Important Questions Class 12 Physics Chapter 3: Current Electricity

Electric current is the rate at which net charge crosses a conductor’s cross-section.
A steady current needs a closed path and a source that maintains potential difference.

Steady current links microscopic electron drift with practical circuit quantities like resistance, voltage, power, and internal resistance. Important Questions Class 12 Physics Chapter 3 help students practise electric current, Ohm’s law, drift velocity, current density, resistivity, temperature effects, cells, Kirchhoff’s rules, and Wheatstone bridge. The CBSE 2026 chapter contains definitions, derivations, graph-based concepts, numerical questions, and circuit analysis from NCERT Current Electricity.

Key Takeaways

• Electric Current: Current is defined as I = ΔQ/Δt for steady charge flow.
• Ohm’s Law: A conductor obeys Ohm’s law when V remains directly proportional to I.
• Drift Velocity: Electron drift speed in metals is small because collisions randomise motion.
• Wheatstone Bridge: A balanced bridge has zero galvanometer current.

Important Questions Class 12 Physics Chapter 3 Structure 2026

Important Questions Class 12 Physics Chapter 3 with Answers

Current Electricity begins with charge flow and then moves into conductor behaviour.
Students should understand both microscopic motion and circuit-level formulas.
These current electricity class 12 important questions follow the NCERT 2026 order.

1. What does Important Questions Class 12 Physics Chapter 3 mainly test?

Important Questions Class 12 Physics Chapter 3 mainly test current, resistance, drift velocity, cells, power, Kirchhoff’s rules, and Wheatstone bridge. The chapter combines conductor physics with circuit analysis.

1. Current Skill: Use I = ΔQ/Δt.
2. Conductor Skill: Use R = ρl/A and j = σE.
3. Cell Skill: Use V = ε − Ir.
4. Circuit Skill: Apply Kirchhoff’s junction and loop rules.
5. Final Result: The chapter tests steady currents and electrical circuits.

2. Why does Current Electricity focus on steady current?

Current Electricity focuses on steady current because many devices use charges moving at a constant rate. Torches and cell-driven clocks are simple examples.

1. Steady Current: Current value remains constant with time.
2. Source Needed: A cell maintains potential difference.
3. Circuit Condition: A closed conducting path is required.
4. Final Result: Steady current needs a closed circuit and source.

3. Why does current stop in a conductor without a source?

Current stops because end charges get neutralised after a short time. A source must keep separating charges.

1. Initial Charges: Positive and negative end charges create electric field.
2. Electron Motion: Electrons move and neutralise end charges.
3. Field Result: Electric field becomes zero.
4. Final Result: Continuous current needs continuous charge separation.

Electric Current Class 12 Physics Questions

Electric current measures net charge crossing a surface per unit time.
Metals carry current through electrons, while electrolytes can use positive and negative ions.
These electric current class 12 physics questions cover definition, direction, and units.

4. What is electric current in Class 12 Physics?

Electric current is the rate of net charge flow across a cross-section. Its SI unit is ampere.

1. Steady Current Formula:
   I = q/t
2. Variable Current Formula:
   I(t) = lim(Δt → 0) ΔQ/Δt
3. SI Unit: Ampere.
4. Final Result: Electric current equals charge flow per unit time.

5. Is electric current a scalar or vector?

Electric current is a scalar quantity. It does not obey vector addition.

1. Definition: Current depends on scalar charge flow rate.
2. Circuit Arrow: The arrow shows chosen direction.
3. Related Vector: Current density is a vector.
4. Final Result: Current is scalar, but current density is vector.

6. What is the conventional direction of current?

Conventional current flows in the direction of positive charge movement. In metals, electrons drift opposite to conventional current.

1. Conventional Direction: Direction of positive charge flow.
2. Electron Direction: Opposite to electric field.
3. Metal Conductors: Electrons carry current.
4. Final Result: Conventional current is opposite to electron drift.

7. Find current when 12 C charge flows in 4 s.

The current is 3 A. Use I = q/t.

1. Given Data:
   q = 12 C
   t = 4 s
2. Formula Used:
   I = q/t
3. Calculation:
   I = 12/4
   I = 3 A
4. Final Result: Current = 3 A.

