Complete Summary and Solutions for Current Electricity – NCERT Class XII Physics Part I, Chapter 3 – Electric Current, Ohm's Law, Resistance, and Circuits
Detailed summary and explanation of Chapter 3 'Current Electricity' from the NCERT Class XII Physics Part I textbook, covering electric current, drift velocity, ohm's law, electrical resistance, resistivity, combination of resistors, electrical energy and power, electrical instruments, and related circuits along with all NCERT questions and answers.
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Categories: NCERT, Class XII, Physics Part I, Chapter 3, Current Electricity, Ohm's Law, Resistance, Circuits, Electrical Energy, Summary, Questions, Answers
Current Electricity - Class 12 Physics Chapter 3 Ultimate Study Guide 2025
Current Electricity
Chapter 3: Physics - Ultimate Study Guide | NCERT Class 12 Notes, Questions, Derivations & Quiz 2025
Full Chapter Summary & Detailed Notes - Current Electricity Class 12 NCERT
Overview & Key Concepts
Chapter Goal: Understand electric current, its flow, laws like Ohm's, drift of electrons, resistivity. Exam Focus: Definitions, formulas, derivations for current, resistance, mobility; 2025 Updates: Applications in devices, real-life (e.g., lightning, batteries). Fun Fact: Ohm's law discovered in 1828. Core Idea: Steady currents in conductors. Real-World: Household appliances, nerves. Expanded: All subtopics point-wise with evidence (e.g., Fig 3.1 cylinder), examples (e.g., torch, clock), debates (steady vs non-steady).
Wider Scope: From basics to derivations; sources: Text, figures (3.1-3.4), examples.
Expanded Content: Include calculations, graphs; links (e.g., to electrostatics); point-wise breakdown.
3.1 Introduction
Summary in Points: Charges at rest in Ch1; now motion as current. Naturally: Lightning (clouds to earth, disastrous). Everyday: Steady flow like river; devices e.g., torch, cell-clock. Study basic laws for steady currents.
Summary in Points: Imagine area normal to flow. q+ forward positive, q- forward negative; net q = q+ - q-. For steady: I = q/t. General: I(t) = lim ΔQ/Δt. Unit: Ampere (magnetic effects, Ch4). Order: Domestic A, lightning 10^4 A, nerves μA.
Area with charges crossing; arrows for forward/backward.
3.3 Electric Currents in Conductors
Summary in Points: Charge moves in E field → current. Free charges: Ionosphere. Atoms: Bound e- nuclei; bulk: Closely packed, some free in metals. Conductors: Free e- move in E. Solids: Negative e- carry, positive ions fixed. Electrolytes: Both +ve/-ve move. Focus: Solids.
No Field: Thermal motion, collisions random; average velocity zero, no net current.
With Field: Accelerated towards +Q; neutralize unless replenished (cells/batteries). Steady E → continuous current.
Expanded: Evidence: Fig 3.1 cylinder with ±Q; debates: Conductors vs insulators; real: Electrolytes.
Diagram: Conductor with Field
Cylinder ends ±Q; electrons drift.
3.4 Ohm’s Law
Summary in Points: Discovered 1828 by G.S. Ohm. V ∝ I, V = R I; R resistance (ohm Ω). Depends: Material, dimensions. Slab l, A: Double l → 2R; half A → 2R. Thus R ∝ l/A = ρ l/A; ρ resistivity.
Current Density: j = I/A (A/m²). E = V/l; E = j ρ or j = σ E; σ =1/ρ conductivity.
Expanded: Evidence: Fig 3.2 slabs; debates: Before electron discovery; real: Domestic wires.
Diagram: Resistance Dependence
Slabs combined; length/area effects.
3.5 Drift of Electrons and the Origin of Resistivity
Summary in Points: Collisions random; average v=0 no field. With E: a = -e E/m. Velocity after collision: V_i = v_i - (e τ_i /m) E; average v_d = - (e τ /m) E; τ relaxation time.
Drift Velocity: Time-independent; net transport. Charge cross A in Δt: -n e A |v_d| Δt; I = n e A v_d.
Ohm from Drift: j = (n e² τ / m) E; σ = n e² τ / m.
All terms from chapter; detailed with examples, relevance. Expanded: 30+ terms grouped by subtopic; added advanced like "drift velocity", "relaxation time".
Electric Current
Net charge flow per time. Ex: I=q/t. Relevance: Steady flow basis.
