accurate than voltmeter.
Potentiometer applications: EMF comparison and internal resistance measurement.
Power: P = I²R = V²/R. Energy: W = Pt = VIt (in joules).
NCERT Exercise
NCERT Exercise Questions — Selected Answers
NCERT 3.1 The storage battery of a car has an EMF of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
ANS Maximum current flows when external resistance R = 0 (short circuit).
Imax = ε / r = 12 / 0.4 = 30 A
In practice this is dangerous — car circuits have fuses to prevent this.
Imax = ε / r = 12 / 0.4 = 30 A
In practice this is dangerous — car circuits have fuses to prevent this.
NCERT 3.2 A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, find the resistance of the resistor. Also find the terminal voltage of the battery.
ANS
Using I = ε / (R + r):
0.5 = 10 / (R + 3) → R + 3 = 20 → R = 17 Ω
Terminal voltage V = ε − Ir = 10 − (0.5 × 3) = 10 − 1.5 = 8.5 V
0.5 = 10 / (R + 3) → R + 3 = 20 → R = 17 Ω
Terminal voltage V = ε − Ir = 10 − (0.5 × 3) = 10 − 1.5 = 8.5 V
NCERT 3.5 At room temperature (27.0°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω? Given: α = 1.70 × 10⁻⁴ °C⁻¹ for nichrome.
ANS
RT = R₀[1 + α(T − T₀)]
117 = 100[1 + 1.70×10⁻⁴ × (T − 27)]
1.17 = 1 + 1.70×10⁻⁴ × (T − 27)
0.17 = 1.70×10⁻⁴ × (T − 27)
T − 27 = 1000 → T = 1027°C
117 = 100[1 + 1.70×10⁻⁴ × (T − 27)]
1.17 = 1 + 1.70×10⁻⁴ × (T − 27)
0.17 = 1.70×10⁻⁴ × (T − 27)
T − 27 = 1000 → T = 1027°C
NCERT 3.11 Determine the equivalent resistance of networks of resistors with three resistors 1 Ω, 2 Ω, 3 Ω in parallel, and this combination in series with 4 Ω.
ANS
1/Rparallel = 1/1 + 1/2 + 1/3 = 6/6 + 3/6 + 2/6 = 11/6
Rparallel = 6/11 Ω
Rtotal = 6/11 + 4 = 6/11 + 44/11 = 50/11 ≈ 4.55 Ω
Rparallel = 6/11 Ω
Rtotal = 6/11 + 4 = 6/11 + 44/11 = 50/11 ≈ 4.55 Ω
NCERT 3.16 The number density of free electrons in a copper conductor is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift from one end of a wire 3 m long to its other end? Current = 3.0 A, area = 2.0 × 10⁻⁶ m².
ANS
vd = I / (nAe) = 3 / (8.5×10²⁸ × 2×10⁻⁶ × 1.6×10⁻¹⁹)
vd = 3 / (2.72×10⁴) ≈ 1.1 × 10⁻⁴ m/s (approximately)
Time = L / vd = 3 / (1.1×10⁻⁴) ≈ 2.7 × 10⁴ s ≈ 7.5 hours!
(This illustrates how slow drift velocity actually is.)
vd = 3 / (2.72×10⁴) ≈ 1.1 × 10⁻⁴ m/s (approximately)
Time = L / vd = 3 / (1.1×10⁻⁴) ≈ 2.7 × 10⁴ s ≈ 7.5 hours!
(This illustrates how slow drift velocity actually is.)
Learning Outcomes
Learning Outcome-Based Questions
LO 1 Explain why drift velocity of electrons is very small, yet the current flows almost instantaneously when a switch is turned on.
ANS Although individual electrons drift slowly (~10⁻⁴ m/s), the electric field that drives them propagates through the circuit at nearly the speed of light (~3 × 10⁸ m/s). As soon as the switch is closed, this field sets all free electrons in the circuit in motion simultaneously. So current flows everywhere in the circuit at essentially the same instant, even though no individual electron travels from switch to bulb quickly.
