NCERT Solutions Class 12 Physics Chapter 3 Current Electricity

Mastering NCERT Solutions Class 12 Physics Chapter 3 Current Electricity is essential for scoring top marks in CBSE Class 12 Physics board examinations and competitive entrance tests like JEE Main, JEE Advanced, and NEET. Unit II (Current Electricity) forms the core foundation of electrodynamics and carries a substantial weightage of 7 to 8 marks in the CBSE Class 12 Physics syllabus. This chapter bridges electrostatic concepts with dynamic charge flows, covering critical topics such as drift velocity, Ohm's law, temperature dependence of resistivity, cell electromotive force (EMF), internal resistance, Kirchhoff's circuit rules, and balanced bridge networks.

This comprehensive guide provides complete, step-by-step NCERT Physics Class 12 solutions with detailed derivations, structured numerical solutions following the official CBSE marking scheme (Given → To Find → Formula → Substitution → Final Answer), conceptual explanations, and common error analyses. Whether you need chapter-wise step-by-step solutions or high-scoring exam tips, this resource is designed to give you clarity and confidence.

Chapter Overview: Current Electricity

In Chapter 3 of Class 12 Physics, we transition from stationary electric charges to charges in steady motion within conducting media. A solid grasp of the underlying microscopic and macroscopic electrical parameters is crucial for both theoretical and numerical questions in board exams.

The key thematic sections covered in this chapter include:

• Electric Current and Current Density: Rate of charge flow, macroscopic current $I = \frac{dq}{dt}$, and microscopic current density vector $\vec{J} = \frac{I}{A}\hat{n}$.
• Drift Velocity and Origin of Resistivity: Microscopic transport of free electrons under an applied electric field, relaxation time ($\tau$), drift speed ($v_d$), and derivation of Ohm's law in vector form ($\vec{J} = \sigma \vec{E}$).
• Ohm's Law, Resistance, and Resistivity: Factors affecting electrical resistance ($R = \rho \frac{L}{A}$), resistivity ($\rho$), electrical conductivity ($\sigma$), and ohmic vs. non-ohmic devices.
• Temperature Dependence of Resistance: Fractional change in resistance with temperature, temperature coefficient of resistance ($\alpha$), and behaviour of metallic conductors, alloys (Manganin, Constantan, Nichrome), and semiconductors.
• Electrical Energy, Power, and Heating Effect: Joule's heating law ($H = I^2Rt$), power dissipation ($P = VI = I^2R = \frac{V^2}{R}$), and maximum power transfer theorem.
• Cells, EMF, and Internal Resistance: Electromotive force ($\varepsilon$), terminal potential difference ($V$), internal resistance ($r$), discharging relation ($V = \varepsilon - Ir$), charging relation ($V = \varepsilon + Ir$), and grouping of cells in series and parallel.
• Kirchhoff's Rules: Kirchhoff's Current Law (KCL / Junction Rule based on charge conservation) and Kirchhoff's Voltage Law (KVL / Loop Rule based on energy conservation).
• Wheatstone Bridge and Measuring Instruments: Balancing condition ($\frac{P}{Q} = \frac{R}{S}$), meter bridge apparatus for unknown resistance determination, and potentiometer principle ($\varepsilon \propto l$) for comparing EMFs and measuring internal resistance.

Important Formulas and Theorems

To solve numerical problems quickly and accurately in CBSE board exams, memorize the following fundamental formulas and standard SI units from NCERT Class 12 Physics Chapter 3:

