The Ohms Law Formula, expressed as V = IR, defines the fundamental relationship between voltage, current, and resistance in an electrical circuit. This formula is a cornerstone of Class 10 Physics (NCERT Chapter 12 — Electricity) and extends into Class 12 Physics for advanced circuit analysis. It is equally critical for JEE Main, JEE Advanced, and NEET aspirants. This article covers the complete formula, its derivation, a comprehensive formula sheet, three progressively challenging solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Ohms Law Formulas at a Glance
Quick reference for the most important Ohm’s Law formulas and related expressions.
- Ohm’s Law: \( V = IR \)
- Current: \( I = \dfrac{V}{R} \)
- Resistance: \( R = \dfrac{V}{I} \)
- Power dissipated: \( P = VI = I^2R = \dfrac{V^2}{R} \)
- Resistivity: \( R = \rho \dfrac{L}{A} \)
- Series resistance: \( R_{\text{eq}} = R_1 + R_2 + R_3 \)
- Parallel resistance: \( \dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \)
What is Ohms Law Formula?
The Ohms Law Formula states that the electric current flowing through a conductor is directly proportional to the potential difference (voltage) across its ends, provided the temperature and other physical conditions remain constant. This relationship was discovered by German physicist Georg Simon Ohm in 1827 and forms the foundation of circuit theory.
In NCERT Class 10 Physics, Chapter 12 (Electricity), Ohm’s Law is introduced as one of the first quantitative laws governing electrical circuits. It is revisited in Class 12 Physics, Chapter 3 (Current Electricity), where students explore its limitations and advanced applications.
Mathematically, Ohm’s Law expresses the proportionality between voltage and current as an equation involving resistance, the constant of proportionality. The formula is written as \( V = IR \), where V is voltage in volts, I is current in amperes, and R is resistance in ohms. Conductors that obey this law are called ohmic conductors. Those that do not — such as diodes and transistors — are called non-ohmic conductors.
Ohms Law Formula — Expression and Variables
The standard form of the Ohms Law Formula is:
\[ V = IR \]
This can be rearranged to find current or resistance:
\[ I = \frac{V}{R} \quad \text{and} \quad R = \frac{V}{I} \]
| Symbol | Quantity | SI Unit | Unit Symbol |
|---|---|---|---|
| V | Potential Difference (Voltage) | Volt | V |
| I | Electric Current | Ampere | A |
| R | Electrical Resistance | Ohm | Ω |
Derivation of Ohms Law Formula
Ohm’s Law is derived from the microscopic model of electron drift in a conductor. When a potential difference V is applied across a conductor of length L, an electric field \( E = V/L \) is established. This field exerts a force on free electrons, giving them a drift velocity \( v_d \). The current is given by \( I = nAev_d \), where n is the number density of electrons, A is the cross-sectional area, and e is the electron charge. Since \( v_d \propto E \propto V \), we get \( I \propto V \). Introducing the constant of proportionality R (resistance), we obtain \( V = IR \). Resistance depends on the material’s resistivity \( \rho \) as \( R = \rho L / A \).
Complete Electricity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Ohm’s Law | \( V = IR \) | V = Voltage, I = Current, R = Resistance | V, A, Ω | Class 10, Ch 12 / Class 12, Ch 3 |
| Electric Power | \( P = VI = I^2R = \dfrac{V^2}{R} \) | P = Power, V = Voltage, I = Current, R = Resistance | Watt (W) | Class 10, Ch 12 |
| Electric Energy | \( E = Pt = VIt \) | E = Energy, P = Power, t = Time | Joule (J) | Class 10, Ch 12 |
| Resistivity | \( R = \rho \dfrac{L}{A} \) | ρ = Resistivity, L = Length, A = Area | Ω·m | Class 12, Ch 3 |
| Series Resistance | \( R_{\text{eq}} = R_1 + R_2 + \cdots + R_n \) | R = Individual resistances | Ω | Class 10, Ch 12 |
| Parallel Resistance | \( \dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots \) | R = Individual resistances | Ω | Class 10, Ch 12 |
| Current Density | \( J = \dfrac{I}{A} = \sigma E \) | J = Current Density, σ = Conductivity, E = Electric Field | A/m² | Class 12, Ch 3 |
| Conductance | \( G = \dfrac{1}{R} \) | G = Conductance, R = Resistance | Siemens (S) | Class 12, Ch 3 |
| Temperature Dependence of Resistance | \( R_T = R_0(1 + \alpha \Delta T) \) | R₀ = Resistance at 0°C, α = Temperature coefficient, ΔT = Change in temperature | Ω | Class 12, Ch 3 |
| Drift Velocity | \( v_d = \dfrac{eE\tau}{m} \) | e = electron charge, E = electric field, τ = relaxation time, m = electron mass | m/s | Class 12, Ch 3 |
Ohms Law Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A resistor is connected to a 12 V battery. The current flowing through the circuit is 2 A. Find the resistance of the resistor.
Given: V = 12 V, I = 2 A, R = ?
Step 1: Write the Ohms Law Formula: \( V = IR \)
Step 2: Rearrange to find R: \( R = \dfrac{V}{I} \)
Step 3: Substitute the values: \( R = \dfrac{12}{2} = 6 \) Ω
Answer
The resistance of the resistor is 6 Ω.
Example 2 (Class 11-12 Level)
Problem: Three resistors of 4 Ω, 6 Ω, and 12 Ω are connected in parallel across a 24 V supply. Find the equivalent resistance and the total current drawn from the supply.