LO 2 A student connects two identical cells in series and then in parallel. In which case will the current through the external resistor be higher, assuming R >> r?
ANS In series: εnet = 2ε, rnet = 2r, Iseries = 2ε/(R+2r) ≈ 2ε/R (when R>>r).
In parallel: εnet = ε, rnet = r/2, Iparallel = ε/(R+r/2) ≈ ε/R.
Since 2ε/R > ε/R, series combination gives more current when R >> r.
In parallel: εnet = ε, rnet = r/2, Iparallel = ε/(R+r/2) ≈ ε/R.
Since 2ε/R > ε/R, series combination gives more current when R >> r.
LO 3 Why is a potentiometer preferred over a voltmeter for measuring EMF of a cell?
ANS A voltmeter, being connected across the cell, draws a small current from it, causing a voltage drop across the internal resistance. So it measures terminal voltage (ε − Ir), not true EMF. The potentiometer works on the null deflection principle — at balance, no current is drawn from the experimental cell. So it measures the true EMF without any disturbance to the circuit.
LO 4 How does the resistance of a conductor change when its length is doubled and its cross-sectional area is halved?
ANS R = ρL/A. New R' = ρ(2L)/(A/2) = 4ρL/A = 4R. The resistance becomes four times the original value.
Advanced Question Types
Assertion–Reason Questions (CBSE Board Pattern)
Instructions: Choose the correct option:
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
AR 1 Assertion (A): The resistivity of a semiconductor decreases with increase in temperature.
Reason (R): With rise in temperature, more electron–hole pairs are generated in a semiconductor, increasing the number of charge carriers.
Reason (R): With rise in temperature, more electron–hole pairs are generated in a semiconductor, increasing the number of charge carriers.
ANS Option (A) — Both A and R are true, and R is the correct explanation of A. As temperature rises, covalent bonds break, releasing more carriers. More carriers → lower resistivity.
AR 2 Assertion (A): Kirchhoff's junction rule follows from conservation of energy.
Reason (R): The algebraic sum of all the currents directed towards a point in a circuit is zero.
Reason (R): The algebraic sum of all the currents directed towards a point in a circuit is zero.
ANS Option (D) — A is false but R is true. Junction rule (KCL) actually follows from the conservation of charge, not energy. Conservation of energy is the basis of the Loop Rule (KVL). The statement of R is correct.
AR 3 Assertion (A): In a Wheatstone bridge, if the battery and galvanometer are interchanged, the balance condition remains the same.
Reason (R): The bridge balance condition P/Q = R/S is independent of battery and galvanometer positions.
Reason (R): The bridge balance condition P/Q = R/S is independent of battery and galvanometer positions.
ANS Option (A) — Both are true and R correctly explains A. The balance condition depends only on the resistances in the four arms, not on the positions of battery and galvanometer. This is a consequence of the reciprocity theorem in network analysis.
AR 4 Assertion (A): Terminal voltage of a cell is always less than its EMF.
Reason (R): Some energy is lost due to internal resistance of the cell.
Reason (R): Some energy is lost due to internal resistance of the cell.
ANS Option (C) — A is false, R is true. Terminal voltage is less than EMF only during discharge. During charging, terminal voltage > EMF (V = ε + Ir). R is correctly stated but it does not support the incorrect assertion A.
CBSE 2023–24 Pattern
Case-Based & HOTS Questions
Case Study: The House Wiring Problem
Riya is checking the wiring in her house. She notices that all electrical appliances — the refrigerator (200 W), TV (150 W), fan (75 W), and lights (100 W total) — are connected in parallel to a 220 V supply. The main fuse is rated 5 A. She wonders if turning on all appliances simultaneously will blow the fuse.
Case Q1 Why are household appliances connected in parallel, not in series?
ANS In parallel: (1) Each appliance receives the full supply voltage (220 V). (2) Each appliance can be switched on/off independently. (3) If one device fails, others continue working. In series, the voltage would be shared and all devices would fail if one broke.