Physical Quantity / Law	Mathematical Formula	Variables and SI Units
Electric Current ($I$)	$I = \frac{q}{t} = \frac{ne}{t}$	$q$ in Coulombs (C), $t$ in seconds (s), $I$ in Amperes (A)
Drift Velocity ($v_d$)	$v_d = \frac{e E \tau}{m} = \frac{e V \tau}{m L}$	$e = 1.6 \times 10^{-19}\text{ C}$, $m = 9.1 \times 10^{-31}\text{ kg}$, $\tau = \text{relaxation time (s)}$
Current in terms of Drift Velocity	$I = n A e v_d$	$n = \text{number density of electrons (m}^{-3}\text{)}$, $A = \text{cross-sectional area (m}^2\text{)}$
Current Density ($J$) & Ohm's Law (Microscopic)	$J = \frac{I}{A} = n e v_d = \sigma E$	$J$ in $\text{A/m}^2$, $\sigma = \text{electrical conductivity (S}\cdot\text{m}^{-1}\text{)}$
Resistivity ($\rho$) in terms of Microscopic Constants	$\rho = \frac{m}{n e^2 \tau} = \frac{1}{\sigma}$	$\rho$ in Ohm-metre ($\Omega\cdot\text{m}$)
Resistance ($R$)	$R = \rho \frac{L}{A} = \frac{m L}{n e^2 \tau A}$	$R$ in Ohms ($\Omega$), $L$ in metres (m), $A$ in square metres ($\text{m}^2$)
Temperature Dependence of Resistance	$R_T = R_0 [1 + \alpha (T - T_0)]$	$\alpha = \text{temperature coefficient of resistance (K}^{-1}\text{ or }^\circ\text{C}^{-1}\text{)}$
Terminal Potential Difference of Cell (Discharging)	$V = \varepsilon - I r = \frac{\varepsilon R}{R + r}$	$\varepsilon = \text{EMF (V)}$, $r = \text{internal resistance (}\Omega\text{)}$, $R = \text{load resistance (}\Omega\text{)}$
Equivalent Cells in Parallel	$\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}$	For identical cells in parallel: $\varepsilon_{\text{eq}} = \varepsilon$, $r_{\text{eq}} = \frac{r}{n}$
Balanced Wheatstone Bridge Condition	$\frac{P}{Q} = \frac{R}{S}$	Galvanometer current $I_g = 0$
Potentiometer Comparison of EMF	$\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}$	$l_1, l_2 = \text{balancing lengths (cm or m)}$

NCERT Solutions and Step-by-Step Exercise Problems

Below are detailed, board-standard solutions for core NCERT and CBSE official question bank problems. Every numerical is solved using the standard four-step CBSE methodology.

Question 1: Drift Velocity Definition and Relation with Current

Question: Define drift velocity. Derive the mathematical relation between electric current flowing through a conductor and the drift velocity of free electrons.

Answer:

Definition: Drift velocity ($v_d$) is defined as the average velocity with which free electrons inside a conductor get drifted towards the positive terminal of the conductor under the influence of an externally applied electric field.

Derivation:

1. Consider a cylindrical conductor of length $L$, uniform cross-sectional area $A$, and free electron number density $n$ (number of free electrons per unit volume).
2. The total volume of the conductor is $\text{Volume} = A \times L$.
3. Total number of free electrons in the conductor = $N = n \times A \times L$.
4. Total electric charge contained in the conductor = $q = N e = n A L e$, where $e$ is the magnitude of electron charge ($1.6 \times 10^{-19}\text{ C}$).
5. When an external electric field $E$ is applied across the conductor, electrons drift with an average drift speed $v_d$. The time taken by electrons to traverse the entire length $L$ is:
   $$t = \frac{L}{v_d}$$
6. By definition, electric current $I$ is the total charge flowing per unit time:
   $$I = \frac{q}{t} = \frac{n A L e}{\frac{L}{v_d}} = n A e v_d$$

Final Expression: $I = n A e v_d$ or current density $J = \frac{I}{A} = n e v_d$.

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Question 2: Kirchhoff's Circuit Rules

Question: State Kirchhoff's laws for electrical networks. Name the conservation principles on which these laws are based.

Answer:

1. Kirchhoff's First Law (Junction Rule / KCL):
   • Statement: In any electrical network, the algebraic sum of currents meeting at any electrical junction is equal to zero. Alternatively, the sum of currents entering a junction equals the sum of currents leaving that junction:
     $$\sum I = 0 \implies \sum I_{\text{in}} = \sum I_{\text{out}}$$
   • Physical Principle: Kirchhoff's Junction Rule is based on the Law of Conservation of Electric Charge (no charge accumulates at a junction in steady state).
2. Kirchhoff's Second Law (Loop Rule / KVL):
   • Statement: Around any closed loop in an electrical network, the algebraic sum of changes in potential (potential drops and EMFs) is equal to zero:
     $$\sum \Delta V = 0 \implies \sum \varepsilon = \sum I R$$
   • Physical Principle: Kirchhoff's Loop Rule is based on the Law of Conservation of Energy (the electrostatic force is conservative, meaning total work done around a closed path is zero).

