Current Electricity — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Current Electricity" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Current Electricity" — 8 important questions with detailed answers for CBSE board exam…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Physics — Current Electricity. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Current Electricity — All 8 Questions Solved
Q1. Define electric current. State its SI unit and symbol.
Electric Current (I): Rate of flow of electric charge through a conductor.
I = Q/t (charge per unit time)
SI Unit: Ampere (A) = Coulomb per second (C/s)
Symbol: I
Direction: Conventional current flows from positive to negative terminal (opposite to electron flow).
One Ampere: When charge of 1 coulomb flows through conductor in 1 second
1 A = 1 C/s
Microscopic definition:
I = nAve
where n = number density of charge carriers
A = cross-sectional area
ve = drift velocity
e = magnitude of charge per carrier
Q2. State Ohm's law. A wire has resistance 5 Ω. If potential difference across it is 10 V, find the current.
Ohm's Law: At constant temperature, current through conductor is directly proportional to applied potential difference.
I ∝ V (at constant temperature)
V = IR
or I = V/R
where V = potential difference (V)
I = current (A)
R = resistance (Ω)
Given: R = 5 Ω, V = 10 V
Step 1: I = V/R = 10/5
Step 2: I = 2 A
Final Answer: Current = 2 A
Note: Ohm's law is valid only for ohmic conductors (metallic conductors at constant temperature). Semiconductors and gases do not follow Ohm's law.
Q3. Define resistance and resistivity. Write formula relating resistance to resistivity.
Resistance (R): Opposition offered by conductor to flow of electric current.
R = V/I (potential difference per unit current)
SI Unit: Ohm (Ω) = V/A
Resistivity (ρ): Inherent property of material that opposes current flow.
SI Unit: Ohm-meter (Ω⋅m)
Symbol: ρ (rho)
Relation between R and ρ:
R = ρL/A
where ρ = resistivity of material
L = length of conductor
A = cross-sectional area
Typical resistivity values at 20°C:
Copper: ρ ≈ 1.7 × 10⁻⁸ Ω⋅m
Aluminum: ρ ≈ 2.6 × 10⁻⁸ Ω⋅m
Nichrome: ρ ≈ 1 × 10⁻⁶ Ω⋅m
Glass: ρ ≈ 10¹¹ Ω⋅m
Q4. A copper wire has length 100 m and cross-sectional area 1 mm². If resistivity of copper is 1.7 × 10⁻⁸ Ω⋅m, find its resistance.
Given: L = 100 m
A = 1 mm² = 1 × 10⁻⁶ m²
ρ = 1.7 × 10⁻⁸ Ω⋅m
Formula: R = ρL/A
Step 1: R = (1.7 × 10⁻⁸) × 100 / (1 × 10⁻⁶)
Step 2: R = 1.7 × 10⁻⁶ / (1 × 10⁻⁶)
Step 3: R = 1.7 Ω
Final Answer: Resistance = 1.7 Ω
Q5. Define emf and internal resistance of a cell. Write expression relating V, emf, and r.
EMF (ε - epsilon): Total energy per unit charge supplied by the cell.
Emf: E = W/Q (work done per unit charge)
SI Unit: Volt (V)
Internal Resistance (r): Resistance of the cell due to electrolyte and electrodes.
SI Unit: Ohm (Ω)
Relation when current I flows through cell:
V = ε - Ir
where V = terminal voltage across cell
ε = emf of cell
I = current flowing through circuit
r = internal resistance
When cell is in open circuit (I = 0):
V = ε (terminal voltage equals emf)
When current flows (I > 0):
V < ε (terminal voltage drops)
The term Ir represents voltage drop across internal resistance
External circuit sees: V = ε - Ir
Total power supplied by cell: P = εI
Power dissipated in external circuit: P_ext = VI = I²R_ext
Power dissipated in internal resistance: P_int = I²r
Q6. Derive expression for combined resistance when resistances are in series.
Series Connection: Resistances connected end-to-end so same current flows through each.
Consider three resistances R₁, R₂, R₃ connected in series across potential difference V.