Given: R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 12 Ω, V = 24 V
Step 1: Apply the parallel resistance formula:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} \]
Step 2: Find the LCM and simplify:
\[ \frac{1}{R_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \]
Step 3: Therefore \( R_{\text{eq}} = 2 \) Ω
Step 4: Apply the Ohms Law Formula to find total current:
\[ I = \frac{V}{R_{\text{eq}}} = \frac{24}{2} = 12 \text{ A} \]
Answer
Equivalent resistance = 2 Ω; Total current = 12 A.
Example 3 (JEE/NEET Level)
Problem: A wire of resistivity \( \rho = 1.6 \times 10^{-8} \) Ω·m has a length of 2 m and a cross-sectional area of \( 4 \times 10^{-6} \) m². It is connected to a 5 V source. Find (a) the resistance of the wire, (b) the current through it, and (c) the power dissipated.
Given: ρ = 1.6 × 10²&sup8; Ω·m, L = 2 m, A = 4 × 10⁻&sup6; m², V = 5 V
Step 1: Calculate resistance using the resistivity formula:
\[ R = \rho \frac{L}{A} = \frac{1.6 \times 10^{-8} \times 2}{4 \times 10^{-6}} \]
\[ R = \frac{3.2 \times 10^{-8}}{4 \times 10^{-6}} = 8 \times 10^{-3} \text{ } \Omega = 0.008 \text{ } \Omega \]
Step 2: Apply the Ohms Law Formula to find current:
\[ I = \frac{V}{R} = \frac{5}{0.008} = 625 \text{ A} \]
Step 3: Calculate power dissipated:
\[ P = I^2 R = (625)^2 \times 0.008 = 390625 \times 0.008 = 3125 \text{ W} \]
Alternatively: \( P = \dfrac{V^2}{R} = \dfrac{25}{0.008} = 3125 \) W
Answer
(a) R = 0.008 Ω (b) I = 625 A (c) P = 3125 W
CBSE Exam Tips 2025-26
- Memorise all three forms: Always write all three rearrangements — \( V = IR \), \( I = V/R \), and \( R = V/I \) — in your answer sheet before solving. This shows conceptual clarity and earns step marks.
- State the law before applying it: CBSE 2025-26 mark schemes award one mark for correctly stating Ohm’s Law before calculation. Never skip this step in 2-mark or 3-mark questions.
- Draw the V-I graph: For questions on ohmic conductors, draw the linear V-I graph through the origin. Label axes correctly. This is a frequent 1-mark sub-question.
- Unit consistency: Always convert milliamperes (mA) to amperes (A) and kilohms (kΩ) to ohms (Ω) before substituting. Unit errors cause unnecessary mark deductions.
- Power formula variants: We recommend memorising \( P = VI \), \( P = I^2R \), and \( P = V^2/R \) together. CBSE often asks you to find power when only two of V, I, R are given.
- Limitations of Ohm’s Law: For Class 12 board exams 2025-26, be prepared to write two limitations — non-ohmic devices (diodes, thermistors) and temperature dependence of resistance.
Common Mistakes to Avoid
- Confusing V and I in the formula: Many students write \( I = VR \) instead of \( I = V/R \). Always remember that higher resistance means lower current for the same voltage.
- Ignoring unit conversion: Using 200 mA instead of 0.2 A in calculations leads to answers that are 1000 times wrong. Always convert to SI units first.
- Applying Ohm’s Law to non-ohmic devices: Ohm’s Law does not apply to diodes, LEDs, or transistors. These devices have a non-linear V-I relationship. Check whether the device is ohmic before applying the formula.
- Forgetting that R depends on temperature: Resistance is not always constant. For metals, R increases with temperature. Assuming R is constant at all temperatures is a common conceptual error in Class 12 questions.
- Mixing up series and parallel rules: In series circuits, add resistances directly. In parallel circuits, add reciprocals. Swapping these two rules is the most frequent error in circuit problems.
JEE/NEET Application of Ohms Law Formula
In our experience, JEE aspirants encounter Ohm’s Law in nearly every Current Electricity problem. The formula rarely appears in isolation at the JEE level. Instead, it is embedded within complex circuit analysis tasks.
Application Pattern 1 — Kirchhoff’s Laws with Ohm’s Law: JEE problems frequently combine Ohm’s Law with Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL). In a multi-loop circuit, you apply \( V = IR \) to each branch and solve simultaneous equations. Mastering this combination is essential for JEE Main 2025 Current Electricity questions.
Application Pattern 2 — Wheatstone Bridge: The Wheatstone Bridge uses the condition \( P/Q = R/S \) for balance. This condition is derived directly from Ohm’s Law applied to each arm. JEE Advanced problems often ask for the galvanometer current when the bridge is slightly unbalanced, requiring careful application of the Ohms Law Formula to each resistor.
Application Pattern 3 — NEET Biological Context: NEET questions sometimes present Ohm’s Law in the context of nerve impulse conduction or bioelectricity. The resistance of biological membranes and the ionic current across them follow the same \( V = IR \) relationship. Our experts suggest practising at least 10 NEET-level MCQs on this topic from previous year papers (2018–2024) to identify recurring patterns.
For JEE Advanced, also study the microscopic form of Ohm’s Law: \( \vec{J} = \sigma \vec{E} \), where \( \vec{J} \) is current density and \( \sigma \) is electrical conductivity. This vector form appears in JEE Advanced problems on non-uniform conductors and varying cross-section wires.
FAQs on Ohms Law Formula
Explore more physics formulas on our Physics Formulas hub. You may also find these related articles useful: Spring Constant Formula, Flow Rate Formula, and Heat of Vaporization Formula. For the official NCERT syllabus and exam guidelines, visit ncert.nic.in.