Last Updated: April 2026
Current Electricity is one of the most important chapters in JEE Main Physics, contributing 2–4 questions per paper consistently across all sessions. It is also directly connected to Magnetism, Electromagnetic Induction, and Semiconductors. This guide provides complete chapter notes, all critical formulas, solved numericals, and 10 MCQs aligned to JEE 2027 pattern.
Importance of Current Electricity in JEE Main 2027
JEE Main Physics carries 30 questions (120 marks). Current Electricity falls under Electricity and Magnetism — one of the highest-weightage units. According to NTA paper analysis (2019–2024), this unit alone contributes 7–10 questions per session. Current Electricity forms the base of this unit, and concepts like Kirchhoff’s laws and Wheatstone bridge appear in complex circuit problems regularly.
Electric Current and Drift Velocity
Electric current (I) is the rate of flow of electric charge:
I = Q/t = nAev_d
Where:
- n = number density of free electrons (electrons per m³)
- A = cross-sectional area of the conductor
- e = charge of electron (1.6 × 10⁻¹⁹ C)
- v_d = drift velocity of electrons
Drift Velocity
Drift velocity is the average velocity of electrons in the direction of electric field (very small — of the order of mm/s or 10⁻⁴ m/s):
v_d = eEτ/m = J/ne
Where τ = relaxation time (average time between successive collisions), m = mass of electron, J = current density.
Ohm’s Law and Resistance
Ohm’s law states that for a metallic conductor at constant temperature, the current is proportional to the applied voltage:
V = IR
Where R = resistance (in Ohms, Ω). Ohm’s law is not a universal law — it is obeyed by metallic conductors at constant temperature.
Resistivity and Factors Affecting It
Resistance of a conductor depends on:
R = ρL/A
Where ρ (rho) = resistivity (unit: Ω·m), L = length, A = cross-sectional area.
| Factor | Effect on Resistance |
|---|---|
| Length (L) increases | R increases (R ∝ L) |
| Cross-section area (A) increases | R decreases (R ∝ 1/A) |
| Temperature increases (metals) | Resistivity increases (ρ = ρ₀(1 + αΔT)) |
| Temperature increases (semiconductors) | Resistivity decreases |
| Material changes | Resistivity changes (intrinsic property) |
Series and Parallel Combinations
| Property | Series | Parallel |
|---|---|---|
| Equivalent Resistance | R_eq = R₁ + R₂ + R₃ + … | 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + … |
| Current | Same through all (I = constant) | Divides (V = constant) |
| Voltage | Divides across each resistor | Same across all |
| R_eq vs individual R | Greater than largest | Less than smallest |
| Two resistors in parallel | — | R_eq = R₁R₂/(R₁+R₂) |
Kirchhoff’s Laws
Kirchhoff’s Current Law (KCL) — Junction Rule
At any junction (node) in a circuit, the sum of currents entering equals the sum of currents leaving:
ΣI_in = ΣI_out (or equivalently: ΣI = 0 at a node)
Based on conservation of charge — charge cannot accumulate at a junction.
Kirchhoff’s Voltage Law (KVL) — Loop Rule
In any closed loop of a circuit, the algebraic sum of all EMFs equals the algebraic sum of all IR voltage drops:
ΣE = ΣIR (or equivalently: ΣV = 0 around a closed loop)
Based on conservation of energy — energy gained from EMF sources equals energy lost in resistors.
Wheatstone Bridge
A Wheatstone Bridge is a circuit for accurate measurement of unknown resistance. It consists of four resistors P, Q, R, S arranged in a diamond pattern with a galvanometer across the middle.
Balance condition (null deflection in galvanometer):
P/Q = R/S
When balanced, no current flows through the galvanometer. The unknown resistance can be calculated once the other three are known.
Meter Bridge: A practical form of Wheatstone bridge where a uniform resistance wire of 100 cm is used. If balance point is at length l: R/S = l/(100−l)
Potentiometer
A potentiometer is a device used to measure EMF accurately (without drawing any current from the cell being measured). Principle: the potential difference across any segment of the wire is proportional to its length.
