Electricity Formulas form the backbone of Physics for CBSE Class 10 (Chapter 12) and Class 12 (Chapters 1–3), covering everything from Ohm’s Law to Kirchhoff’s rules. These formulas are expressed as mathematical relationships between current, voltage, resistance, power, and charge. Students preparing for JEE Main, JEE Advanced, and NEET will encounter electricity-based problems in nearly every paper. This article provides a complete electricity formula sheet, step-by-step solved examples, common mistakes, and CBSE exam tips for 2025-26.
Key Electricity Formulas at a Glance
Quick reference for the most important electricity formulas used in CBSE and competitive exams.
- Ohm’s Law: \( V = IR \)
- Electric Current: \( I = Q/t \)
- Electric Power: \( P = VI = I^2R = V^2/R \)
- Resistances in Series: \( R_s = R_1 + R_2 + R_3 \)
- Resistances in Parallel: \( 1/R_p = 1/R_1 + 1/R_2 + 1/R_3 \)
- Electrical Energy: \( E = Pt = VIt \)
- Resistivity: \( R = \rho L / A \)
What are Electricity Formulas?
Electricity Formulas are mathematical expressions that describe the behaviour of electric charges, currents, voltages, and energy in a circuit. They allow us to calculate unknown quantities when certain values are known. These formulas are introduced in NCERT Class 10 Physics (Chapter 12 — Electricity) and extended significantly in NCERT Class 12 Physics (Chapter 1 — Electric Charges and Fields, Chapter 2 — Electrostatic Potential and Capacitance, and Chapter 3 — Current Electricity).
At the Class 10 level, students learn Ohm’s Law, resistance combinations, and electric power. At Class 12, the scope expands to include Kirchhoff’s Laws, the Wheatstone bridge, and the potentiometer. Understanding these electricity formulas is essential for scoring well in CBSE board exams. It is equally critical for JEE and NEET, where numerical problems test deep conceptual understanding. Every electricity formula connects a measurable physical quantity to others through a precise relationship.
Electricity Formulas — Expressions and Variables
Ohm’s Law
\[ V = IR \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| V | Electric Potential Difference (Voltage) | Volt (V) |
| I | Electric Current | Ampere (A) |
| R | Resistance | Ohm (Ω) |
Derivation of Ohm’s Law
Ohm’s Law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant. Mathematically, \( V \propto I \). Introducing a constant of proportionality R (resistance), we get \( V = IR \). The resistance R depends on the material, length, and cross-sectional area of the conductor. This relationship was experimentally established by Georg Simon Ohm in 1827 and is the most fundamental of all electricity formulas.
Electric Current Formula
\[ I = \frac{Q}{t} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | Electric Current | Ampere (A) |
| Q | Electric Charge | Coulomb (C) |
| t | Time | Second (s) |
Resistivity Formula
\[ R = \frac{\rho L}{A} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | Resistance | Ohm (Ω) |
| ρ | Resistivity | Ohm·metre (Ω·m) |
| L | Length of conductor | Metre (m) |
| A | Cross-sectional area | Square metre (m²) |
Electric Power Formula
\[ P = VI = I^2 R = \frac{V^2}{R} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Electric Power | Watt (W) |
| V | Voltage | Volt (V) |
| I | Current | Ampere (A) |
| R | Resistance | Ohm (Ω) |
Complete Electricity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Electric Current | \( I = Q/t \) | I = current, Q = charge, t = time | A | Class 10, Ch 12 |
| Ohm’s Law | \( V = IR \) | V = voltage, I = current, R = resistance | V | Class 10, Ch 12 |
| Resistivity | \( R = \rho L / A \) | ρ = resistivity, L = length, A = area | Ω | Class 10, Ch 12 |
| Series Resistance | \( R_s = R_1 + R_2 + R_3 \) | R₁, R₂, R₃ = individual resistances | Ω | Class 10, Ch 12 |
| Parallel Resistance | \( 1/R_p = 1/R_1 + 1/R_2 + 1/R_3 \) | R₁, R₂, R₃ = individual resistances | Ω | Class 10, Ch 12 |
| Electric Power | \( P = VI = I^2R = V^2/R \) | P = power, V = voltage, I = current | W | Class 10, Ch 12 |