Case Q2 Calculate the total current drawn and determine if the fuse will blow.
ANS
Total power = 200 + 150 + 75 + 100 = 525 W
Total current I = P/V = 525/220 ≈ 2.39 A
Since 2.39 A < 5 A (fuse rating), the fuse will NOT blow if all appliances run simultaneously.
Total current I = P/V = 525/220 ≈ 2.39 A
Since 2.39 A < 5 A (fuse rating), the fuse will NOT blow if all appliances run simultaneously.
HOTS 1 A wire of resistance R is stretched uniformly until its length becomes double. What will be the new resistance?
ANS
Volume of wire remains constant: A₁L₁ = A₂L₂
New length L₂ = 2L₁ → New area A₂ = A₁/2
New R' = ρL₂/A₂ = ρ(2L₁)/(A₁/2) = 4ρL₁/A₁ = 4R
Resistance becomes 4 times the original.
New length L₂ = 2L₁ → New area A₂ = A₁/2
New R' = ρL₂/A₂ = ρ(2L₁)/(A₁/2) = 4ρL₁/A₁ = 4R
Resistance becomes 4 times the original.
HOTS 2 In the potentiometer experiment, if the driver cell's EMF decreases (its charge is running low), what happens to the null point?
ANS
Potential gradient k = εdriver / L (for the whole wire). If εdriver decreases, k decreases. Since εcell = kℓ (at null point), a smaller k requires a larger ℓ to balance the same EMF. So the null point shifts further from the starting end. If k becomes so small that k×L < εcell, no null point is found at all.
Competency Q A student measures the resistance of an unknown resistor using a meter bridge and finds the balance point at 60 cm from the left with R = 10 Ω. When the positions of R and S are interchanged, where will the new balance point be?
ANS
S = R(100−ℓ)/ℓ = 10 × (40/60) = 6.67 Ω
After interchange: new ℓ' satisfies 6.67/10 = ℓ'/(100−ℓ')
6.67(100−ℓ') = 10ℓ' → 667 = 16.67ℓ' → ℓ' = 40 cm
(Balance point shifts to 40 cm — as expected, since 40 + 60 = 100.)
After interchange: new ℓ' satisfies 6.67/10 = ℓ'/(100−ℓ')
6.67(100−ℓ') = 10ℓ' → 667 = 16.67ℓ' → ℓ' = 40 cm
(Balance point shifts to 40 cm — as expected, since 40 + 60 = 100.)
Multiple Choice
MCQs — With Answers and Explanations
1. The drift velocity of free electrons in a conductor is of the order of:
✓ Answer: C | Drift velocity is very small (~10⁻⁴ m/s), unlike thermal velocities (~10⁶ m/s).
2. Which of the following does NOT obey Ohm's Law?
✓ Answer: C | A p-n junction diode has a non-linear I–V curve and is non-ohmic.
3. Three resistors 2 Ω, 3 Ω and 6 Ω are connected in parallel. Their equivalent resistance is:
✓ Answer: B | 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 → R = 1 Ω
4. Kirchhoff's Voltage Law is based on conservation of:
✓ Answer: B | KVL (loop rule) is based on conservation of energy; KCL (junction rule) is based on conservation of charge.
5. The internal resistance of a cell is measured using a potentiometer. If null lengths are ℓ₁ = 75 cm (open circuit) and ℓ₂ = 60 cm (with R = 10 Ω), what is the internal resistance?
✓ Answer: B | r = R(ℓ₁−ℓ₂)/ℓ₂ = 10×(75−60)/60 = 10×15/60 = 2.5 Ω
6. When a metal wire is heated, its resistance:
✓ Answer: A | Metals have positive temperature coefficient (α > 0). Heating increases lattice vibrations, reducing relaxation time τ, thus increasing resistivity and resistance.
7. In a balanced Wheatstone bridge, the galvanometer shows zero deflection because:
✓ Answer: C | At balance, points B and D are at equal potential, so no current flows through the galvanometer (BD arm).
