Current Electricity Class 12 Case Based Questions 2026-27 | CBSE Physics

Current Electricity Class 12 Physics — Case-Based Questions with Answers

This practice page contains original passage-based questions for CBSE Class 12 Physics 2026–27, Chapter 3: Current Electricity. Each case connects a realistic circuit situation with four short questions testing calculation, interpretation and concept application.

Exam-format note: CBSE's Physics sample-paper framework uses a dedicated case-study section, and the 2025–26 official Physics SQP specified two case-study questions of 4 marks each. The current 2026–27 CBSE Academics page provides the Class XII Physics SQP and marking scheme for the new session. These questions are original CBSE-style practice, not official CBSE questions or PYQs.

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1. Case 1: Current, Drift Velocity and Current Density
2. Case 2: Ohm's Law and Resistance
3. Case 3: Electrical Power and Temperature
4. Case 4: Cell, EMF and Internal Resistance
5. Case 5: Cells in Series and Parallel
6. Case 6: Kirchhoff's Rules
7. Case 7: Wheatstone Bridge
8. Case 8: V–I Characteristics and Conductivity
9. Quick Answer Key

1. Case Study — Current, Drift Velocity and Current Density

Passage: A copper wire of cross-sectional area 2.0 × 10⁻⁶ m² carries a steady current of 3.2 A. The number density of free electrons in the wire is 8.0 × 10²⁸ m⁻³. Take the magnitude of electronic charge as 1.6 × 10⁻¹⁹ C. A student wants to understand why a measurable current can exist even though the average drift speed of electrons is very small.

(i) The current density in the wire is:

A. 1.6 × 10⁶ A m⁻²

B. 1.6 × 10⁷ A m⁻²

C. 6.4 × 10⁶ A m⁻²

D. 6.4 × 10⁷ A m⁻²

Answer: A
J = I/A = 3.2/(2.0 × 10⁻⁶) = 1.6 × 10⁶ A m⁻²

(ii) The drift velocity is approximately:

A. 1.25 × 10⁻⁴ m s⁻¹

B. 1.25 × 10⁻³ m s⁻¹

C. 2.5 × 10⁻⁴ m s⁻¹

D. 2.5 × 10⁻³ m s⁻¹

Answer: A
v_d = I/(neA) = 3.2/[(8×10²⁸)(1.6×10⁻¹⁹)(2×10⁻⁶)] = 1.25 × 10⁻⁴ m s⁻¹

(iii) If the current is doubled while the carrier density and area remain unchanged, the drift velocity will:

A. Become half

B. Remain unchanged

C. Double

D. Become four times

Answer: C
From I = neAv_d, v_d is directly proportional to I when n and A are fixed.

(iv) The conventional current direction in the metal is:

A. Same as electron drift

B. Opposite to electron drift

C. Perpendicular to electron drift

D. Random

Answer: B
Electrons are negatively charged, so their drift direction is opposite to conventional current.

2. Case Study — Ohm's Law and Resistance

Passage: A student connects a resistor to a variable DC supply and records the voltage across it and the current through it. Over the measured range, the V–I graph is a straight line passing through the origin. The student then replaces the resistor with a filament-type device whose resistance changes significantly as its temperature changes.

(i) The first resistor is behaving as:

A. An ohmic conductor over the measured range

B. A capacitor

C. A source of emf

D. A conductor with zero resistance

Answer: A
A straight-line V–I relation through the origin is consistent with Ohm's law under the stated conditions.

(ii) On a V-versus-I graph, the slope represents:

A. Conductance

B. Resistance

C. Current density

D. Mobility

Answer: B
slope = ΔV/ΔI = R

(iii) If the resistor has R = 5 Ω and V = 20 V, its current is:

A. 0.25 A

B. 2 A

C. 4 A

D. 100 A

Answer: C
I = V/R = 20/5 = 4 A

(iv) Which statement about the filament-type device is most appropriate?

A. V = IR must describe its complete V–I curve with one constant R.

B. It may show non-ohmic behaviour because its resistance changes with operating conditions.

C. Its resistance must always be zero.

D. Ohm's law says no electrical device can be non-ohmic.

Answer: B
Ohm's law does not describe the complete behaviour of every device under all conditions.

3. Case Study — Electrical Power and Temperature

Passage: A 100 Ω resistor is connected to a 200 V supply. It is used for a fixed time, and the electrical energy converted into heat is studied. The resistor has a positive temperature coefficient and its resistance increases approximately linearly over the temperature range considered.