| Electrical Energy | \( E = Pt = VIt \) | E = energy, P = power, t = time | J | Class 10, Ch 12 |
| Coulomb’s Law | \( F = k q_1 q_2 / r^2 \) | k = 9×10⁹ N·m²/C², q = charges, r = distance | N | Class 12, Ch 1 |
| Electric Field | \( E = F/q = kQ/r^2 \) | E = field, F = force, q = test charge | N/C or V/m | Class 12, Ch 1 |
| Electric Potential | \( V = kQ/r \) | V = potential, Q = source charge, r = distance | V | Class 12, Ch 2 |
| Capacitance | \( C = Q/V \) | C = capacitance, Q = charge, V = voltage | Farad (F) | Class 12, Ch 2 |
| Energy in Capacitor | \( U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} \) | U = energy, C = capacitance, V = voltage | J | Class 12, Ch 2 |
| Kirchhoff’s Current Law | \( \sum I_{in} = \sum I_{out} \) | I = currents at a junction | A | Class 12, Ch 3 |
| Kirchhoff’s Voltage Law | \( \sum V = 0 \) | V = voltage drops around a closed loop | V | Class 12, Ch 3 |
| Drift Velocity | \( v_d = I / (nAe) \) | n = number density, A = area, e = electron charge | m/s | Class 12, Ch 3 |
| EMF and Terminal Voltage | \( V = \varepsilon – Ir \) | ε = EMF, I = current, r = internal resistance | V | Class 12, Ch 3 |
Electricity Formulas — Solved Examples
Example 1 (Class 10 Level)
Problem: A bulb has a resistance of 100 Ω. It is connected to a 220 V power supply. Find the current through the bulb and the power consumed.
Given: R = 100 Ω, V = 220 V
Step 1: Apply Ohm’s Law to find current: \( I = V/R \)
Step 2: Substitute values: \( I = 220/100 = 2.2 \) A
Step 3: Calculate power using \( P = VI \): \( P = 220 \times 2.2 = 484 \) W
Step 4: Verify using \( P = V^2/R \): \( P = (220)^2 / 100 = 48400/100 = 484 \) W ✓
Answer
Current through the bulb = 2.2 A; Power consumed = 484 W
Example 2 (Class 11-12 Level)
Problem: Three resistors of 6 Ω, 3 Ω, and 2 Ω are connected in parallel. This combination is connected in series with a 4 Ω resistor and a 12 V battery of negligible internal resistance. Find the total current drawn from the battery and the current through each parallel resistor.
Given: R₁ = 6 Ω, R₂ = 3 Ω, R₃ = 2 Ω (parallel), R₄ = 4 Ω (series), V = 12 V
Step 1: Find the equivalent resistance of the parallel combination:
\( \frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{6}{6} = 1 \)
So \( R_p = 1 \) Ω
Step 2: Find total resistance in the circuit:
\( R_{total} = R_p + R_4 = 1 + 4 = 5 \) Ω
Step 3: Find total current from the battery using Ohm’s Law:
\( I = V / R_{total} = 12 / 5 = 2.4 \) A
Step 4: Find voltage across the parallel combination:
\( V_p = I \times R_p = 2.4 \times 1 = 2.4 \) V
Step 5: Find current through each parallel branch:
\( I_1 = V_p / R_1 = 2.4/6 = 0.4 \) A
\( I_2 = V_p / R_2 = 2.4/3 = 0.8 \) A
\( I_3 = V_p / R_3 = 2.4/2 = 1.2 \) A
Verification: \( I_1 + I_2 + I_3 = 0.4 + 0.8 + 1.2 = 2.4 \) A = I ✓
Answer
Total current = 2.4 A; I₁ = 0.4 A, I₂ = 0.8 A, I₃ = 1.2 A
Example 3 (JEE/NEET Level)
Problem: A battery of EMF 10 V and internal resistance 1 Ω is connected to an external resistance of 4 Ω. Calculate the terminal voltage of the battery, the power delivered to the external resistance, and the power lost in the internal resistance.
Given: ε = 10 V, r = 1 Ω, R = 4 Ω
Step 1: Find current using the EMF equation:
\( I = \frac{\varepsilon}{R + r} = \frac{10}{4 + 1} = \frac{10}{5} = 2 \) A
Step 2: Find terminal voltage of the battery:
\( V_{terminal} = \varepsilon – Ir = 10 – (2)(1) = 10 – 2 = 8 \) V
Step 3: Find power delivered to external resistance:
\( P_R = I^2 R = (2)^2 \times 4 = 4 \times 4 = 16 \) W
Step 4: Find power lost in internal resistance:
\( P_r = I^2 r = (2)^2 \times 1 = 4 \) W
Step 5: Verify total power equals total EMF power:
\( P_{total} = \varepsilon I = 10 \times 2 = 20 \) W = \( P_R + P_r = 16 + 4 = 20 \) W ✓
Answer
Terminal voltage = 8 V; Power to external resistance = 16 W; Power lost internally = 4 W
CBSE Exam Tips 2025-26
- Memorise all three forms of the power formula. The CBSE board frequently asks you to calculate power given different combinations of V, I, and R. Know \( P = VI \), \( P = I^2R \), and \( P = V^2/R \) by heart.