8. The unit of resistivity is:
✓ Answer: B | From R = ρL/A → ρ = RA/L = Ω × m² / m = Ω·m
Board Exam Prep
Previous-Year CBSE & State Board Questions
CBSE 2023 3 Marks State Kirchhoff's laws of electrical networks. Apply the junction rule to the junction A in the given circuit where I₁ = 2A, I₂ = 3A are flowing into junction A, and I₃ flows out.
ANS
KCL: The algebraic sum of all currents at any junction in a network is zero: ΣI = 0.
KVL: The algebraic sum of all potential differences around any closed loop is zero: ΣV = 0.
Applying KCL at A: I₁ + I₂ = I₃ → 2 + 3 = I₃ = 5 A
KVL: The algebraic sum of all potential differences around any closed loop is zero: ΣV = 0.
Applying KCL at A: I₁ + I₂ = I₃ → 2 + 3 = I₃ = 5 A
CBSE 2022 5 Marks With the help of a circuit diagram, explain the working of a potentiometer to compare the EMFs of two cells. Derive the formula used.
ANS
Circuit: A driver cell E is connected across a uniform wire AB of length L. The two cells (ε₁ and ε₂) are connected through a two-way key between A and the jockey J, with a galvanometer in series.
Working: For cell ε₁, jockey is moved till galvanometer shows zero deflection at length ℓ₁.
ε₁ = k·ℓ₁ where k = potential gradient = V/L
Similarly, for cell ε₂: ε₂ = k·ℓ₂
Dividing: ε₁/ε₂ = ℓ₁/ℓ₂
Working: For cell ε₁, jockey is moved till galvanometer shows zero deflection at length ℓ₁.
ε₁ = k·ℓ₁ where k = potential gradient = V/L
Similarly, for cell ε₂: ε₂ = k·ℓ₂
Dividing: ε₁/ε₂ = ℓ₁/ℓ₂
CBSE 2020 2 Marks Write two differences between EMF and terminal voltage of a cell.
ANS
| EMF (ε) | Terminal Voltage (V) |
|---|---|
| Energy provided per unit charge by cell's chemical energy | Actual potential difference across terminals during current flow |
| Constant for a given cell (independent of current) | Depends on current drawn (V = ε − Ir) |
| Measured in open circuit (I = 0) | Always less than EMF during discharge |
State Board 3 Marks Explain the principle of a Wheatstone bridge. Draw the circuit and derive the balance condition.
ANS
Principle (Null Method): Four resistors P, Q, R, S are arranged in a diamond circuit. A battery is connected across one diagonal (AC) and a galvanometer across the other (BD). When the bridge is balanced, no current flows through the galvanometer.
Derivation using KVL:
At balance, IG = 0, so current I₁ flows through P and Q; current I₂ flows through R and S.
Applying KVL: VAB = VAD (since B and D are at equal potential)
I₁P = I₂R ... (i)
I₁Q = I₂S ... (ii)
Dividing (i) by (ii): P/Q = R/S
Derivation using KVL:
At balance, IG = 0, so current I₁ flows through P and Q; current I₂ flows through R and S.
Applying KVL: VAB = VAD (since B and D are at equal potential)
I₁P = I₂R ... (i)
I₁Q = I₂S ... (ii)
Dividing (i) by (ii): P/Q = R/S
CBSE 2019 3 Marks A carbon resistor has colour bands: Yellow, Violet, Brown, Gold. Find its resistance and tolerance.