(i) The initial current through the resistor is:

A. 0.5 A

B. 2 A

C. 20 A

D. 200 A

Answer: B
I = V/R = 200/100 = 2 A

(ii) The initial power consumed is:

A. 200 W

B. 400 W

C. 20,000 W

D. 100 W

Answer: B
P = V²/R = 200²/100 = 400 W

(iii) If the resistor operates at the same voltage but its resistance increases to 200 Ω, the power becomes:

A. 100 W

B. 200 W

C. 400 W

D. 800 W

Answer: B
P = V²/R = 200²/200 = 200 W

(iv) The commercial unit of electrical energy is:

A. watt

B. joule/second

C. kilowatt-hour

D. volt

Answer: C
Electrical energy used commercially is commonly measured in kWh.

4. Case Study — EMF and Internal Resistance

Passage: A real cell has an emf of 12 V and internal resistance 2 Ω. It is connected to an external resistance of 4 Ω. The cell supplies current to the external circuit. A student measures the terminal voltage across the cell while it is operating.

(i) The current supplied by the cell is:

A. 1 A

B. 2 A

C. 3 A

D. 6 A

Answer: B
I = ε/(R+r) = 12/(4+2) = 2 A

(ii) The terminal voltage is:

A. 4 V

B. 6 V

C. 8 V

D. 12 V

Answer: C
V = ε − Ir = 12 − (2)(2) = 8 V

(iii) The potential drop inside the cell is:

A. 2 V

B. 4 V

C. 8 V

D. 10 V

Answer: B
Ir = 2×2 = 4 V

(iv) If the same cell is being charged with current I, the corresponding terminal-voltage relation is:

A. V = ε − Ir

B. V = ε + Ir

C. V = ε/I

D. V = Ir/ε

Answer: B
For charging, the terminal voltage is greater than the emf under the usual sign convention: V = ε + Ir.

5. Case Study — Cells in Series and Parallel

Passage: Three identical cells are available. Each cell has emf 2 V and internal resistance 0.5 Ω. The cells can be connected either in series aiding or in parallel. An external resistor of 4.5 Ω is used to compare the resulting current.

(i) When the three cells are connected in series aiding, their equivalent emf is:

A. 0.67 V

B. 2 V

C. 4 V

D. 6 V

Answer: D
ε_eq = nε = 3×2 = 6 V

(ii) Their equivalent internal resistance in series is:

A. 0.5 Ω

B. 1.0 Ω

C. 1.5 Ω

D. 2.0 Ω

Answer: C
r_eq = nr = 3×0.5 = 1.5 Ω

(iii) The current with the cells in series is:

A. 0.5 A

B. 1 A

C. 1.5 A

D. 2 A

Answer: B
I = 6/(4.5+1.5) = 1 A

(iv) If the same three identical cells are connected in parallel, their equivalent internal resistance is:

A. 1.5 Ω

B. 1 Ω

C. 0.5 Ω

D. 0.167 Ω

Answer: D
r_eq = r/n = 0.5/3 ≈ 0.167 Ω

6. Case Study — Kirchhoff's Rules

Passage: A multi-loop circuit contains several resistors and cells. Since the circuit cannot be reduced completely into simple series and parallel combinations, a student assigns branch currents and writes equations at junctions and around independent loops.

(i) Kirchhoff's junction rule is based on conservation of:

A. Energy

B. Charge

C. Momentum

D. Mass only

Answer: B
The algebraic sum of currents at a junction is zero, expressing conservation of charge.

(ii) Kirchhoff's loop rule is based on conservation of:

A. Charge

B. Energy

C. Current

D. Resistance

Answer: B
The algebraic sum of potential changes around a closed loop is zero, expressing energy conservation.

(iii) A resistor of 3 Ω carries an assumed current of 2 A. The potential change across it in the direction of the assumed current is:

A. +6 V

B. −6 V

C. +1.5 V

D. −1.5 V

Answer: B
Moving through a resistor in the direction of assumed current gives a drop: −IR = −(2)(3) = −6 V.

(iv) A calculated branch current is −0.75 A. What does this indicate?

A. No current flows.

B. The resistance is negative.

C. The actual current is 0.75 A opposite to the assumed direction.

D. The cell has zero emf.

Answer: C
The negative sign reverses the initially chosen reference direction.

