The EMF Formula gives the total electromotive force of a source as EMF = V + Ir, where V is the terminal voltage, I is the current, and r is the internal resistance. This concept is central to NCERT Class 12 Physics, Chapter 3 (Current Electricity), and appears regularly in CBSE board exams as well as JEE Main, JEE Advanced, and NEET. In this article, we cover the EMF formula in full detail — its definition, derivation, variables, a complete formula sheet, three solved examples, exam tips, and the most common mistakes students make.
Key EMF Formulas at a Glance
Quick reference for the most important EMF and related circuit formulas.
- EMF (basic): \( \mathcal{E} = V + Ir \)
- Terminal voltage: \( V = \mathcal{E} – Ir \)
- Current in circuit: \( I = \dfrac{\mathcal{E}}{R + r} \)
- EMF (work done): \( \mathcal{E} = \dfrac{W}{Q} \)
- Cells in series: \( \mathcal{E}_{\text{total}} = \mathcal{E}_1 + \mathcal{E}_2 + \cdots \)
- Cells in parallel: \( \mathcal{E}_{\text{eq}} = \dfrac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2} \)
- Faraday's induced EMF: \( \mathcal{E} = -N\dfrac{d\Phi}{dt} \)
What is the EMF Formula?
The EMF Formula defines the electromotive force (EMF) of an electrical source such as a battery or generator. EMF is not a force in the mechanical sense. It is the energy supplied per unit charge by the source to drive current around a complete circuit. The SI unit of EMF is the Volt (V), identical to the unit of potential difference.
In NCERT Class 12 Physics, Chapter 3 (Current Electricity), EMF is introduced as the open-circuit voltage of a cell. When a cell drives current through an external resistance R, some voltage drops across its own internal resistance r. The EMF formula captures this relationship precisely. Understanding EMF is essential not just for board exams but also for analysing real-world circuits, batteries, generators, and electromagnetic induction problems in JEE and NEET.
There are two common contexts for the EMF formula. The first is circuit EMF, which relates EMF to terminal voltage and internal resistance. The second is induced EMF, given by Faraday's law, which relates EMF to the rate of change of magnetic flux. Both are covered in NCERT Class 12 and tested in competitive exams.
EMF Formula — Expression and Variables
1. Circuit EMF Formula (Current Electricity)
The primary EMF formula relating electromotive force, terminal voltage, and internal resistance is:
\[ \mathcal{E} = V + Ir \]
Rearranging for terminal voltage:
\[ V = \mathcal{E} – Ir \]
And for the current delivered to the circuit:
\[ I = \frac{\mathcal{E}}{R + r} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \mathcal{E} \) | Electromotive Force (EMF) | Volt (V) |
| \( V \) | Terminal Voltage | Volt (V) |
| \( I \) | Current through the circuit | Ampere (A) |
| \( r \) | Internal resistance of the cell | Ohm (Ω) |
| \( R \) | External resistance | Ohm (Ω) |
2. EMF as Energy per Unit Charge
\[ \mathcal{E} = \frac{W}{Q} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( W \) | Work done by the source | Joule (J) |
| \( Q \) | Charge moved through the circuit | Coulomb (C) |
| \( \mathcal{E} \) | Electromotive Force | Volt (V) |
3. Faraday's Law — Induced EMF Formula
\[ \mathcal{E} = -N\frac{d\Phi_B}{dt} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( N \) | Number of turns in the coil | Dimensionless |
| \( \Phi_B \) | Magnetic flux through one turn | Weber (Wb) |
| \( d\Phi_B/dt \) | Rate of change of magnetic flux | Weber per second (Wb/s) |
| \( \mathcal{E} \) | Induced EMF | Volt (V) |
Derivation of the Circuit EMF Formula
Consider a cell of EMF \( \mathcal{E} \) and internal resistance \( r \) connected to an external resistance \( R \). By Kirchhoff's Voltage Law, the total EMF equals the sum of all voltage drops in the loop.
Step 1: The current in the circuit is \( I \). The voltage drop across the external resistance is \( V = IR \). The voltage drop across the internal resistance is \( Ir \).
Step 2: Applying KVL around the loop: \( \mathcal{E} = IR + Ir \).
Step 3: Since \( V = IR \) (terminal voltage), we get \( \mathcal{E} = V + Ir \).
Step 4: Solving for current: \( I = \dfrac{\mathcal{E}}{R + r} \). This is the standard EMF formula used in all NCERT problems.