Ohm’s Law Class 12 Physics Important Questions

Ohm’s law connects voltage and current for conductors that show linear V-I behaviour.
The law holds only when physical conditions, especially temperature, remain fixed.
These Ohm’s Law Class 12 Physics Important Questions cover resistance and V-I graphs.

8. State Ohm’s law.

Ohm’s law states that current through a conductor is directly proportional to potential difference. Temperature must remain constant.

1. Statement: V ∝ I.
2. Formula Used:
   V = RI
3. Here: R is resistance.
4. Final Result: Ohm’s law gives V = RI.

9. What is resistance in Current Electricity?

Resistance is the opposition offered by a conductor to electric current. It equals voltage divided by current.

1. Formula Used:
   R = V/I
2. SI Unit: Ohm.
3. Unit Relation:
   1 Ω = 1 V/A
4. Final Result: Resistance = V/I.

10. What is the relation between resistance, length, and area?

Resistance is directly proportional to length and inversely proportional to area. The formula is R = ρl/A.

1. Length Relation: R ∝ l.
2. Area Relation: R ∝ 1/A.
3. Formula Used:
   R = ρl/A
4. Final Result: Longer wires have larger resistance.

11. What is resistivity in Class 12 Physics?

Resistivity is a material property that measures opposition to current. It does not depend on conductor dimensions.

1. Formula Used:
   ρ = RA/l
2. SI Unit: Ω m.
3. Depends On: Material and temperature.
4. Final Result: Resistivity is a property of the material.

12. Find resistance of a wire with ρ = 2 × 10^-8 Ω m, l = 4 m, and A = 1 × 10^-6 m².

The resistance is 0.08 Ω. Use R = ρl/A.

1. Given Data:
   ρ = 2 × 10^-8 Ω m
   l = 4 m
   A = 1 × 10^-6 m²
2. Formula Used:
   R = ρl/A
3. Calculation:
   R = (2 × 10^-8 × 4)/(1 × 10^-6)
   R = 8 × 10^-2 Ω
4. Final Result: R = 0.08 Ω.

13. Why does Ohm’s law fail for some devices?

Ohm’s law fails when V and I do not remain proportional. Diodes and GaAs show non-ohmic behaviour.

1. Linear Case: V-I graph is a straight line.
2. Non-linear Case: V-I graph bends.
3. Direction Dependence: Current can depend on voltage polarity.
4. Final Result: Non-ohmic devices do not follow V ∝ I.

Drift Velocity Class 12 Questions

Thermal motion alone gives no net current because electron directions stay random.
An electric field adds a small average drift to this random motion.
These drift velocity class 12 questions explain the microscopic origin of current.

14. What is drift velocity in Class 12 Physics?

Drift velocity is the average velocity acquired by free electrons under an electric field. It acts opposite to the field.

1. Charge Carrier: Electron.
2. Electric Field Direction: From higher to lower potential.
3. Drift Direction: Opposite to electric field.
4. Final Result: Electron drift creates current in metals.

15. What is the formula for drift velocity?

The drift velocity formula is vd = eEτ/m in magnitude. Here τ is relaxation time.

1. Electric Force: eE.
2. Acceleration: a = eE/m.
3. Average Collision Time: τ.
4. Formula Used:
   vd = eEτ/m
5. Final Result: Drift velocity depends on E and τ.

16. How is current related to drift velocity?

Current is related to drift velocity by I = neAvd. Here n is free electron number density.

1. Number Density: n.
2. Area: A.
3. Electron Charge: e.
4. Formula Used:
   I = neAvd
5. Final Result: Current increases with drift velocity.

17. Why is drift speed small but current large?

Drift speed is small, but electron number density is enormous. Metals contain nearly 10^29 free electrons per cubic metre.

1. Drift Speed: Very small in ordinary wires.
2. Electron Density: Very large in metals.
3. Formula Link: I = neAvd.
4. Final Result: Large n gives measurable current.

18. Estimate drift speed for I = 1.5 A, n = 8.5 × 10^28 m^-3, and A = 1 × 10^-7 m².

The drift speed is 1.1 × 10^-3 m/s. Use vd = I/(neA).

