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Tamil Nadu Class 12 Physics — Current Electricity Notes, Formulae & Important Questions

Complete Samacheer Kalvi Class 12 Physics Current Electricity notes covering Ohm's law, Kirchhoff's laws, Wheatstone bridge, metre bridge, potentiometer, and worked numericals for board exam and NEET.

Updated 22 May 20269 min read

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Board Exam Weight: Current Electricity carries 8–10 marks in Tamil Nadu Class 12 Physics board exams (Volume 1, Chapter 2). NEET typically has 2–4 questions from this chapter.


1. Electric Current and Drift Velocity

Electric current I is the rate of flow of charge:

I = Q / t (Amperes)

Drift velocity v_d is the average velocity of free electrons in the direction opposite to the electric field:

I = nAev_d

where n = number density of free electrons, A = cross-sectional area, e = 1.6×10⁻¹⁹ C.


2. Ohm's Law — Two Forms

Macroscopic Form

V = IR

Valid when temperature is constant and the conductor is ohmic (linear V-I graph).

Microscopic Form (Ohm's Law in terms of field quantities)

J = σE

where:

  • J = current density = I/A (A/m²)
  • σ = electrical conductivity (Ω⁻¹·m⁻¹ or S/m)
  • E = electric field inside conductor (V/m)

Resistivity ρ = 1/σ, so: J = E/ρ → E = ρJ

Resistance from resistivity:

R = ρL / A

where L = length, A = cross-sectional area.

Temperature Dependence of Resistance

For most metals, resistance increases with temperature:

R_T = R₀(1 + αT)

where α is the temperature coefficient of resistance (unit: K⁻¹ or °C⁻¹).


3. Kirchhoff's Laws

KCL — Kirchhoff's Current Law (Junction Rule)

"The algebraic sum of currents at any junction is zero."

ΣI_in = ΣI_out

Based on conservation of charge.

KVL — Kirchhoff's Voltage Law (Loop Rule)

"The algebraic sum of all potential differences (EMF and IR drops) around any closed loop is zero."

ΣE = ΣIR

Based on conservation of energy.

Sign conventions for KVL:

  • Traversing a resistor in the direction of assumed current: voltage drop = −IR
  • Traversing a battery from − to +: gain = +E
  • Traversing a battery from + to −: drop = −E

4. Wheatstone Bridge

Four resistors P, Q, R, S arranged in a bridge circuit with a galvanometer connecting the midpoints.

Balanced condition (no current through galvanometer):

P/Q = R/S → PS = QR

To find unknown resistance S:

S = QR / P

Uses: Finding unknown resistance with high accuracy. Metre bridge is a practical form.


5. Metre Bridge

A practical form of the Wheatstone bridge using a 1-metre uniform resistance wire.

Principle:

P/Q = R/S

If balancing length is l cm from left end (for resistance R in left gap and unknown S in right gap):

R/S = l / (100 − l) S = R(100 − l) / l

Precautions:

  • Jockey should be tapped gently — not dragged
  • Null point should be near the middle of wire (40–60 cm) for accuracy
  • Connections should be tight to avoid contact resistance

6. Potentiometer

A long uniform resistance wire used to make precise voltage and EMF measurements.

Principle: When no current is drawn from the cell, the potential drop across a length l of wire is proportional to l:

V ∝ l → V = φl

where φ = potential gradient = V_total / L_wire (V/m).

Applications

1. Comparing two EMFs:
E₁ / E₂ = l₁ / l₂
2. Finding internal resistance r of a cell:
Answer

First balance with external resistance R connected (length l₂), then disconnect R (length l₁):

r = R(l₁ − l₂) / l₂

Advantage over voltmeter: Potentiometer draws no current from the cell being tested, so it measures true EMF. A voltmeter has finite resistance and draws some current.


7. EMF and Internal Resistance

Every real battery has internal resistance r. When current I flows:

E = V + Ir → V = E − Ir (discharging)
V = E + Ir (charging — terminal voltage exceeds EMF)

Power delivered to external circuit:

P = VI = I²R

Power dissipated internally:

P_internal = I²r

Maximum power theorem: Power delivered to external R is maximum when R = r (internal resistance).


8. Worked Numericals

Numerical 1 — KVL Circuit Problem

Problem: A circuit has a 12 V battery (negligible internal resistance) with three resistors 3 Ω, 4 Ω, and 5 Ω connected in series. Find the current and voltage across each resistor.

Solution
Total resistance R = 3 + 4 + 5 = 12 Ω Current I = V/R = 12/12 = 1 A Voltage across 3Ω: V₁ = IR₁ = 1 × 3 = 3 V Voltage across 4Ω: V₂ = IR₂ = 1 × 4 = 4 V Voltage across 5Ω: V₃ = IR₃ = 1 × 5 = 5 V Check (KVL): 3 + 4 + 5 = 12 V ✓

Numerical 2 — Wheatstone Bridge

Problem: In a Wheatstone bridge, P = 10 Ω, Q = 15 Ω, R = 20 Ω. Find the value of S for balance.