ANS
Yellow = 4, Violet = 7, Brown = multiplier 10¹, Gold = ±5%
Resistance = 47 × 10 = 470 Ω ± 5%
Range: 446.5 Ω to 493.5 Ω
Resistance = 47 × 10 = 470 Ω ± 5%
Range: 446.5 Ω to 493.5 Ω
Competitive Exam Prep
JEE / NEET / CUET / NDA Exam Questions
JEE Main 2023 In a circuit, a cell of EMF 2V and internal resistance 1 Ω is connected to two resistors of 4 Ω and 12 Ω in parallel. The current through the 4 Ω resistor is:
ANS
Parallel equivalent of 4 Ω and 12 Ω: Rp = (4×12)/(4+12) = 48/16 = 3 Ω
Total resistance = 3 + 1 = 4 Ω
Total current I = 2/4 = 0.5 A
Voltage across parallel combination = 0.5 × 3 = 1.5 V
Current through 4 Ω = 1.5/4 = 0.375 A
Total resistance = 3 + 1 = 4 Ω
Total current I = 2/4 = 0.5 A
Voltage across parallel combination = 0.5 × 3 = 1.5 V
Current through 4 Ω = 1.5/4 = 0.375 A
NEET 2022 A potentiometer wire has a resistance of 4 Ω/m and a length of 5 m. It is connected to a battery of EMF 2 V and internal resistance 2 Ω. What is the potential gradient?
ANS
Total resistance of wire = 4 × 5 = 20 Ω
Circuit: I = 2 / (20 + 2) = 2/22 = 1/11 A
Voltage across wire = I × Rwire = (1/11) × 20 = 20/11 V
Potential gradient k = (20/11) / 5 = 4/11 ≈ 0.364 V/m
Circuit: I = 2 / (20 + 2) = 2/22 = 1/11 A
Voltage across wire = I × Rwire = (1/11) × 20 = 20/11 V
Potential gradient k = (20/11) / 5 = 4/11 ≈ 0.364 V/m
JEE Main 2021 The equivalent resistance between terminals A and B of the following network: 3 resistors of R each forming a triangle (∆) with A and B at two vertices, and C at the third. One resistor connects A to B directly, two connect A-C and C-B.
ANS
Path 1 (direct A-B): R
Path 2 (A-C-B): R + R = 2R (two resistors in series)
Req = R || 2R = (R × 2R)/(R + 2R) = 2R²/3R = 2R/3
Path 2 (A-C-B): R + R = 2R (two resistors in series)
Req = R || 2R = (R × 2R)/(R + 2R) = 2R²/3R = 2R/3
CUET 2023 Which of the following materials has the lowest temperature coefficient of resistance, making it ideal for standard resistors?
ANS Manganin (an alloy of copper, manganese, and nickel) has an almost zero temperature coefficient of resistance (~±2 × 10⁻⁵ °C⁻¹). This makes its resistance nearly independent of temperature, ideal for precision standard resistors used in Wheatstone bridges and laboratory measurements. Constantan also has similar properties.
NDA / SSC A bulb rated "100 W, 220 V" and another rated "60 W, 220 V" are connected in series to a 220 V source. Which bulb glows brighter?
ANS
R₁₀₀ = V²/P = 220²/100 = 484 Ω
R₆₀ = 220²/60 = 806.7 Ω
In series, same current flows. P ∝ R. Since R₆₀ > R₁₀₀, the 60 W bulb dissipates more power and glows brighter in series!
Key insight: In series, higher-rated-resistance (lower-wattage) bulb glows brighter. In parallel, the opposite is true.
R₆₀ = 220²/60 = 806.7 Ω
In series, same current flows. P ∝ R. Since R₆₀ > R₁₀₀, the 60 W bulb dissipates more power and glows brighter in series!
Key insight: In series, higher-rated-resistance (lower-wattage) bulb glows brighter. In parallel, the opposite is true.
Exam Strategy
Common Mistakes Students Make — And How to Fix Them
❌ Mistake 1: Confusing EMF with Terminal Voltage
Students write V = ε even when current is flowing, ignoring the internal resistance drop.
✅ Fix: Always check if the circuit is open (I = 0 → V = ε) or closed (I ≠ 0 → V = ε − Ir).
❌ Mistake 2: Wrong sign convention in KVL
Students randomly add or subtract IR and ε terms without a consistent sign rule.
✅ Fix: Always decide a loop direction first. Mark all IR drops and EMF signs before writing the equation. Stick to one convention throughout the problem.