1. Given Data:
   I = 1.5 A
   n = 8.5 × 10^28 m^-3
   e = 1.6 × 10^-19 C
   A = 1 × 10^-7 m²
2. Formula Used:
   vd = I/(neA)
3. Calculation:
   vd = 1.5/(8.5 × 10^28 × 1.6 × 10^-19 × 1 × 10^-7)
   vd = 1.1 × 10^-3 m/s
4. Final Result: vd = 1.1 mm/s.

19. Why does current start almost instantly if electrons drift slowly?

Current starts almost instantly because the electric field establishes through the circuit quickly. Electrons everywhere start local drift together.

1. Drift Speed: Very small.
2. Field Propagation: Nearly at electromagnetic wave speed.
3. Current Establishment: Local drift begins throughout the wire.
4. Final Result: Current does not wait for one electron to cross the wire.

Current Density Class 12 Physics Questions

Current density gives current per unit area inside a conductor.
It helps express Ohm’s law in local vector form using electric field.
These current density class 12 physics questions connect I, A, E, and conductivity.

20. What is current density in Class 12 Physics?

Current density is current flowing per unit cross-sectional area. It is a vector quantity.

1. Formula Used:
   j = I/A
2. SI Unit: A/m².
3. Direction: Along conventional current.
4. Final Result: Current density = current per unit area.

21. What is vector form of Ohm’s law?

The vector form of Ohm’s law is j = σE. It relates current density to electric field.

1. Current Density: j.
2. Conductivity: σ.
3. Electric Field: E.
4. Formula Used:
   j = σE
5. Final Result: Current density is proportional to electric field.

22. What is the relation between resistivity and conductivity?

Conductivity is the reciprocal of resistivity. The relation is σ = 1/ρ.

1. Resistivity: ρ.
2. Conductivity: σ.
3. Formula Used:
   σ = 1/ρ
4. Final Result: Higher resistivity means lower conductivity.

23. What is mobility in Current Electricity?

Mobility is drift speed per unit electric field. It measures how easily charge carriers move.

1. Formula Used:
   μ = |vd|/E
2. Electron Relation:
   μ = eτ/m
3. SI Unit: m² V^-1 s^-1.
4. Final Result: Mobility depends on relaxation time.

Resistivity Class 12 Questions

Resistivity separates the material’s nature from the wire’s dimensions.
Metals, semiconductors, and insulators differ widely in resistivity values.
These resistivity class 12 questions support material comparison and numerical practice.

24. How are materials classified by resistivity?

Materials are classified as conductors, semiconductors, and insulators by resistivity. Metals have the lowest resistivity.

1. Conductors: Low resistivity.
2. Semiconductors: Intermediate resistivity.
3. Insulators: Very high resistivity.
4. Final Result: Resistivity decides electrical classification.

25. What is the resistivity range of metals?

Metals have resistivity in the range 10^-8 Ω m to 10^-6 Ω m. This low range makes them good conductors.

1. Material Type: Metals.
2. Range: 10^-8 Ω m to 10^-6 Ω m.
3. Examples: Copper and silver.
4. Final Result: Metals have low resistivity.

26. Why do semiconductors differ from metals in resistivity?

Semiconductors differ because their carrier number increases with temperature. Their resistivity decreases when temperature rises.

1. Metals: Carrier number stays nearly constant.
2. Semiconductors: Carrier number increases with temperature.
3. Result: Resistivity decreases in semiconductors.
4. Final Result: Semiconductors show negative temperature behaviour.

27. Find resistivity if l = 15 m, A = 6 × 10^-7 m², and R = 5 Ω.

The resistivity is 2 × 10^-7 Ω m. Use ρ = RA/l.

1. Given Data:
   R = 5 Ω
   A = 6 × 10^-7 m²
   l = 15 m
2. Formula Used:
   ρ = RA/l
3. Calculation:
   ρ = 5 × 6 × 10^-7/15
   ρ = 2 × 10^-7 Ω m
4. Final Result: ρ = 2 × 10^-7 Ω m.

Temperature Dependence of Resistivity Class 12 Questions

Temperature changes resistivity by changing collision frequency and charge carrier number.
Metals and semiconductors show opposite trends because their carrier behaviour differs.
These temperature dependence of resistivity class 12 questions follow NCERT graphs and examples.

28. What is the temperature dependence of resistivity for metals?

Metal resistivity increases with temperature over a limited range. The relation is ρT = ρ0[1 + α(T − T0)].