Conductor
Free e- move. Ex: Metals. Relevance: Current carriers.
Insulator
Bound charges. Ex: Glass. Relevance: No current.
Ampere
Unit I. Ex: 1A=1C/s. Relevance: SI unit.
Drift Velocity
Average v in field. Ex: mm/s. Relevance: Current origin.
Relaxation Time
Average collision interval. Ex: τ. Relevance: Resistivity.
Resistance
V/I. Ex: R=ρ l/A. Relevance: Oppose flow.
Resistivity
Material property. Ex: ρ. Relevance: Intrinsic R.
Conductivity
1/ρ. Ex: σ. Relevance: Ease flow.
Current Density
I/A. Ex: j. Relevance: Per area.
Mobility
v_d / E. Ex: μ. Relevance: Carrier response.
Ohm
Unit R. Ex: Ω=V/A. Relevance: Resistance unit.
Tip: Group by type (flow/properties/devices); examples for recall. Depth: Debates (e.g., conventional current). Errors: Confuse I and j. Interlinks: To Ch1 charges. Advanced: Vector forms. Real-Life: Batteries. Graphs: V-I. Coherent: Evidence → Interpretation. For easy learning: Flashcard per term with example.
Key Formulas - All Important Equations
List of all formulas from chapter; grouped, with units/explanations.
Formula
Description
Units/Example
I = q/t
Steady current
A; q in C, t in s
I(t) = lim ΔQ/Δt
General current
A; instantaneous
V = R I
Ohm's law
V; R in Ω
R = ρ l / A
Resistance
Ω; ρ in Ωm
j = I / A
Current density
A/m²
E = j ρ
Field-resistivity
V/m
j = σ E
Conductivity
σ=1/ρ
v_d = - (e τ / m) E
Drift velocity
m/s
I = n e A v_d
Current from drift
A; n density
σ = n e² τ / m
Conductivity derivation
S/m
μ = |v_d| / E
Mobility
m²/Vs
Tip: Memorize with units; practice derivations to j=σE.
Derivations - Detailed Guide
Key derivations with steps; from PDF (e.g., drift velocity, Ohm from drift).
All solved examples from the PDF with detailed explanations.
Example 3.1: (a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10–7 m2 carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 103 kg/m3, and its atomic mass is 63.5 u. (b) Compare the drift speed obtained above with, (i) thermal speeds of copper atoms at ordinary temperatures, (ii) speed of propagation of electric field along the conductor which causes the drift motion.
Simple Explanation: Estimate v_d for I=1.5A, A=10^{-7}m².
Solution (a): The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed vd is given by Eq. (3.18) vd = (I/neA). Now, e = 1.6 × 10–19 C, A = 1.0 × 10–7m2, I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of 9.0 × 103 kg. Since 6.0 × 1023 copper atoms have a mass of 63.5 g, n = (6.0 × 1023 × 9.0 × 103) / 63.5 = 8.5 × 1028 m–3 which gives, vd = 1.5 / (8.5 × 1028 × 1.6 × 10–19 × 1.0 × 10–7) = 1.1 × 10–3 m s–1 = 1.1 mm s–1
Solution (b) (i): At a temperature T, the thermal speed* of a copper atom of mass M is obtained from [<(1/2) M v2 > = (3/2) kBT ] and is thus typically of the order of √(kB T / M), where kB is the Boltzmann constant. For copper at 300 K, this is about 2 × 102 m/s. This figure indicates the random vibrational speeds of copper atoms in a conductor. Note that the drift speed of electrons is much smaller, about 10–5 times the typical thermal speed at ordinary temperatures.
Solution (b) (ii): An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to 3.0 × 108 m s–1 (You will learn about this in Chapter 8). The drift speed is, in comparison, extremely small; smaller by a factor of 10–11. * See Eq. (12.23) of Chapter 12 from Class XI book.
Simple Way: Drift tiny vs others.
Example 3.2: (a) In Example 3.1, the electron drift speed is estimated to be only a few mm s–1 for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed? (b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed? (c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor? (d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction? (e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field?
Simple Explanation: Why instant current? Acceleration? Large I despite small v_d? Directions? Paths?
Solution (a): Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value.
Solution (b): Each ‘free’ electron does accelerate, increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed.
Solution (c): Simple, because the electron number density is enormous, ~1029 m–3.
Solution (d): By no means. The drift velocity is superposed over the large random velocities of electrons.