7. Case Study — Wheatstone Bridge

Passage: A Wheatstone bridge contains four resistive arms P, Q, R and S. A galvanometer connects the two middle junctions. The bridge is adjusted until the galvanometer shows no deflection. At that point the two junctions connected to the galvanometer are at the same potential.

(i) At balance, the galvanometer current is:

A. Maximum

B. Zero

C. Equal to the battery current

D. Infinite

Answer: B
Equal potential at the galvanometer terminals gives zero potential difference and hence zero current.

(ii) The standard balance condition is:

A. P + Q = R + S

B. P/Q = R/S

C. P/R = Q/S

D. PQ = RS

Answer: B
The standard balance relation is P/Q = R/S, equivalently PS = QR.

(iii) If P = 2 Ω, Q = 4 Ω and R = 3 Ω, the value of S at balance is:

A. 1.5 Ω

B. 4 Ω

C. 6 Ω

D. 8 Ω

Answer: C
2/4 = 3/S ⇒ S = 6 Ω

(iv) If the bridge is not balanced, which statement is correct?

A. The balance relation may be used automatically.

B. The galvanometer must carry zero current.

C. A potential difference exists across the galvanometer, so current may flow through it.

D. All four resistances must be equal.

Answer: C
When the junction potentials differ, the galvanometer has a potential difference across it and can carry current.

8. Case Study — V–I Behaviour, Resistivity and Conductivity

Passage: Two wires A and B are made of different materials. Their lengths and cross-sectional areas are known. Wire A has resistivity 2 × 10⁻⁸ Ω m and wire B has resistivity 8 × 10⁻⁸ Ω m. Both are considered at the same specified temperature. The student compares their resistance and conductivity.

(i) If both wires have identical dimensions, which has the greater resistance?

A. A

B. B

C. Both are equal

D. Cannot be compared even with identical dimensions

Answer: B
For identical L and A, R is proportional to ρ. Wire B has four times the resistivity, so it has four times the resistance.

(ii) The conductivity of wire A is:

A. 2 × 10⁻⁸ S m⁻¹

B. 5 × 10⁷ S m⁻¹

C. 2 × 10⁸ S m⁻¹

D. 0.5 × 10⁸ S m⁻¹

Answer: B
σ = 1/ρ = 1/(2×10⁻⁸) = 5×10⁷ S m⁻¹

(iii) If the length of wire A is doubled while its area remains unchanged, its resistance becomes:

A. R/2

B. R

C. 2R

D. 4R

Answer: C
R = ρL/A, so doubling L doubles R.

(iv) Which statement correctly distinguishes resistivity from resistance?

A. Both depend only on the length of the wire.

B. Resistance depends on sample geometry; resistivity is a material property at specified physical conditions.

C. Resistivity is always measured in ohms.

D. Resistance is independent of material.

Answer: B
Resistance depends on material and geometry; resistivity characterises the material under specified conditions.

9. Quick Answer Key

Case	Answers
Case 1	i-A, ii-A, iii-C, iv-B
Case 2	i-A, ii-B, iii-C, iv-B
Case 3	i-B, ii-B, iii-B, iv-C
Case 4	i-B, ii-C, iii-B, iv-B
Case 5	i-D, ii-C, iii-B, iv-D
Case 6	i-B, ii-B, iii-B, iv-C
Case 7	i-B, ii-B, iii-C, iv-C
Case 8	i-B, ii-B, iii-C, iv-B

Self-test method: First read only each passage and question. Write your four answers without opening the explanations. Then check both the option and the reasoning. A correct option reached by incorrect reasoning should be treated as a concept that needs revision.

10. Continue Current Electricity Preparation

Current Electricity Complete Notes Current Electricity Important Questions Current Electricity MCQs Current Electricity Numericals Chapter 1: Electric Charges and Fields Chapter 2: Electrostatic Potential and Capacitance

Only published and verified Learn Revise Hub resources are linked here.

11. Official CBSE Resources

• CBSE Physics Curriculum 2026–27
• CBSE Class XII Sample Question Papers & Marking Schemes 2026–27
• CBSE Question Paper Archive

Source note: The Chapter 3 scope and Unit II marks were checked against the official CBSE 2026–27 Physics curriculum. The current CBSE Class XII 2026–27 SQP/MS page is live. Case-based practice patterns were also reviewed across current educational resources; all questions and solutions on this page are original Learn Revise Hub material.