Solution
Balanced condition: P/Q = R/S S = QR / P S = (15 × 20) / 10 S = 300 / 10 S = 30 Ω

The unknown resistance S = 30 Ω.


Numerical 3 — Internal Resistance

Problem: A battery of EMF 6 V and internal resistance 0.5 Ω is connected to an external resistance of 11.5 Ω. Find: (a) current in the circuit, (b) terminal voltage of the battery.

Solution
(a) Total resistance = R + r = 11.5 + 0.5 = 12 Ω Current I = E / (R + r) = 6 / 12 = 0.5 A (b) Terminal voltage V = E − Ir V = 6 − (0.5 × 0.5) V = 6 − 0.25 = 5.75 V

Terminal voltage (5.75 V) is less than EMF (6 V) because of the voltage drop across internal resistance.


Numerical 4 — Potentiometer (Comparing EMFs)

Problem: In a potentiometer experiment, a standard cell of 2.0 V gives a balancing length of 400 cm. An unknown cell gives a balancing length of 300 cm. Find the EMF of the unknown cell.

Solution
Using the principle: E₁/E₂ = l₁/l₂ E₁ = 2.0 V, l₁ = 400 cm, l₂ = 300 cm E₂ = E₁ × (l₂/l₁) E₂ = 2.0 × (300/400) E₂ = 2.0 × 0.75 E₂ = 1.5 V

The EMF of the unknown cell is 1.5 V.


9. Board Exam Model Questions

2-Mark Questions

Q1. State Kirchhoff's voltage law and give its physical basis. KVL: The algebraic sum of all EMFs and potential drops around any closed loop in a circuit equals zero. Physical basis: conservation of energy — a charge completing a closed loop returns to the same potential, so net work done is zero.

Q2. Define potential gradient in a potentiometer. Give its SI unit. Potential gradient is the potential drop per unit length of the potentiometer wire: φ = V/L. It is uniform along the wire if the wire has uniform cross-section and material. SI unit: V/m (or V/cm in practice).

Q3. Why is a potentiometer preferred over a voltmeter for measuring EMF? A potentiometer draws no current from the cell at the null point (balance condition), so it measures the true EMF without any terminal voltage drop. A voltmeter, having finite resistance, draws current and measures terminal voltage (V = E − Ir), which is less than the true EMF.

5-Mark Questions

Q4. Describe the construction and working of a metre bridge. Derive the formula for unknown resistance.
Answer

A metre bridge consists of a 1-metre uniform resistance wire AC, with a galvanometer connecting the jockey (D) to a junction B between two resistances R (known) and S (unknown) placed in a Wheatstone bridge arrangement.

The balance condition of the Wheatstone bridge gives:

R/S = P/Q

where P and Q are the resistances of the two segments of the wire AD and DC.

Since the wire is uniform, P/Q = l/(100 − l), giving:

R/S = l/(100 − l) S = R(100 − l)/l
Q5. Explain how a potentiometer can be used to determine the internal resistance of a cell.
Answer

Circuit: Cell of EMF E and internal resistance r, with a switch and external resistance R in series. Potentiometer wire connected to a driver battery.

Step 1: Open switch (no current from cell). Balance length = l₁. Here V = E, so E ∝ l₁.

Step 2: Close switch (current through R). Balance length = l₂. Here V = E − Ir = ER/(R+r), so V ∝ l₂.

Dividing: E/V = l₁/l₂ → (R+r)/R = l₁/l₂

r = R(l₁ − l₂)/l₂

10. NEET Relevance

TopicNEET Frequency
Ohm's law — resistivity, drift velocityHigh
Kirchhoff's laws — circuit solvingVery High
Wheatstone bridgeHigh
Potentiometer — comparing EMFsMedium
Internal resistance — terminal voltageVery High
Combination of resistorsVery High

NEET Tip: Practice multi-loop KVL problems. NEET often gives a circuit with two loops and asks for current through a specific element — apply KVL systematically to each loop.


11. Key Formulae Summary

QuantityFormula
CurrentI = Q/t = nAev_d
Ohm's lawV = IR, J = σE
ResistivityR = ρL/A
KCLΣI_in = ΣI_out
KVLΣE = ΣIR (around loop)
Wheatstone balanceP/Q = R/S
Metre bridgeS = R(100−l)/l
Potentiometer (EMFs)E₁/E₂ = l₁/l₂
Internal resistanceV = E − Ir
Internal r (potentiometer)r = R(l₁−l₂)/l₂

Constants: e = 1.6×10⁻¹⁹ C

Frequently Asked Questions

EMF (E) is the maximum voltage a battery can provide when no current flows (open circuit). Terminal voltage (V) is the actual voltage across the battery terminals when current is flowing: V = E − Ir, where r is internal resistance and I is the current. Terminal voltage is always less than EMF when current is drawn.