Complete Electricity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| EMF (Circuit) | \( \mathcal{E} = V + Ir \) | V = terminal voltage, I = current, r = internal resistance | Volt (V) | Class 12, Ch 3 |
| Terminal Voltage | \( V = \mathcal{E} – Ir \) | ℰ = EMF, I = current, r = internal resistance | Volt (V) | Class 12, Ch 3 |
| Current in Circuit | \( I = \dfrac{\mathcal{E}}{R + r} \) | ℰ = EMF, R = external resistance, r = internal resistance | Ampere (A) | Class 12, Ch 3 |
| EMF as Work per Charge | \( \mathcal{E} = \dfrac{W}{Q} \) | W = work done, Q = charge | Volt (V) | Class 12, Ch 3 |
| Cells in Series (EMF) | \( \mathcal{E}_{s} = \mathcal{E}_1 + \mathcal{E}_2 \) | ℰ₁, ℰ₂ = individual EMFs | Volt (V) | Class 12, Ch 3 |
| Cells in Parallel (EMF) | \( \mathcal{E}_{p} = \dfrac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2} \) | r₁, r₂ = internal resistances | Volt (V) | Class 12, Ch 3 |
| Induced EMF (Faraday's Law) | \( \mathcal{E} = -N\dfrac{d\Phi_B}{dt} \) | N = turns, Φ₂ = magnetic flux | Volt (V) | Class 12, Ch 6 |
| Motional EMF | \( \mathcal{E} = Bvl \) | B = magnetic field, v = velocity, l = length of conductor | Volt (V) | Class 12, Ch 6 |
| EMF of a Rotating Coil | \( \mathcal{E} = NBA\omega\sin(\omega t) \) | N = turns, B = field, A = area, ω = angular frequency | Volt (V) | Class 12, Ch 7 |
| Ohm's Law | \( V = IR \) | V = voltage, I = current, R = resistance | Volt (V) | Class 12, Ch 3 |
EMF Formula — Solved Examples
Example 1 (Class 10-11 Level)
Problem: A battery has an EMF of 12 V and an internal resistance of 0.5 Ω. It is connected to an external resistance of 5.5 Ω. Find the current in the circuit and the terminal voltage of the battery.
Given: \( \mathcal{E} = 12 \) V, \( r = 0.5 \) Ω, \( R = 5.5 \) Ω
Step 1: Write the current formula: \( I = \dfrac{\mathcal{E}}{R + r} \)
Step 2: Substitute values: \( I = \dfrac{12}{5.5 + 0.5} = \dfrac{12}{6} = 2 \) A
Step 3: Find terminal voltage using \( V = \mathcal{E} – Ir \).
Step 4: Substitute: \( V = 12 – (2)(0.5) = 12 – 1 = 11 \) V
Answer
Current in the circuit = 2 A. Terminal voltage = 11 V.
Example 2 (Class 12 Level)
Problem: Two cells, each of EMF 1.5 V and internal resistance 0.2 Ω, are connected in series with an external resistance of 2.6 Ω. Calculate the current through the circuit and the terminal voltage across each cell.
Given: \( \mathcal{E}_1 = \mathcal{E}_2 = 1.5 \) V, \( r_1 = r_2 = 0.2 \) Ω, \( R = 2.6 \) Ω
Step 1: Total EMF for cells in series: \( \mathcal{E}_{\text{total}} = 1.5 + 1.5 = 3.0 \) V
Step 2: Total internal resistance: \( r_{\text{total}} = 0.2 + 0.2 = 0.4 \) Ω
Step 3: Current: \( I = \dfrac{\mathcal{E}_{\text{total}}}{R + r_{\text{total}}} = \dfrac{3.0}{2.6 + 0.4} = \dfrac{3.0}{3.0} = 1 \) A
Step 4: Voltage drop across each cell's internal resistance: \( Ir = 1 \times 0.2 = 0.2 \) V
Step 5: Terminal voltage of each cell: \( V = 1.5 – 0.2 = 1.3 \) V
Answer
Current = 1 A. Terminal voltage of each cell = 1.3 V.
Example 3 (JEE/NEET Level)
Problem: A cell of EMF \( \mathcal{E} \) and internal resistance \( r \) is connected to a variable external resistance \( R \). The power delivered to the external resistance is maximum when \( R = r \). If the maximum power delivered externally is 9 W and the internal resistance is 1 Ω, find the EMF of the cell.