- Draw circuit diagrams neatly. In Class 10 and Class 12 board exams, marks are awarded for correctly labelled circuit diagrams. Always mark current direction and component labels.
- Apply Kirchhoff’s Laws systematically. For Class 12, write KCL at each junction and KVL for each loop before solving. We recommend practising at least 10 Kirchhoff problems before the 2025-26 board exam.
- Check units at every step. A common source of lost marks is unit mismatch. Always convert milliamperes to amperes and kilowatts to watts before substituting.
- Learn the condition for maximum power transfer. For JEE and Class 12 exams, remember that maximum power is delivered to the external resistance when \( R = r \) (internal resistance).
- Practice the Wheatstone bridge condition. The balanced bridge condition \( P/Q = R/S \) appears frequently in both CBSE and JEE papers. Our experts suggest solving it from scratch each time rather than memorising the result.
Common Mistakes to Avoid with Electricity Formulas
- Confusing series and parallel formulas. Many students accidentally add reciprocals in series or add directly in parallel. Remember: resistances add directly in series (\( R_s = R_1 + R_2 \)) and reciprocals add in parallel (\( 1/R_p = 1/R_1 + 1/R_2 \)).
- Ignoring internal resistance. In problems involving a real battery, always include internal resistance r in the total circuit resistance. The terminal voltage is NOT equal to the EMF unless the current is zero.
- Mixing up electric field and electric potential. Electric field E is a vector (N/C or V/m). Electric potential V is a scalar (Volt). Students often apply the wrong formula or forget that \( E = -dV/dr \) for a non-uniform field.
- Using the wrong power formula. When resistance R is constant, use \( P = I^2R \). When voltage V is constant, use \( P = V^2/R \). Applying the wrong form leads to incorrect answers in parallel/series circuit problems.
- Forgetting to convert units. Charge is often given in microcoulombs (μC) and distance in centimetres (cm) in Coulomb’s Law problems. Always convert to SI units (C and m) before substituting into \( F = kq_1q_2/r^2 \).
JEE/NEET Application of Electricity Formulas
In our experience, JEE aspirants encounter electricity problems in almost every paper, making this one of the highest-weightage topics in Physics. NEET also includes 3–5 questions from current electricity and electrostatics combined. Here are the key application patterns to master:
Pattern 1: Complex Resistance Networks
JEE Main and Advanced frequently present ladder networks or Wheatstone bridge circuits. The key approach is to identify symmetry first. If the circuit is symmetric about a central axis, the potential at symmetric nodes is equal and no current flows through the connecting wire. This simplifies the network significantly. Always reduce the network step by step using series and parallel combinations.
Pattern 2: Battery with Internal Resistance
Both JEE and NEET test the concept of terminal voltage and maximum power transfer. The formula \( V = \varepsilon – Ir \) is central to these problems. A key JEE Advanced concept is that when multiple batteries are connected, you must apply Kirchhoff’s Laws carefully. The condition for maximum power delivered to external resistance R is \( R = r \), giving \( P_{max} = \varepsilon^2 / (4r) \).
Pattern 3: Capacitor Circuits
NEET and JEE Main regularly include capacitors in series and parallel. The energy stored in a capacitor \( U = \frac{1}{2}CV^2 \) is a standard formula. For capacitors in series, \( 1/C_s = 1/C_1 + 1/C_2 \). For parallel, \( C_p = C_1 + C_2 \). Note that this is the opposite pattern compared to resistors. Students who confuse the two patterns lose easy marks. In our experience, practising at least 20 capacitor problems before the exam eliminates this confusion entirely.
FAQs on Electricity Formulas
For a deeper understanding of related topics, explore our complete guide on the Electric Field Formula, which covers field intensity and Gauss’s Law in detail. Students studying capacitors should also read our article on the Capacitance Formula for series and parallel capacitor combinations. For a broader overview of all Physics topics, visit our Physics Formulas hub page. For official NCERT textbook content, refer to the NCERT official website.