“Cells,EMF and Terminal Potential difference:Derivations & Diagrams”

Cells,EMF and Terminal Potential difference

Introduction:

Every electrical circuit, from a simple flashlight to sophisticated electronic devices, depends on a source of electrical energy. This energy is supplied by an electric cell, which converts chemical energy into electrical energy and drives charges through a circuit.To understand how cells power electrical systems, it is essential to study two fundamental concepts: Electromotive Force (EMF) and Terminal Potential Difference (TPD).

Although these quantities are both measured in volts, they represent different physical ideas. The EMF of a cell describes the maximum energy supplied per unit charge by the cell, while the terminal potential difference represents the actual voltage available across the cell terminals when current flows through an external circuit. The difference between them arises because every real cell possesses an internal resistance,which causes part of the supplied energy to be dissipated within the cell itself.

In this article, we will develop a clear understanding of electric cells, EMF, internal resistance, and terminal potential difference through step-by-step derivations, circuit diagrams, graphical interpretations, and numerical applications.

Whether you are preparing for CBSE Class 12 Physics, NEET, JEE Main, or other competitive examinations, mastering these concepts will strengthen your understanding of Current Electricity and help solve a wide range of conceptual and numerical problems.

Core Concepts:

Cells,EMF and Terminal Potential difference:Derivations & Diagrams

Differences Between EMF and Potential Difference

Relation Between Internal Resistance,EMF, and Terminal Potential Difference

Three Cells in Series – Physical Connection

Parallel Combination of Two Cells

Mixed Grouping of Cells: The ‘mxn’ array

INTERNAL RESISTANCE-IMAGE

Internal Resistance – Formula

How to Calculate the Internal Resistance of a Cell?Battery?

Frequently Asked Questions (FAQs) – Internal  resistance of a cell / battery

1. What is internal resistance in a battery?
2. Why is internal resistance important?
3.  How can internal resistance be measured?
4. What factors affect internal resistance?
5. Can internal resistance be reduced?

Answers for the Frequently Asked Questions (FAQs)- Internal resistance of a cell/battery

1. Internal resistance of a cell/battery is the resistance offered by the electrode and electrolyte which oppose the current flow inside the cell. The higher the internal resistance, the less current the battery is capable of giving.                                                                             (OR)Internal resistance is the opposition to the flow of current within the battery itself, which causes a voltage drop when current flows.
2. Internal resistance is important because it affects the battery’s efficiency, performance, and lifespan. Higher internal resistance leads to more energy loss as heat.
3. It can be measured using specialized equipment like a battery analyzer or by using the voltage drop method under load.
4. Factors include battery chemistry, temperature, age, and state of charge.
5. While it cannot be eliminated, using high-quality materials, proper maintenance, and optimal operating conditions can minimize it.

NEET/JEE One-Liner Tricks

Lost volts is heat: Power wasted inside = I^2r = I \times V_{lost}. That’s why old batteries get hot.
Good cell = small r: Car batteries have r\approx 0.01\Omega → tiny lost volts even at 100A.

Torch cells have r\approx 1\Omega → huge loss if you try 1A.
Charging case: V = E + Ir → Here “lost volts” adds to EMF. Terminal PD > EMF. You’re forcing current backwards, so you need extra voltage to overcome internal resistance + EMF.
Rule of thumb: If V < E → cell discharging, lost volts = E-V.
If V > E → cell charging, “gained volts” = V-E = Ir.

Questions:
1. Define electromotive force (EMF) of a cell.
2. A cell has EMF 2 V and internal resistance 0.5 Ω. What is the terminal voltage when it delivers a current of 2 A?
3. When a cell is connected to a 2 Ω resistor, the terminal voltage is 1.8 V. If the EMF is 2 V, find internal resistance.
4. Two cells of EMF 1.5 V and 2 V with internal resistances 0.3 Ω and 0.2 Ω are connected in series. Find total EMF and total internal resistance.
5. Two identical cells each of EMF 2 V and internal resistance 1 Ω are connected in parallel. Find equivalent EMF and internal resistance.
6. Four cells of EMF 1.5 V and internal resistance 0.2 Ω each are connected in parallel. Find total EMF and internal resistance.
7. A cell of EMF 2 V and internal resistance 1 Ω is connected to an external resistor R. If terminal voltage is 1.5 V, find R.
8. If the external resistance equals internal resistance of a cell, what is terminal voltage in terms of EMF?
9. When a cell is short-circuited, what is the current?
10.12 identical cells (E=2V, r=1Ω) are to be arranged to send max current through a 3Ω resistor. Find m and n.
11.Two cells of EMF 1 V, 2 V and internal resistances 2 Ω, 1 Ω are connected in parallel with like terminals together. Find equivalent EMF.
12.Why is EMF not a force?
13.For a cell, when R=2Ω, I=0.5A; when R=5Ω, I=0.25A. Find E and r.
14.What is the condition for zero terminal voltage?
15.A cell of EMF 1.5 V gives 0.2 A current through a 5 Ω resistor. Find internal resistance.

Answers:
1. EMF is the potential difference across the terminals of a cell when no current is drawn from it. It is the energy supplied by the cell per unit charge.
2. V = E – Ir = 2 – (2 × 0.5) = 2 – 1 = 1 V.
3. I = V/R = 1.8/2 = 0.9 A, r = (E-V)/I = (2-1.8)/0.9 = 0.2/0.9 = 0.222 Ω.
4. 3.5 V and O.5 Ω
5. E=2V and r=0.5 Ω
6. E=1.5V and r=0.05 Ω
7. I=0.5 A nd R=3 Ω
8. V=E/2
9. I=E/2
10. m=6 n=2
11 5/3 V
12.It is energy per unit charge (joules/coulomb = volt), not a mechanical force.
13.r=1Ω, E=1.5V.
14.Short circuit (R=0) or when cell is completely discharged (E=0).
15.2.5 Ω.

Conclusion:

The concepts of EMF, internal resistance, and terminal potential difference form the foundation of electrical circuit analysis.While the EMF represents the total energy supplied by a cell per unit charge, the terminal potential difference represents the usefulenergy available to the external circuit. The presence of internal resistance explains why the terminal voltage decreases when current is drawn from the cell.

A strong grasp of these principles enables students to analyze real electrical circuits, understand battery performance,solve numerical problems efficiently, and build a solid conceptual bridge to advanced topics such as Kirchhoff’s laws,network analysis, and electrical power systems. By combining theoretical understanding with circuit diagrams and graphical representations,
students can approach examination questions with greater confidence and accuracy.