1. Formula Used:
   ρT = ρ0[1 + α(T − T0)]
2. For Metals: α is positive.
3. Reason: Collision frequency increases with temperature.
4. Final Result: Metal resistivity rises with temperature.

29. Why does nichrome suit heating elements?

Nichrome suits heating elements because its resistance changes weakly with temperature. It also withstands high operating temperatures.

1. Material: Nichrome.
2. Property: Weak temperature dependence.
3. Use: Heating element in toasters.
4. Final Result: Nichrome gives stable heating resistance.

30. A heating element has R1 = 100 Ω at 27°C and R2 = 117 Ω. Find temperature if α = 1.70 × 10^-4 °C^-1.

The temperature is 1027°C. Use R2 = R1[1 + α(T2 − T1)].

1. Given Data:
   R1 = 100 Ω
   R2 = 117 Ω
   T1 = 27°C
   α = 1.70 × 10^-4 °C^-1
2. Formula Used:
   R2 = R1[1 + α(T2 − T1)]
3. Calculation:
   117/100 = 1 + α(T2 − 27)
   0.17 = 1.70 × 10^-4(T2 − 27)
   T2 − 27 = 1000
4. Final Result: T2 = 1027°C.

31. Why does semiconductor resistivity decrease with temperature?

Semiconductor resistivity decreases because carrier number increases with temperature. This increase dominates the decrease in relaxation time.

1. Temperature Rise: More charge carriers appear.
2. Carrier Density: n increases.
3. Formula Link: ρ = m/(ne²τ).
4. Final Result: Semiconductor resistivity decreases with temperature.

Electrical Energy and Power Class 12 Questions

Electric power gives the rate of electrical energy conversion in a circuit.
Inside a resistor, charge carriers share gained energy with ions through collisions.
These electrical energy and power class 12 questions use P = VI, I²R, and V²/R.

32. What is electric power in a resistor?

Electric power in a resistor is the energy dissipated per unit time. Its formula is P = VI.

1. Potential Difference: V.
2. Current: I.
3. Formula Used:
   P = VI
4. Final Result: Electric power = VI.

33. What are the three formulas for power in a resistor?

The formulas are P = VI, P = I²R, and P = V²/R. They follow from Ohm’s law.

1. Basic Formula: P = VI.
2. Using V = IR: P = I²R.
3. Using I = V/R: P = V²/R.
4. Final Result: Power formulas depend on given data.

34. Why is power transmitted at high voltage?

Power is transmitted at high voltage to reduce line loss. Cable loss varies as I²Rc.

1. Delivered Power: P = VI.
2. For Same P: Higher V gives lower I.
3. Line Loss: Pc = I²Rc.
4. Final Result: High voltage reduces transmission loss.

35. Find power in a 20 Ω resistor carrying 3 A current.

The power is 180 W. Use P = I²R.

1. Given Data:
   I = 3 A
   R = 20 Ω
2. Formula Used:
   P = I²R
3. Calculation:
   P = 3² × 20
   P = 180 W
4. Final Result: P = 180 W.

Cells and Internal Resistance Class 12 Questions

A cell maintains potential difference using chemical energy.
Its internal resistance causes a voltage drop when current flows.
These cells and internal resistance class 12 questions cover emf, terminal voltage, and maximum current.

36. What is emf of a cell?

Emf is the open-circuit potential difference between cell terminals. It is not a mechanical force.

1. Symbol: ε.
2. Meaning: Work done per unit charge by the source.
3. Open Circuit: No current flows.
4. Final Result: Emf is measured in volt.

37. What is internal resistance of a cell?

Internal resistance is resistance offered by the electrolyte and internal parts of a cell. It causes voltage drop inside the cell.

1. Symbol: r.
2. Current: I.
3. Internal Drop: Ir.
4. Final Result: Internal resistance reduces terminal voltage.

38. What is terminal voltage during discharge?

The terminal voltage during discharge is V = ε − Ir. It is less than emf.

1. Emf: ε.
2. Internal Drop: Ir.
3. Formula Used:
   V = ε − Ir
4. Final Result: Terminal voltage during discharge is ε − Ir.

39. A car battery has ε = 12 V and r = 0.4 Ω. Find maximum current.

The maximum current is 30 A. It occurs when external resistance is zero.