Given: \( P_{\text{max}} = 9 \) W, \( r = 1 \) Ω, condition: \( R = r \) for maximum power
Step 1: At maximum power, \( R = r = 1 \) Ω. Write the current: \( I = \dfrac{\mathcal{E}}{R + r} = \dfrac{\mathcal{E}}{2r} \)
Step 2: Power delivered to R: \( P = I^2 R = \left(\dfrac{\mathcal{E}}{2r}\right)^2 \times r = \dfrac{\mathcal{E}^2}{4r} \)
Step 3: Set equal to maximum power: \( \dfrac{\mathcal{E}^2}{4r} = 9 \)
Step 4: Substitute \( r = 1 \): \( \dfrac{\mathcal{E}^2}{4} = 9 \implies \mathcal{E}^2 = 36 \implies \mathcal{E} = 6 \) V
Answer
The EMF of the cell = 6 V.
CBSE Exam Tips 2025-26
- Always distinguish EMF from terminal voltage. EMF is the open-circuit voltage. Terminal voltage is always less than EMF when current flows. Many students confuse the two and lose marks.
- State the formula before substituting. In CBSE board exams, writing \( \mathcal{E} = V + Ir \) before substituting values earns a formula mark even if the final answer has a calculation error.
- Check units carefully. Resistance must be in ohms, current in amperes, and EMF in volts. We recommend converting all units before substituting.
- For cells in series, add both EMFs and both internal resistances. For cells in parallel with equal EMFs, the equivalent EMF equals the individual EMF but the equivalent internal resistance is halved.
- In 3-mark questions on induced EMF, always write Faraday's law first, then substitute. Mention the negative sign and explain it using Lenz's law for full marks.
- Practise the maximum power transfer condition \( R = r \). This concept appeared in CBSE 2023 and 2024 board papers and is expected again in 2025-26.
Common Mistakes to Avoid
- Mistake 1 — Treating EMF as a force: EMF is energy per unit charge (Joules per Coulomb), not a mechanical force. Writing its unit as Newton (N) is incorrect. The correct unit is Volt (V).
- Mistake 2 — Ignoring internal resistance: Many students use \( V = \mathcal{E} \) directly. This is only valid when the internal resistance is zero or negligible. Always check whether internal resistance is given.
- Mistake 3 — Wrong sign in Faraday's law: The induced EMF is \( \mathcal{E} = -N\dfrac{d\Phi_B}{dt} \). The negative sign is not optional. It indicates the direction of the induced current opposes the change in flux (Lenz's law).
- Mistake 4 — Incorrect combination of cells: When cells are in series, students sometimes add only the EMFs but forget to add the internal resistances. Both must be added for a correct answer.
- Mistake 5 — Confusing motional EMF formula: The motional EMF is \( \mathcal{E} = Bvl \), valid only when B, v, and l are mutually perpendicular. If the angle between v and B is \( \theta \), the formula becomes \( \mathcal{E} = Bvl\sin\theta \).
JEE/NEET Application of the EMF Formula
In our experience, JEE aspirants encounter the EMF formula in multiple forms across different chapters. Mastering all three versions — circuit EMF, Faraday's induced EMF, and motional EMF — is essential for scoring in Physics.
Application Pattern 1 — Internal Resistance and Power Problems (JEE Main)
JEE Main frequently tests the relationship between EMF, internal resistance, and power. A typical question gives you the EMF and asks for the value of external resistance that maximises power transfer. The condition is \( R = r \), and the maximum power is \( P_{\text{max}} = \dfrac{\mathcal{E}^2}{4r} \). Students must derive this from the EMF formula and the power expression \( P = I^2 R \).
Application Pattern 2 — Kirchhoff's Laws with Multiple Cells (JEE Advanced)
JEE Advanced problems often involve complex circuits with multiple cells of different EMFs and internal resistances. The EMF formula \( \mathcal{E} = V + Ir \) is applied loop by loop using Kirchhoff's Voltage Law. Our experts suggest practising at least 15 multi-loop problems before the exam. Pay attention to the polarity of each cell when writing the KVL equation.
Application Pattern 3 — Electromagnetic Induction (NEET and JEE)
NEET tests Faraday's law almost every year. A coil rotating in a uniform magnetic field generates an EMF given by \( \mathcal{E} = NBA\omega\sin(\omega t) \). The peak EMF is \( \mathcal{E}_0 = NBA\omega \). Questions ask for the instantaneous EMF at a given time, or the frequency needed to produce a certain peak EMF. Knowing the derivation of this formula from \( \mathcal{E} = -N\dfrac{d\Phi_B}{dt} \) gives you a significant advantage in NEET Physics.
FAQs on EMF Formula
For more related physics concepts, explore our detailed guide on the Electric Field Formula, which connects closely with EMF in electrostatics. You may also find our article on the Capacitance Formula useful for understanding energy storage in circuits. For a broader revision, visit the complete Physics Formulas hub covering all NCERT Class 11 and Class 12 topics. For official NCERT textbook content, refer to the NCERT official website.