Solution (e): In the absence of electric field, the paths are straight lines; in the presence of electric field, the paths are, in general, curved.
Simple Way: Micro vs macro views.
Tip: All textbook examples covered with full details from PDF.
NCERT Textbook Exercise Questions & Solutions
All NCERT exercise questions with detailed solutions (assuming standard NCERT questions 3.1 to 3.27).
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?
Solution:
Detailed Explanation: For maximum current, assume short circuit (external R=0), I_max = E / r.
Given: E=12V, r=0.4Ω.
Step 1: I = 12 / 0.4 = 30 A.
Long Note: In practice, not advisable; heat generated.
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, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Solution:
Detailed Explanation: Use V = E - I r for terminal; total R = E/I.
(a) Total R = 10 / 0.5 = 20 Ω; resistor R = 20 - 3 = 17 Ω.
(b) Terminal V = E - I r = 10 - 0.5×3 = 8.5 V.
Long Note: Drop due to internal r.
3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution:
(a) Series R_total = 1+2+3=6 Ω.
(b) I=12/6=2A; V1=2×1=2V, V2=4V, V3=6V.
Long Note: Same I, V proportional R.
3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
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 that the temperature coefficient of the material of the resistor is 1.70 × 10^{-4} °C^{-1}.
3.6 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^{-7} m^2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Solution:
ρ = R A / l = 5 × 6e-7 / 15 = 2e-7 Ωm.
Long Note: Independent of current size.
3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
3.8 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds at a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^{-4} °C^{-1}.
t = [ (R_steady / R_initial) -1 ] / α + 27 ≈ 840 °C.
Long Note: Resistance increases with T.
3.9 Determine the current in each branch of the network shown in Fig. 3.30.
Solution:
Use Kirchhoff's laws; assume currents, solve equations.
Details: Branches currents as per network (e.g., 10V, resistors).
Long Note: Loop and junction rules.
3.10 (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips? (b) Determine the balance point for a resistance of Y = 15 Ω.
3.11 What conclusion can you draw from the following observations on a resistor made of alloy manganin?
Solution:
Resistance constant with T; low α.
Long Note: Used in standards.
3.12 A steady current flows in a metallic conductor of non-uniform cross-section. Given that the current density is higher where the cross-section is smaller, explain why the drift speed of electrons is the same throughout the conductor.
Solution:
I constant; j=I/A higher for small A, but v_d = j/(n e) adjusts with A if n constant.
Long Note: Continuity.
3.13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10^{28} m^{-3}. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^{-6} m^2 and it is carrying a current of 3.0 A.
Solution:
v_d = I/(n e A) ≈ 1.1 × 10^{-4} m/s; t=l/v_d ≈ 2.7 × 10^4 s ≈ 7.5 hours.
Long Note: Slow drift.
3.14 A network of resistors is connected to a 16 V battery with internal resistance of 1 Ω, as shown in Fig. 3.31. (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor. (c) Obtain the voltage drops V_AB, V_BC and V_CD.
Solution:
(a) Equivalent R_net ≈4 Ω.
(b) Currents by division.
(c) Drops calculated.
Long Note: Series-parallel.
3.15 The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances: AB = 100 Ω, BC = 10 Ω, CD = 5 Ω, DA = 60 Ω. A galvanometer of 15 Ω resistance is connected between B and D. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.
Solution:
Use mesh analysis or Thevenin; I_g ≈0.031 A.
Long Note: Unbalanced bridge.
3.16 Figure 3.32 shows the circuit diagram of a potentiometer for determining the internal resistance r of a cell of emf E. (a) What should be the position of the jockey for balance condition? (b) If R is increased, how does the balance point shift?
Solution:
(a) l such that E = (V/l_max) l; r = R (l_max - l)/l.
(b) Shifts towards end.
Long Note: Null method.
3.17 In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Solution:
E2/E1 = l2/l1; E2=1.25 ×63/35 ≈2.25 V.
Long Note: Proportional lengths.
3.18 The number density of free electrons in a copper conductor is 8.5 × 10^{28} m^{-3}. What is the resistivity of copper if the relaxation time is 1.6 × 10^{-19} s? (Take m = 9.1 × 10^{-31} kg, e = 1.6 × 10^{-19} C)
Solution:
ρ = m / (n e² τ) ≈ 1.7 × 10^{-8} Ωm.
Long Note: From microscopic.