1. Given Data:
   ε = 12 V
   r = 0.4 Ω
2. Formula Used:
   Imax = ε/r
3. Calculation:
   Imax = 12/0.4
   Imax = 30 A
4. Final Result: Maximum current = 30 A.

40. A 10 V battery with r = 3 Ω gives current 0.5 A. Find external resistance.

The external resistance is 17 Ω. Use I = ε/(R + r).

1. Given Data:
   ε = 10 V
   r = 3 Ω
   I = 0.5 A
2. Formula Used:
   I = ε/(R + r)
3. Calculation:
   R + r = ε/I = 10/0.5 = 20 Ω
   R = 20 − 3
   R = 17 Ω
4. Final Result: External resistance = 17 Ω.

Cells in Series and Parallel Class 12 Questions

Cell combinations change effective emf and internal resistance.
Series combinations raise voltage, while parallel combinations lower effective internal resistance.
These cells in series and parallel class 12 questions use equivalent cell formulas.

41. What is equivalent emf of cells in series?

The equivalent emf of cells in series is the algebraic sum of their emfs. Same-direction cells add.

1. Same Direction: εeq = ε1 + ε2.
2. Opposite Direction: εeq = ε1 − ε2.
3. Internal Resistance: req = r1 + r2.
4. Final Result: Series cells add emfs algebraically.

42. What is equivalent internal resistance of cells in parallel?

Equivalent internal resistance of two cells in parallel is req = r1r2/(r1 + r2). It matches parallel resistance form.

1. Formula Used:
   1/req = 1/r1 + 1/r2
2. Two Cells:
   req = r1r2/(r1 + r2)
3. Final Result: Parallel cells reduce internal resistance.

43. When should cells be connected in series?

Cells should be connected in series when a higher emf is needed. Their internal resistances also add.

1. Need: Higher voltage.
2. Effect: εeq increases.
3. Limitation: req also increases.
4. Final Result: Series combination increases emf.

44. When should cells be connected in parallel?

Cells should be connected in parallel when lower internal resistance is needed. This combination can supply larger current safely.

1. Need: Lower effective internal resistance.
2. Effect: req decreases.
3. Use: Better current supply.
4. Final Result: Parallel combination lowers internal resistance.

Kirchhoff Law Class 12 Physics Questions

Kirchhoff’s rules solve circuits that cannot reduce to simple series or parallel forms.
The junction rule uses charge conservation, and the loop rule uses energy conservation.
These Kirchhoff law class 12 physics questions focus on equations and sign logic.

45. State Kirchhoff’s junction rule.

Kirchhoff’s junction rule states that total current entering a junction equals total current leaving it. It follows conservation of charge.

1. Rule: ΣIin = ΣIout.
2. Reason: Charge does not accumulate at a junction.
3. Use: Multi-branch circuits.
4. Final Result: Incoming current equals outgoing current.

46. State Kirchhoff’s loop rule.

Kirchhoff’s loop rule states that the algebraic sum of potential changes around a closed loop is zero. It follows conservation of energy.

1. Rule: ΣΔV = 0.
2. Resistor Drop: −IR along current direction.
3. Cell Rise: +ε from negative to positive terminal.
4. Final Result: Net potential change in a loop is zero.

47. Why are Kirchhoff’s rules needed?

Kirchhoff’s rules are needed for complex circuits with multiple loops and branches. Simple series-parallel reduction may fail.

1. Series Formula: Works for a single path.
2. Parallel Formula: Works for common endpoints.
3. Complex Network: Needs junction and loop equations.
4. Final Result: Kirchhoff’s rules solve general circuits.

48. What is the physical basis of Kirchhoff’s rules?

The junction rule comes from charge conservation, and the loop rule comes from energy conservation. These principles apply to steady circuits.

1. Junction Rule: Charge conservation.
2. Loop Rule: Energy conservation.
3. Condition: Steady current.
4. Final Result: Kirchhoff’s rules express conservation laws.

Wheatstone Bridge Class 12 Important Questions

A Wheatstone bridge compares four resistances using a null deflection condition.
At balance, no current flows through the galvanometer arm.
These Wheatstone bridge class 12 important questions cover balance condition and unknown resistance.