Same current I flows through all: I = I₁ = I₂ = I₃
Voltage distribution:
V₁ = IR₁
V₂ = IR₂
V₃ = IR₃
Total voltage:
V = V₁ + V₂ + V₃ = IR₁ + IR₂ + IR₃ = I(R₁ + R₂ + R₃)
Let equivalent resistance be R_s:
V = IR_s
Comparing: R_s = R₁ + R₂ + R₃
For n resistances in series:
R_s = R₁ + R₂ + R₃ + ... + Rₙ
Properties of series connection:
- Equivalent resistance is sum of individual resistances
- R_s > maximum of individual resistances
- Voltage divides among resistances
- Current same through all
- Total power = sum of individual powers
Q7. Derive expression for combined resistance when resistances are in parallel.
Parallel Connection: Resistances connected between same two points so same voltage appears across each.
Consider three resistances R₁, R₂, R₃ connected in parallel across potential difference V.
Same voltage V across all:
V = V₁ = V₂ = V₃
Currents through each:
I₁ = V/R₁
I₂ = V/R₂
I₃ = V/R₃
Total current:
I = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃)
Let equivalent resistance be R_p:
I = V/R_p
Comparing: 1/R_p = 1/R₁ + 1/R₂ + 1/R₃
For n resistances in parallel:
1/R_p = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/Rₙ
Properties of parallel connection:
- Reciprocal of equivalent resistance = sum of reciprocals
- R_p < minimum of individual resistances
- Current divides among resistances
- Voltage same across all
- Reciprocals are additive (conductances are additive)
Q8. State Kirchhoff's laws. Apply them to analyze a circuit with two cells and multiple resistances.
Kirchhoff's First Law (Junction Rule): At any junction, sum of currents entering = sum of currents leaving.
∑I_in = ∑I_out
Or: ∑I = 0 (algebraic sum of currents at junction is zero)
Basis: Conservation of electric charge
Kirchhoff's Second Law (Loop Rule): In any closed loop, algebraic sum of emfs = algebraic sum of potential drops.
∑ε = ∑IR
Or: ∑(ε - IR) = 0 around any closed loop
Basis: Conservation of energy
Sign Convention:
- Current direction: Choose arbitrarily; positive if assumed direction, negative if opposite
- Emf: Positive when going from negative to positive terminal through cell
- Potential drop IR: Positive in direction of assumed current
Application to circuit with two cells:
1. Identify all junctions and loops
2. Assume directions for currents
3. Apply first law at junctions (gives equations for current relationships)
4. Apply second law to independent loops (gives emf-resistance equations)
5. Solve simultaneous equations
6. Positive result confirms assumed direction; negative means opposite direction
Number of independent equations needed = Number of unknown currents
Number of equations from junctions = (n-1) where n = number of junctions
Number of equations from loops = (m-n+1) where m = number of branches
Current Electricity — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A wire of resistance R is stretched to twice its original length. Assuming the volume remains constant, what will be its new resistance?
- A. R
- B. 2R
- C. 4R
- D. R/2
Answer: C. 4R — When a wire is stretched, its volume (V = AL) remains constant. If length becomes 2L, the area must become A/2. Since resistance R = ρL/A, the new resistance R' = ρ(2L)/(A/2) = 4ρL/A = 4R.
MCQ 2. Kirchhoff's first law (current law) is a consequence of the conservation of which quantity?
- A. Energy
- B. Charge
- C. Momentum
- D. Mass
Answer: B. Charge — Kirchhoff's current law states that the algebraic sum of currents entering a junction is zero. This implies that no charge accumulates at the junction, hence it's based on the conservation of charge.
MCQ 3. If the drift velocity of electrons in a copper wire is v_d, what happens to the drift velocity if the current is doubled and the cross-sectional area of the wire is halved?
- A. v_d remains the same
- B. v_d becomes 2v_d
- C. v_d becomes 4v_d
- D. v_d becomes v_d/4
Answer: C. v_d becomes 4v_d — The relation between current (I), drift velocity (v_d), and cross-sectional area (A) is I = neAv_d. Thus, v_d is proportional to I/A. If I' = 2I and A' = A/2, then v_d' = (2I)/(ne(A/2)) = 4I/(neA) = 4v_d.
MCQ 4. Two cells of EMF E₁ and E₂ and internal resistances r₁ and r₂ respectively are connected in parallel. What is the equivalent EMF of the combination?