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Question 3: Wire Stretching Problem

Question: A wire of resistance $R$ is stretched uniformly to double its original length. What is the new resistance of the wire?

Solution:

1. Given: Initial length = $L_1 = L$, Initial cross-sectional area = $A_1 = A$, Initial resistance = $R_1 = R = \rho \frac{L}{A}$, Final length = $L_2 = 2L$.

2. Concept: When a wire is stretched uniformly, its mass and volume remain constant, and the material resistivity ($\rho$) remains unchanged.

3. Step-by-Step Calculation:

1. Equating initial volume and final volume:
   $$\text{Volume}_1 = \text{Volume}_2 \implies A_1 L_1 = A_2 L_2$$
   $$A \times L = A_2 \times (2L) \implies A_2 = \frac{A}{2}$$
2. The new resistance $R_2$ is given by:
   $$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{4L}{A} = 4 \left(\rho \frac{L}{A}\right) = 4R$$

Final Answer: The new resistance of the stretched wire is $R' = 4R$ (it increases by a factor of 4).

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Question 4: Temperature Coefficient of Resistance and Temperature Variation

Question: What is the temperature coefficient of resistance ($\alpha$)? How does the resistance of (i) metallic conductors, (ii) semiconductors, and (iii) alloys vary with temperature?

Answer:

Definition: The temperature coefficient of resistance ($\alpha$) is defined as the fractional increase in electrical resistance per unit original resistance per degree rise in temperature.

$$\alpha = \frac{R_2 - R_1}{R_1 (T_2 - T_1)} = \frac{\Delta R}{R_1 \Delta T}$$

SI unit of $\alpha$ is $\text{K}^{-1}$ or $^\circ\text{C}^{-1}$.

Variation with Temperature:

• Metallic Conductors (e.g., Cu, Al, Ag): As temperature rises, thermal agitation of lattice ions increases. Free electrons collide more frequently, causing relaxation time ($\tau$) to decrease ($\rho = \frac{m}{ne^2\tau}$). Since free electron density ($n$) remains nearly constant, resistance increases with temperature ($\alpha > 0$).
• Semiconductors (e.g., Silicon, Germanium): With increasing temperature, covalent bonds break and release new electron-hole pairs, causing electron density ($n$) to increase exponentially. This increase dominates over the decrease in $\tau$, so resistance decreases exponentially with temperature ($\alpha < 0$, negative temperature coefficient).
• Standard Alloys (e.g., Constantan, Manganin, Nichrome): These alloys possess very high resistivity and an extremely small temperature coefficient of resistance ($\alpha \approx 0$). Their resistance remains virtually unchanged with moderate temperature changes, making them ideal for standard resistance coils and meter bridge wires.

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Question 5: Cell EMF, Terminal Voltage, and Power Dissipation

Question: A cell of EMF $2\text{ V}$ and internal resistance $0.5\ \Omega$ is connected across an external resistor of resistance $4.5\ \Omega$. Calculate: (a) the electric current in the circuit, (b) the terminal voltage across the cell, and (c) the electrical power dissipated in the external resistor.

