Q.Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ (with $\varepsilon_2 > \varepsilon_1$) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel between two common terminals $A$ and $B$: one branch is the first battery $\varepsilon_1$ in series with $r_1$, and the other branch is the second battery $\varepsilon_2$ in series with $r_2$, both branches joining the same two terminals $A$ and $B$ with their positive terminals on the same side. Which of the following statements about the equivalent emf $\varepsilon_{eq}$ of this combination is correct?

Internal Resistance – The Battery That Fights Itself

Imagine you have a bucket of water with a tap at the bottom. When you open the tap fully, water flows out freely. But if the pipe is narrow or clogged, the flow is weaker even though the bucket is full. A battery behaves the same way: it has a "full bucket" of electrical energy (its emf, or electromotive force), but inside the battery, there is always some "clogged pipe" that resists the flow of charge. That clog is called internal resistance.

The Intuition

Every real battery is not a perfect source of voltage. If you connect a small bulb to a fresh battery, it glows brightly. But if you connect a powerful motor that draws a large current, the same battery might struggle — the voltage at its terminals drops, and the motor runs slower. Why? Because inside the battery, the chemical reactions and the materials themselves have a natural opposition to the flow of charge. This opposition is internal resistance, denoted by $r$ (or sometimes $R_{\text{int}}$).

Think of it this way: the battery has two "battles" to fight. First, it must push charge through the external circuit (the bulb, the motor, the wires). Second, it must push charge through its own internal structure. The harder it has to push (i.e., the larger the current), the more voltage it "loses" inside itself.

The Precise Statement

Internal resistance is the opposition to the flow of electric current offered by the materials and chemical processes inside a source of emf (like a battery, cell, or generator). It is measured in ohms ($\Omega$).

For a cell or battery, the relationship between the emf ($\mathcal{E}$), the terminal voltage ($V$), the current ($I$), and the internal resistance ($r$) is given by:

This is the single most important equation to remember.

Here:

• $\mathcal{E}$ is the electromotive force — the maximum voltage the battery can provide when no current flows (open circuit). Think of it as the "ideal" voltage.
• $V$ is the terminal voltage — the actual voltage you measure across the battery's terminals when it is delivering current.
• $I$ is the current flowing through the circuit (and therefore through the battery itself).
• $r$ is the internal resistance.

The term $I r$ is the voltage drop inside the battery. It is the "price" the battery pays for delivering current.

What Happens in Different Situations?

Why Does Internal Resistance Matter?

1. Power loss inside the battery: The power dissipated as heat inside the battery is $P_{\text{loss}} = I^2 r$. This is why batteries get warm when delivering large currents.
2. Maximum power transfer: There is a famous theorem (Maximum Power Transfer Theorem) that says a source delivers maximum power to an external load when the load resistance equals the internal resistance ($R_{\text{load}} = r$). But this is inefficient — half the power is wasted inside the source.
3. Battery health: As a battery ages, its internal resistance increases. Measuring $r$ is a common way to test if a battery is still good.

A Simple Example

A cell has an emf of $1.5\ \text{V}$ and an internal resistance of $0.2\ \Omega$. It is connected to a $3.0\ \Omega$ resistor. Find the current and the terminal voltage.

Solution:

The total resistance in the circuit is $R_{\text{total}} = R_{\text{load}} + r = 3.0 + 0.2 = 3.2\ \Omega$.

Using Ohm's law for the whole circuit: $I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{1.5}{3.2} = 0.46875\ \text{A}$.

Terminal voltage: $V = \mathcal{E} - I r = 1.5 - (0.46875 \times 0.2) = 1.5 - 0.09375 = 1.40625\ \text{V}$.

Notice that the terminal voltage ($1.41\ \text{V}$) is less than the emf ($1.5\ \text{V}$). The difference is small here because the current is modest. If you short-circuited the cell ($R_{\text{load}} = 0$), the current would be $I = \frac{1.5}{0.2} = 7.5\ \text{A}$, and the terminal voltage would drop to zero.

The Big Picture

Internal resistance is not a flaw — it is a fundamental property of every real voltage source. It explains why batteries have limits, why they heat up, and why you cannot get infinite current from them. Whenever you see a battery symbol in a circuit diagram, remember that there is always a tiny resistor hiding inside it, silently opposing the flow.

Internal resistance of a cell and the terminal-voltage equation V = ε − Ir form a core part of the NCERT Class 12 Physics chapter on current electricity, tested extensively in CBSE board numericals, JEE Main and NEET. Students revising "internal resistance of a cell formula numericals class 12 physics" will find this emf-versus-terminal-voltage explanation matches the NCERT derivation.