❌ Mistake 3: Assuming current is same in parallel branches
Students incorrectly apply series-circuit thinking to parallel circuits.
✅ Fix: In parallel, voltage is equal across all branches. Current divides inversely as resistance: I₁/I₂ = R₂/R₁.
❌ Mistake 4: Confusing resistivity with resistance
Students say "resistivity of a wire doubles when its length doubles" — WRONG!
✅ Fix: Resistivity ρ is a material property — it does NOT depend on the shape or size of the conductor. Only resistance R = ρL/A changes with dimensions.
❌ Mistake 5: Forgetting that potentiometer needs balance condition
Students think the potentiometer always works regardless of driver cell voltage.
✅ Fix: The driver cell's EMF must be greater than the EMF being measured. If εdriver < εcell, no balance point exists anywhere on the wire.
❌ Mistake 6: Using incorrect Meter Bridge formula
Students write S = Rℓ/(100−ℓ) instead of S = R(100−ℓ)/ℓ.
✅ Fix: Remember P/Q = R/S where P = ℓ (left arm), Q = (100−ℓ) (right arm). So S = R × Q/P = R × (100−ℓ)/ℓ. R is on the left arm side, S on the right side.
Last-Minute Revision
Quick Revision Notes — Formula Sheet
| Quantity | Symbol | Formula | SI Unit |
|---|---|---|---|
| Current | I | Q/t = nevdA | Ampere (A) |
| Resistance | R | V/I = ρL/A | Ohm (Ω) |
| Resistivity | ρ | m/(ne²τ) | Ω·m |
| Conductivity | σ | 1/ρ = ne²τ/m | S/m or Ω⁻¹m⁻¹ |
| Current density | J | I/A = σE | A/m² |
| Drift velocity | vd | eEτ/m | m/s |
| Power | P | VI = I²R = V²/R | Watt (W) |
| Energy | W | VIt = I²Rt | Joule (J) |
| Terminal voltage | V | ε − Ir (discharge) | Volt (V) |
| Potential gradient | k | V/L | V/m |
Memory Tricks
- "VIR" triangle: Cover what you want to find → V = IR, I = V/R, R = V/I
- Series = Same Current (S.S.C.) | Parallel = Same Voltage (P.S.V.)
- Resistors in parallel always give LESS resistance than the smallest individual resistor
- EMF → chemical energy; Terminal V → electrical energy available externally
- Wheatstone bridge: "Products of opposite arms are equal" → PS = QR at balance
Visual Learning
Chapter Mind Map (Text Format)
⚡ CURRENT ELECTRICITY
├── 🔵 CURRENT & CHARGE
│ ├── I = Q/t (definition)
│ ├── Conventional current: + to −
│ └── Electrons flow: − to +
├── 🔵 DRIFT VELOCITY
│ ├── vd = eEτ/m (very small ~ 10⁻⁴ m/s)
│ ├── I = nevdA
│ └── τ = relaxation time
├── 🔵 OHM'S LAW
│ ├── V = IR (ohmic conductors)
│ ├── J = σE (microscopic form)
│ └── Non-ohmic: diodes, thermistors
├── 🔵 RESISTANCE & RESISTIVITY
│ ├── R = ρL/A
│ ├── ρ = material property (not shape)
│ ├── Temperature: R = R₀[1+α(T−T₀)]
│ └── Metals: +α; Semiconductors: −α
├── 🔵 COMBINATIONS
│ ├── Series: Req = R₁+R₂+... (same I)
│ └── Parallel: 1/Req = 1/R₁+1/R₂+... (same V)
├── 🔵 CELLS & EMF
│ ├── ε = EMF; r = internal resistance
│ ├── V = ε−Ir (discharge)
│ ├── n cells series: nε, nr
│ └── n cells parallel: ε, r/n
├── 🔵 KIRCHHOFF'S LAWS
│ ├── KCL: ΣI = 0 (conservation of charge)
│ └── KVL: ΣV = 0 (conservation of energy)
├── 🔵 WHEATSTONE & METER BRIDGE
│ ├── Balance: P/Q = R/S (IG = 0)
│ └── Meter Bridge: S = R(100−ℓ)/ℓ
└── 🔵 POTENTIOMETER
├── Null method → draws no current
├── k = V/L (potential gradient)
├── EMF comparison: ε₁/ε₂ = ℓ₁/ℓ₂
└── Internal resistance: r = R(ℓ₁−ℓ₂)/ℓ₂
Frequently Asked
FAQs — Current Electricity
What is the difference between electric current and electron flow?