Solution:

1. Given Data:

• Electromotive force of cell ($\varepsilon$) = $2\text{ V}$
• Internal resistance of cell ($r$) = $0.5\ \Omega$
• External load resistance ($R$) = $4.5\ \Omega$

2. Formulas:

• Current: $I = \frac{\varepsilon}{R + r}$
• Terminal Voltage: $V = \varepsilon - I r$ or $V = I R$
• Power Dissipated: $P = I^2 R$

3. Step-by-Step Calculation:

1. Part (a) - Circuit Current:
   $$I = \frac{2}{4.5 + 0.5} = \frac{2}{5.0} = 0.4\text{ A}$$
2. Part (b) - Terminal Voltage ($V$):
   $$V = \varepsilon - I r = 2 - (0.4 \times 0.5) = 2 - 0.20 = 1.8\text{ V}$$
   Verification: $V = I R = 0.4 \times 4.5 = 1.8\text{ V}$.
3. Part (c) - Power Dissipated in External Resistor ($P$):
   $$P = I^2 R = (0.4)^2 \times 4.5 = 0.16 \times 4.5 = 0.72\text{ W}$$

Final Answers:

• (a) Current flowing in the circuit = $0.4\text{ A}$
• (b) Terminal voltage of the cell = $1.8\text{ V}$
• (c) Power dissipated in the resistor = $0.72\text{ W}$

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Question 6: Balanced Wheatstone Bridge Derivation

Question: State the principle of a Wheatstone bridge. Derive the balancing condition for a Wheatstone bridge using Kirchhoff's loop rules. How is it applied in a Meter Bridge to determine an unknown resistance?

Answer:

Principle: A Wheatstone bridge consists of four resistors $P, Q, R, S$ arranged in a closed loop $ABCD$ with a galvanometer connected between junctions $B$ and $D$, and a battery connected between $A$ and $C$. The bridge is balanced when the potential at junction $B$ equals the potential at junction $D$ ($V_B = V_D$), resulting in zero galvanometer deflection ($I_g = 0$).

Derivation using Kirchhoff's Rules:

1. Let currents through branches $AB$ and $AD$ be $I_1$ and $I_2$.
2. At balance ($I_g = 0$), by KCL at junction $B$, current through branch $BC$ is also $I_1$. Similarly, by KCL at junction $D$, current through branch $DC$ is $I_2$.
3. Apply Kirchhoff's Voltage Law (KVL) to closed loop $ABDA$:
   $$-I_1 P - I_g R_g + I_2 R = 0$$ Since $I_g = 0$:
   $$I_1 P = I_2 R \quad \text{--- (Equation 1)}$$
4. Apply KVL to closed loop $BCDB$:
   $$-I_1 Q + I_2 S + I_g R_g = 0$$ Since $I_g = 0$:
   $$I_1 Q = I_2 S \quad \text{--- (Equation 2)}$$
5. Dividing Equation (1) by Equation (2):
   $$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \mathbf{\frac{P}{Q} = \frac{R}{S}}$$

Application in Meter Bridge: In a practical Meter Bridge, a uniform wire of length $100\text{ cm}$ replaces the ratio arms $P$ and $Q$. If the null point is obtained at balancing length $l\text{ cm}$ from end $A$, then $P \propto l$ and $Q \propto (100 - l)$. The unknown resistance $X$ is given by:

$$\frac{R}{X} = \frac{l}{100 - l} \implies \mathbf{X = R \left(\frac{100 - l}{l}\right)}$$

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Question 7: Potentiometer Principle and Comparison of EMFs

Question: Explain the working principle of a potentiometer. Explain with the help of a circuit diagram how a potentiometer is used to compare the EMFs of two primary cells. Why is a potentiometer preferred over a standard voltmeter for measuring EMF?

Answer:

Principle of Potentiometer: The potential fall across any portion of a wire of uniform cross-sectional area and uniform composition carrying a steady current is directly proportional to the length of that portion:

$$V \propto l \implies V = k l \implies \mathbf{k = \frac{V}{l}}$$

where $k$ is the potential gradient of the potentiometer wire (potential drop per unit length, in $\text{V/m}$ or $\text{V/cm}$).