- A. (E₁r₁ + E₂r₂)/(r₁ + r₂)
- B. (E₁r₂ + E₂r₁)/(r₁ + r₂)
- C. (E₁r₂ + E₂r₁)/(r₁r₂)
- D. (E₁ + E₂)/2
Answer: B. (E₁r₂ + E₂r₁)/(r₁ + r₂) — For two cells connected in parallel, the equivalent EMF (E_eq) is given by E_eq = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂) = (E₁r₂ + E₂r₁)/(r₁ + r₂).
MCQ 5. A potentiometer wire of length 100 cm has a resistance of 10 Ω. It is connected in series with a resistance and an accumulator of EMF 2V and negligible internal resistance. If a source of EMF 40 mV is balanced by 40 cm of the potentiometer wire, what is the resistance connected in series with the potentiometer wire?
- A. 90 Ω
- B. 99 Ω
- C. 100 Ω
- D. 190 Ω
Answer: D. 190 Ω — The potential gradient (k) = E_source / L_balance = 40 mV / 40 cm = 1 mV/cm = 0.1 V/m. The total potential drop across the wire (V_wire) = k * L_wire = 0.1 V/m * 1 m = 0.1 V. Also, V_wire = (E_accumulator * R_wire) / (R_series + R_wire). So, 0.1 = (2 * 10) / (R_series + 10). Solving gives R_series = 190 Ω.
MCQ 6. A cell of EMF E and internal resistance r is connected across an external resistance R. If the terminal voltage across the cell is V, which of the following is true?
- A. V = E
- B. V = E + Ir
- C. V = E - Ir
- D. V = Ir
Answer: C. V = E - Ir — When a current I flows from a cell, there is a voltage drop (Ir) across its internal resistance. The terminal voltage V across the external resistance is the EMF (E) minus this internal drop, so V = E - Ir.
MCQ 7. Which of the following expressions represents the power dissipated in a resistor R when a current I flows through it?
- A. I²R
- B. IR²
- C. V/I
- D. V²I
Answer: A. I²R — The power dissipated in a resistor is given by the product of the square of the current and the resistance, P = I²R. Other equivalent forms are P = VI and P = V²/R.
MCQ 8. In a balanced Wheatstone bridge, if the positions of the galvanometer and the battery are interchanged, what happens to the balance condition?
- A. The bridge becomes unbalanced.
- B. The bridge remains balanced.
- C. The galvanometer shows deflection.
- D. The current through the resistors changes significantly.
Answer: B. The bridge remains balanced. — The balance condition of a Wheatstone bridge (P/Q = R/S) is independent of the positions of the galvanometer and the battery. Interchanging them does not affect the condition for zero current through the galvanometer.
MCQ 9. The resistance of a conductor increases with temperature primarily because:
- A. The number of free electrons increases.
- B. The drift velocity of electrons increases.
- C. The relaxation time of electrons decreases.
- D. The density of the conductor decreases.
Answer: C. The relaxation time of electrons decreases. — With increasing temperature, the thermal vibrations of the lattice atoms increase, leading to more frequent collisions with free electrons. This reduces the average time between collisions (relaxation time), which in turn increases the resistance.
MCQ 10. Kirchhoff's second law (voltage law) is a consequence of the conservation of which quantity?
- A. Charge
- B. Energy
- C. Momentum
- D. Mass
Answer: B. Energy — Kirchhoff's voltage law states that the algebraic sum of potential differences around any closed loop in a circuit is zero. This implies that the total work done by electrostatic forces on a charge completing a closed loop is zero, which is a statement of the conservation of energy.
MCQ 11. In a meter bridge experiment, a null point is obtained at 40 cm from the left end when a resistance of 10 Ω is placed in the left gap. What resistance should be placed in the right gap?
- A. 15 Ω
- B. 20 Ω
- C. 25 Ω
- D. 30 Ω
Answer: A. 15 Ω — For a meter bridge, the balance condition is R_left / L_left = R_right / L_right. Given R_left = 10 Ω, L_left = 40 cm. Then L_right = 100 cm - 40 cm = 60 cm. So, 10 / 40 = R_right / 60. R_right = (10 / 40) * 60 = 15 Ω.
MCQ 12. Which of the following materials is an example of a non-ohmic conductor?
- A. Copper wire
- B. Germanium semiconductor
- C. Nichrome wire
- D. Silver wire
Answer: B. Germanium semiconductor — Ohmic conductors follow Ohm's law (V∝I) under constant physical conditions. Semiconductors like Germanium and diodes are non-ohmic, as their resistance changes with voltage or current.