49. What is a Wheatstone bridge?

A Wheatstone bridge is a circuit of four resistors with a galvanometer across one diagonal. It measures unknown resistance.

1. Four Arms: R1, R2, R3, R4.
2. Battery Arm: One diagonal.
3. Galvanometer Arm: Other diagonal.
4. Final Result: Wheatstone bridge measures resistance by balance.

50. What is the balanced condition of Wheatstone bridge?

The balanced condition is R2/R1 = R4/R3. At balance, galvanometer current is zero.

1. Condition: Ig = 0.
2. Loop Result: I1R1 = I2R2.
3. Second Loop: I1R3 = I2R4.
4. Final Result: R2/R1 = R4/R3.

51. How do you find unknown resistance using Wheatstone bridge?

Unknown resistance is found by adjusting one known resistance until galvanometer deflection becomes zero. Then use the balance condition.

1. Unknown Arm: Let R4 be unknown.
2. Balance Formula:
   R4 = R3(R2/R1)
3. Condition: Ig = 0.
4. Final Result: Unknown resistance follows from bridge balance.

52. Find R4 if R1 = 100 Ω, R2 = 10 Ω, and R3 = 50 Ω at balance.

The unknown resistance is 5 Ω. Use R4 = R3(R2/R1).

1. Given Data:
   R1 = 100 Ω
   R2 = 10 Ω
   R3 = 50 Ω
2. Formula Used:
   R4 = R3(R2/R1)
3. Calculation:
   R4 = 50 × (10/100)
   R4 = 5 Ω
4. Final Result: R4 = 5 Ω.

NCERT Class 12 Physics Chapter 3 Questions

NCERT numericals combine resistivity, temperature coefficient, cells, and drift speed.
Students should convert area, temperature, and current units before substitution.
These NCERT Class 12 Physics Chapter 3 questions follow the 2026 exercise pattern.

53. A battery of emf 8 V and r = 0.5 Ω is charged by 120 V supply through 15.5 Ω. Find terminal voltage.

The terminal voltage during charging is 11.5 V. The charging current is 7 A.

1. Given Data:
   Supply voltage = 120 V
   Battery emf = 8 V
   Total resistance = 15.5 Ω + 0.5 Ω = 16 Ω
2. Current:
   I = (120 − 8)/16
   I = 7 A
3. Terminal Voltage During Charging:
   V = ε + Ir
   V = 8 + 7 × 0.5
   V = 11.5 V
4. Final Result: Terminal voltage = 11.5 V.

54. Why is a series resistor used while charging a battery?

A series resistor limits charging current. It protects the battery from excessive current.

1. Without Resistor: Current can become very high.
2. With Resistor: Total circuit resistance increases.
3. Effect: Charging current stays controlled.
4. Final Result: Series resistor prevents battery damage.

55. How long does an electron take to drift through 3 m copper wire carrying 3 A?

The drift time is about 2.7 × 10^4 s. Use vd = I/(neA) and t = l/vd.

1. Given Data:
   n = 8.5 × 10^28 m^-3
   A = 2.0 × 10^-6 m²
   I = 3.0 A
   l = 3.0 m
2. Drift Speed:
   vd = I/(neA)
   vd = 3/(8.5 × 10^28 × 1.6 × 10^-19 × 2.0 × 10^-6)
   vd = 1.1 × 10^-4 m/s
3. Time:
   t = l/vd
   t = 3/(1.1 × 10^-4)
   t ≈ 2.7 × 10^4 s
4. Final Result: Drift time ≈ 7.5 hours.

56. Find temperature coefficient if silver wire resistance changes from 2.1 Ω at 27.5°C to 2.7 Ω at 100°C.

The temperature coefficient is 3.94 × 10^-3 °C^-1. Use R = R0[1 + αΔT].

1. Given Data:
   R1 = 2.1 Ω
   R2 = 2.7 Ω
   T1 = 27.5°C
   T2 = 100°C
2. Formula Used:
   α = (R2 − R1)/(R1(T2 − T1))
3. Calculation:
   α = 0.6/(2.1 × 72.5)
   α = 3.94 × 10^-3 °C^-1
4. Final Result: α = 3.94 × 10^-3 °C^-1.

Class 12 Physics Chapter-Wise Important Questions