Comparison of EMF of Two Cells:

1. A primary circuit maintains a constant current through potentiometer wire $AB$ using a storage accumulator ($\varepsilon_0$), rheostat ($Rh$), and key ($K$).
2. Two primary cells of EMFs $\varepsilon_1$ and $\varepsilon_2$ are connected such that their positive terminals join the common point $A$, while their negative terminals connect to a two-way key.
3. When cell $\varepsilon_1$ is connected into the circuit and a null point is obtained at length $l_1$:
   $$\varepsilon_1 = k l_1 \quad \text{--- (Equation 1)}$$
4. When cell $\varepsilon_2$ is connected and a null point is obtained at length $l_2$:
   $$\varepsilon_2 = k l_2 \quad \text{--- (Equation 2)}$$
5. Dividing Equation (1) by Equation (2):
   $$\mathbf{\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}}$$

Why Potentiometer is Preferred over a Voltmeter:

• A standard voltmeter draws a small finite current from the cell to produce deflection, thereby measuring terminal potential difference ($V = \varepsilon - Ir$) rather than true EMF ($\varepsilon$).
• A potentiometer operates on the null deflection method, drawing zero current from the cell under test at the balance point. Therefore, it acts as an ideal infinite-resistance voltmeter and measures the accurate, true open-circuit EMF ($\varepsilon$).

Common Mistakes and Tips

CBSE evaluators frequently spot recurring student errors in Current Electricity numericals and circuit derivations. Pay close attention to these critical exam tips:

• Mistake 1: Forgetting Volume Conservation when a Wire is Stretched or Compressed
  Incorrect: Assuming resistance simply doubles when length doubles ($R \propto L \implies R' = 2R$).
  Correct: When a wire is stretched, cross-sectional area decreases proportionally ($A' = A/n$ when $L' = nL$). Therefore, $R' = n^2 R$. If length is doubled, $R' = (2)^2 R = 4R$.
• Mistake 2: Confusing Terminal Potential Difference ($V$) with Cell EMF ($\varepsilon$)
  During Discharging (normal circuit use): $V = \varepsilon - I r$ ($V < \varepsilon$).
  During Charging (current forced into positive terminal): $V = \varepsilon + I r$ ($V > \varepsilon$).
  Open Circuit ($I = 0$): $V = \varepsilon$.
• Mistake 3: Sign Convention Errors in Kirchhoff's Loop Rule
  Always fix a loop traversal direction (clockwise or counter-clockwise) before writing equations:
  • Traversing along current direction through resistor: $\Delta V = -IR$ (potential drop).
  • Traversing opposite to current through resistor: $\Delta V = +IR$ (potential rise).
  • Traversing from negative to positive terminal of cell: $\Delta V = +\varepsilon$.
  • Traversing from positive to negative terminal of cell: $\Delta V = -\varepsilon$.
• Mistake 4: Missing SI Unit Conversions
  Always convert resistance temperature coefficients from $^\circ\text{C}^{-1}$ to $\text{K}^{-1}$ (or ensure $\Delta T$ is in degrees Celsius/Kelvin), balancing lengths from $\text{cm}$ to $\text{m}$, and wire diameters from $\text{mm}$ to radii in $\text{m}$ before calculating area ($A = \pi r^2$).

Board Exam Relevance

In the CBSE Class 12 Physics board examination, Chapter 3 (Current Electricity) accounts for approximately 7 to 8 marks out of the total 70 theory marks. The question distribution typically spans all standard CBSE paper sections:

• Section A (1-Mark MCQs and Assertion-Reason): Questions testing drift velocity relations, temperature dependence graphs of resistivity ($\rho$ vs $T$ for Cu, Nichrome, and Semiconductor), and basic cell formulas.
• Section B (2-Mark Short Answer): Definitions of relaxation time, mobility ($\mu = v_d / E$), drift velocity derivations, or two-cell parallel combination calculations.
• Section C (3-Mark Derivations & Numericals): Kirchhoff's rule network calculations, Wheatstone bridge balanced condition proof, or cell EMF and internal resistance word problems.
• Section D / E (4-Mark Case-Based or 5-Mark Long Answer): Comprehensive questions covering the principle and working of the meter bridge or potentiometer, paired with multi-part numericals.

More NCERT Solutions and Practice

Consistent numerical problem-solving and structured answer writing are key to scoring a perfect 70/70 in CBSE Class 12 Physics. To test your conceptual understanding, generate customized mock question papers, and practice chapter-wise CBSE previous years' questions (PYQs) with step-by-step marking schemes, visit QPTool at theorify.in today.