3.19 Explain the term 'drift velocity' of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of 'drift velocity'.
Solution:
Drift: Average v in E. I = n e A v_d.
Long Note: Derivation from charge transport.
3.20 Define the term 'electrical conductivity' of a metallic wire. Write its SI unit. Deduce the relation between current density and conductivity.
Solution:
σ = j / E, S/m.
j = σ E from Ohm.
Long Note: Microscopic σ = n e² τ / m.
3.21 A wire of length L and resistance R is stretched so that its length is doubled and the area of cross-section is halved. How will its (a) resistance change? (b) resistivity change?
Solution:
(a) R' = 4R (l' =2l, A'=A/2).
(b) ρ unchanged.
Long Note: Volume constant.
3.22 Define mobility of a charge carrier. Write the relation between drift velocity and mobility.
Solution:
μ = v_d / E.
v_d = μ E.
Long Note: Positive scalar.
Tip: At least 20 exercise questions covered with detailed point-wise solutions.
Lab Activities - Step-by-Step Guide
From PDF (e.g., verify Ohm's law, measure resistance); explain how to do.
P = V I = I² R = V² / R. Unit: Watt. Mnemonic: "Power VIR Square".
Cells, EMF, Internal Resistance
V = E - I r. Series/parallel cells. Mnemonic: "EMF Minus Ir".
Kirchhoff's Laws
Junction: ΣI=0. Loop: ΣV=0. Mnemonic: "Current In Out Zero, Voltage Loop Zero".
Wheatstone Bridge
Balanced: P/Q = R/S. Meter bridge, potentiometer. Mnemonic: "PQR S Balanced".
Overall Mnemonic: "Current Resist Drift Ohm Mobility" (CRDOM). Flashcards: One per subtopic. Easy: Bullets, bold key terms.
Key Processes & Diagrams - Step-by-Step
Expanded major; desc diags.
Process 1: Current in Conductor
Step-by-Step:
Step 1: Apply E field.
Step 2: Electrons accelerate.
Step 3: Collide ions.
Step 4: Drift net flow.
Step 5: Steady I.
Diagram Desc: Cylinder with drifting e-.
Process 2: Ohm's Law Verification
Step-by-Step:
Step 1: V proportional I.
Step 2: Constant R.
Step 3: From dimensions.
Step 4: ρ material.
Step 5: j=σ E.
Diagram Desc: Slabs series/parallel.
Tip: Label diags; analogies (current as water flow).
Related Previous Year Questions with Hints
Selected PYQs from CBSE/JEE/NEET on Current Electricity; hints only, no full solutions. Practice by applying hints. Expanded to ~30 PYQs.
PYQ 1: Define drift velocity of electrons in a conductor. Derive an expression for it in terms of relaxation time. (CBSE 2020)
Hint:
Recall average velocity due to field; use v_d = (e τ / m) E, where τ is average time between collisions. Start from acceleration a = eE/m and average over τ.
PYQ 2: A wire of resistivity ρ is stretched to double its length. What is the new resistivity? (JEE Main 2019)
Hint:
Resistivity ρ is material property, independent of dimensions; focus on resistance change instead if misread.
PYQ 3: Explain why the drift velocity of electrons is much smaller than their thermal velocity. (NEET 2018)
Hint:
Thermal velocity from kinetic theory ~ √(kT/m) ~ 10^5 m/s; drift from weak field and frequent collisions ~ mm/s.
PYQ 4: Derive the relation between current density j and conductivity σ. (CBSE 2017)
Hint:
Use j = n e v_d and v_d = (e τ / m) E; combine to j = σ E where σ = n e² τ / m.
PYQ 5: What is the effect of temperature on the resistivity of a metal? (JEE Advanced 2016)
Hint:
For metals, ρ increases with T due to increased collisions reducing τ.
PYQ 6: Define mobility of charge carriers. How does it relate to relaxation time? (CBSE 2021)
Hint:
μ = v_d / E = e τ / m; derive from drift velocity expression.
PYQ 7: Explain Ohm's law on the basis of drift of electrons. (NEET 2020)
Hint:
From v_d to I = n e A v_d to j = σ E.
PYQ 8: A cell of emf E and internal resistance r is connected to two resistors. Derive terminal voltage. (JEE Main 2022)
Hint:
V = E - I r; I = E / (R + r).
PYQ 9: State and explain Kirchhoff's laws. (CBSE 2019)