Conventional electric current flows from positive to negative terminal (high to low potential). Electrons, being negatively charged, flow in the opposite direction — from negative to positive. The concept of conventional current was established before electrons were discovered, and we still use it today in circuit analysis.
Why does resistance increase with temperature for metals?
As temperature rises, the atoms in the metallic lattice vibrate with greater amplitude. This increases the frequency of collisions between drifting electrons and the lattice atoms, reducing the relaxation time (τ). Since ρ = m/(ne²τ), a smaller τ means higher resistivity and hence higher resistance.
Can Kirchhoff's laws be applied to AC circuits?
Yes! Kirchhoff's laws are universally applicable to both DC and AC circuits at any instant of time. In AC circuits, we apply the instantaneous values. For practical calculations, we use phasors and impedances, but the fundamental KCL and KVL still hold.
Why is the Wheatstone bridge said to be a "null method"?
Because the measurement is made under the condition of zero (null) reading on the galvanometer. Null methods are very accurate because you don't need to calibrate the galvanometer — you only need to know when its reading is zero. The measurement depends only on the resistance values, not on the galvanometer's sensitivity or calibration.
What happens if we use a very sensitive galvanometer in a Wheatstone bridge?
A more sensitive galvanometer helps detect even tiny imbalances in the bridge, making it possible to find the balance point more precisely. However, if the bridge is very unbalanced, a highly sensitive galvanometer can be damaged by large currents. So a protective resistor (or we start with a less sensitive instrument) is used initially, then switch to the sensitive one near balance.
What is the significance of relaxation time (τ) in current electricity?
Relaxation time τ is the average time interval between two successive collisions of an electron with the lattice atoms. It determines the drift velocity: vd = eEτ/m. A larger τ means fewer collisions, higher drift velocity, and lower resistivity. At higher temperatures, τ decreases (more frequent collisions), hence resistance increases for metals.
Which carries more current — a thick wire or a thin wire of the same material and length, at the same voltage?
A thick wire (larger cross-sectional area A). Since R = ρL/A, a larger A gives smaller R, and by Ohm's Law (I = V/R), smaller R means larger current. Physically, a thicker wire has more free electrons per unit length and offers less opposition to flow.
🎯 Conclusion & Exam Preparation Tips
Current Electricity is one of those chapters where understanding the concept makes all the numericals easy. Once you internalize what current, drift velocity, and Kirchhoff's laws really mean — the formulas follow naturally.
- Master the three forms of Power: P = VI, I²R, and V²/R — choose the right one based on what's given.
- For KVL problems: draw the circuit neatly, assign current directions first, then apply equations.
- In Wheatstone/Meter Bridge problems, always identify which arm is P (left), Q (right), R, and S.
- In potentiometer problems, check if εdriver > εcell before proceeding.
- Remember: resistivity is a material property; resistance depends on shape, size, AND material.
- For board exams: learn the circuit diagrams of potentiometer carefully — they carry 2–3 marks on their own.
- For JEE/NEET: practice loop analysis, multi-cell circuits, and star-delta transformations for complex networks.
- Revise this chapter 2 days before the exam — the formulas are short and high-yield.
📚 Next Chapter → Moving Charges and Magnetism (Chapter 4)
The electric currents you studied here create magnetic fields — that's the bridge to the next chapter!
⚡ Current Electricity · Class 12 Physics · NCERT Chapter 3
All formulas, facts, and explanations are aligned with the latest NCERT syllabus (2024–25).
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