V ∝ L → V = k·L (where k = potential gradient in V/m)
Applications:
- Comparison of EMFs: E₁/E₂ = L₁/L₂
- Measurement of internal resistance: r = R(L₁−L₂)/L₂
- Measurement of very small EMFs (thermocouples)
EMF and Internal Resistance
Every real cell has an internal resistance r. The terminal voltage V (actual voltage available) is less than EMF E when current flows:
V = E − Ir (during discharging)
V = E + Ir (during charging)
- When no current flows (open circuit): V = E
- Short circuit current: I_max = E/r
- Maximum power transferred to external R: when R = r (matched load condition)
Master Formula Table — Current Electricity
| Concept | Formula | Unit / Note |
|---|---|---|
| Ohm’s Law | V = IR | Ω = V/A |
| Current | I = Q/t = nAev_d | Ampere (A) |
| Drift velocity | v_d = I/(nAe) | m/s (very small) |
| Resistance | R = ρL/A | Ohm (Ω) |
| Temp dependence | ρ = ρ₀(1 + αΔT) | α = temp coefficient |
| Series R | R_eq = R₁ + R₂ + … | R_eq > each R |
| Parallel R | 1/R_eq = 1/R₁ + 1/R₂ + … | R_eq < smallest R |
| Two in parallel | R_eq = R₁R₂/(R₁+R₂) | Product/Sum formula |
| KCL | ΣI_in = ΣI_out | At junction |
| KVL | ΣE = ΣIR | Around loop |
| Wheatstone (balanced) | P/Q = R/S | No current in galvanometer |
| Terminal voltage | V = E − Ir | During discharge |
| Power | P = VI = I²R = V²/R | Watt (W) |
| Max power transfer | P_max when R_ext = r | P_max = E²/4r |
| Potentiometer | E₁/E₂ = L₁/L₂ | Comparison of EMFs |
Solved Numericals
Numerical 1: Find current in a circuit
Problem: A battery of EMF 12V and internal resistance 2Ω is connected to an external resistance of 10Ω. Find the current and terminal voltage.
Solution:
Total resistance = R + r = 10 + 2 = 12 Ω
Current I = E/(R+r) = 12/12 = 1 A
Terminal voltage V = E − Ir = 12 − 1×2 = 10 V
Or: V = IR = 1×10 = 10 V ✓
Numerical 2: Wheatstone Bridge — Unknown Resistance
Problem: In a Wheatstone bridge, P = 10Ω, Q = 20Ω, R = 15Ω, and the bridge is balanced. Find S.
Solution:
Balanced condition: P/Q = R/S
10/20 = 15/S
S = 15 × 20/10 = 30 Ω
Numerical 3: Series-Parallel Combination
Problem: Three resistors 6Ω, 6Ω, and 3Ω are connected — first two in series, their combination in parallel with the third. Find equivalent resistance.
Solution:
Series: R₁₂ = 6 + 6 = 12 Ω
Parallel with 3Ω: R_eq = (12 × 3)/(12 + 3) = 36/15 = 2.4 Ω
Practice MCQs — JEE Main Physics 2027
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Frequently Asked Questions (FAQs)
Q1: What is the difference between EMF and potential difference?
EMF (E) is the work done per unit charge by the cell to move charge through the entire circuit (including internal resistance). It is a property of the source. Potential Difference (V) is the work done per unit charge between two points in the external circuit. V = E − Ir. When no current flows, V = E. The key distinction: EMF is a source property; potential difference is between circuit points.
Q2: When does Ohm’s law fail?
Ohm’s law (V ∝ I) fails in: (1) Semiconductors — V-I relationship is non-linear; (2) Diodes — current flows only in one direction; (3) Electrolytes — conductivity varies with concentration; (4) Thyristors and other non-ohmic devices. At very high electric fields, even metals deviate from Ohm’s law. In JEE, any question mentioning “non-ohmic” means Ohm’s law does not apply.
Q3: How do I approach complex circuits with Kirchhoff’s Laws?
Follow these steps: (1) Label all junctions and assign current variables to each branch; (2) Apply KCL at junctions to reduce variables; (3) Choose independent loops and apply KVL — assign voltage drops as positive when traversing in the direction of current; (4) Solve the simultaneous equations. For a circuit with n nodes and b branches: you need (n−1) KCL equations and (b−n+1) KVL equations. Practice 2-loop and 3-loop circuits until you can set up equations in under 2 minutes.
Q4: What is the principle of a potentiometer and why is it preferred over a voltmeter?
A potentiometer works on the null deflection principle — at balance, no current is drawn from the cell being measured. This means it measures the true EMF without any error due to internal resistance. A voltmeter, even with very high resistance, always draws some current and thus measures terminal voltage (E − Ir), not true EMF. This makes the potentiometer far more accurate for EMF measurement.
Master JEE Physics at JEE Gurukul
Current Electricity is a chapter that rewards systematic understanding over rote learning. Master Kirchhoff’s laws with sufficient practice, understand the physical meaning behind every formula, and never skip numericals. Explore our complete JEE Physics course at JEE Gurukul Courses for structured chapter videos, daily practice sheets, and full-syllabus mock tests.
JEE Tip: Wheatstone bridge and potentiometer are asked almost every year — master the balance condition (P/Q = R/S) and the potentiometer null point method thoroughly!