MCQ 13. The relationship between current density (J), conductivity (σ), and electric field (E) is given by:
- A. J = σE
- B. J = E/σ
- C. J = σ/E
- D. J = σ²E
Answer: A. J = σE — This is the microscopic form of Ohm's Law. Current density J is directly proportional to the electric field E, with conductivity σ as the constant of proportionality.
MCQ 14. Three identical resistors, each of resistance R, are connected to form an equilateral triangle. What is the equivalent resistance between any two vertices of the triangle?
- A. R
- B. 2R
- C. 3R/2
- D. 2R/3
Answer: D. 2R/3 — Between any two vertices (say A and B), one resistor (R) is directly connected. The other two resistors are in series (R+R=2R) forming the alternative path. These two paths (R and 2R) are in parallel. So, R_eq = (R * 2R) / (R + 2R) = 2R²/3R = 2R/3.
MCQ 15. A potentiometer is used to measure the internal resistance (r) of a cell. When the cell is in open circuit, the balancing length is L₁. When a resistance R is connected across the cell, the balancing length is L₂. The internal resistance 'r' is given by:
- A. r = R (L₁ / L₂ - 1)
- B. r = R (L₂ / L₁ - 1)
- C. r = R (L₁ - L₂) / L₁
- D. r = R L₂ / L₁
Answer: A. r = R (L₁ / L₂ - 1) — In open circuit, the EMF E = kL₁. When resistance R is connected, terminal voltage V = kL₂. We know E/V = (R+r)/R. Substituting, L₁/L₂ = (R+r)/R, which simplifies to r = R (L₁/L₂ - 1).
MCQ 16. Superconductors are materials that have:
- A. Infinite resistance at high temperatures.
- B. Zero resistance at very low temperatures.
- C. High resistance at room temperature.
- D. Negative resistance.
Answer: B. Zero resistance at very low temperatures. — Superconductors are materials that exhibit exactly zero electrical resistance when cooled below a critical temperature, a phenomenon known as superconductivity.
MCQ 17. A 100 W, 220 V bulb is connected to a 110 V supply. The power consumed by the bulb will be:
- A. 100 W
- B. 50 W
- C. 25 W
- D. 75 W
Answer: C. 25 W — First, find the resistance of the bulb: R = V_rated² / P_rated = (220)² / 100 = 484 Ω. When connected to a 110 V supply, the power consumed P' = V_supply² / R = (110)² / 484 = 12100 / 484 = 25 W.
MCQ 18. Two cells with EMFs E₁ and E₂ (E₁ > E₂) and internal resistances r₁ and r₂ respectively are connected in series, but with their positive terminals connected together. What is the equivalent EMF of this combination?
- A. E₁ + E₂
- B. E₁ - E₂
- C. (E₁ + E₂)/2
- D. |E₁ - E₂| / (r₁ + r₂)
Answer: B. E₁ - E₂ — When cells are connected in series with opposing polarities (e.g., positive to positive, or negative to negative), their EMFs subtract. The equivalent EMF will be E_eq = E₁ - E₂ (assuming E₁ > E₂). The total internal resistance will still be r₁ + r₂.
MCQ 19. If a wire of resistance R is cut into three equal pieces and these pieces are connected in parallel, what is the equivalent resistance of the combination?
- A. R/3
- B. R/9
- C. 3R
- D. R/27
Answer: B. R/9 — If a wire of resistance R is cut into three equal pieces, each piece has a resistance of R/3. When these three pieces are connected in parallel, their equivalent resistance R_eq is given by 1/R_eq = 1/(R/3) + 1/(R/3) + 1/(R/3) = 3/R + 3/R + 3/R = 9/R. Thus, R_eq = R/9.
MCQ 20. A metallic wire has a resistance of 10 Ω at 0°C and 20 Ω at 100°C. What is the temperature coefficient of resistance of the material?
- A. 0.001 °C⁻¹
- B. 0.01 °C⁻¹
- C. 0.005 °C⁻¹
- D. 0.05 °C⁻¹
Answer: B. 0.01 °C⁻¹ — The resistance at temperature T is given by R_T = R₀(1 + αT). Given R₀ = 10 Ω, R_T = 20 Ω at T = 100°C. So, 20 = 10(1 + α * 100). This simplifies to 2 = 1 + 100α, so 1 = 100α, which means α = 0.01 °C⁻¹.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.