A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing t

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Concept understanding — Internal Resistance

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:

V = \mathcal{E} - I r

This is the single most important equation to remember.

V = \mathcal{E} - I r

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?

Condition	Current $I$	Terminal Voltage $V$	Why?
Open circuit (no load)	$I = 0$	$V = \mathcal{E}$	No current means no internal drop.
Small load (e.g., a dim bulb)	Small $I$	$V \approx \mathcal{E}$	The $I r$ term is tiny.
Large load (e.g., a powerful motor)	Large $I$	$V < \mathcal{E}$ significantly	The $I r$ drop becomes noticeable.
Short circuit (wires directly across terminals)	Very large $I$	$V \approx 0$	Almost all voltage is lost inside the battery; the battery heats up and may be damaged.

Watch out

A common mistake is to think that internal resistance is a "bad" thing that can be eliminated. It cannot — every real source has some internal resistance. Even a brand new AA battery has a small $r$ (typically $0.1$ to $0.5\ \Omega$ ). Old or weak batteries have much larger $r$ , which is why they fail to power high-current devices.

Why Does Internal Resistance Matter?

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.
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. …

Why this formula?

Internal Resistance: Why the Key Formulas Hold

Internal resistance is a fundamental concept in real-world circuits. No battery is perfect — every practical source has some internal resistance ( $r$ ) that opposes the flow of current inside the source itself.

1. The Core Idea: A Real Battery = An Ideal Source + A Resistor

Think of a real battery as:

An ideal EMF source ( $\mathcal{E}$ ) — provides constant voltage with zero internal resistance.
A small resistor ( $r$ ) — connected in series inside the battery.

Why series? Because the current that leaves the battery must first pass through its internal material (electrolyte, electrodes), which offers resistance.

2. The Terminal Voltage Formula

When the battery delivers current $I$ to an external circuit:

The ideal source produces $\mathcal{E}$ .
The internal resistor $r$ drops some voltage: $V_{\text{drop}} = I r$ (Ohm's law).
The voltage available at the terminals ( $V$ ) is what's left:

V = \mathcal{E} - I r

Why this makes sense:

If $I = 0$ (open circuit), $V = \mathcal{E}$ — you measure the full EMF.
If $I$ increases, $V$ decreases — the internal drop grows.
If $I$ is huge (short circuit), $V \to 0$ and all voltage is lost inside.

3. The Short-Circuit Current Formula

If you connect the terminals directly (external resistance $R = 0$ ):

The only resistance in the circuit is $r$ .
By Ohm's law: $I_{\text{short}} = \frac{\mathcal{E}}{r}$

Why this holds:

The entire EMF is now dropped across $r$ alone.
This is the maximum current the battery can deliver — limited by its internal resistance.
In practice, this can damage the battery (heating, chemical damage).

4. Power Delivered to External Load

When the battery is connected to an external load $R$ :

Total circuit resistance: $R + r$
Current: $I = \frac{\mathcal{E}}{R + r}$
Power delivered to the external load ( $P_{\text{out}}$ ):

P_{\text{out}} = I^2 R = \left( \frac{\mathcal{E}}{R + r} \right)^2 R

Why this matters:

Power is not simply $\mathcal{E}^2 / R$ — because $r$ limits current. …

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The terminal voltage (P.D. across $R$ ) increases with $R$ , starting from zero when $R=0$ and asymptotically approaching the emf $E$ as $R\to\infty$ . The graph is a smooth, increasing curve that saturates at $E$ .

Why this problem matters — and the intuition

When you connect a real cell (with internal resistance $r$ ) to an external resistor $R$ , the voltage you actually measure across $R$ is not the cell's emf $E$ . Some voltage is "lost" inside the cell itself, across its internal resistance. The bigger the current drawn, the bigger this internal drop.

The key insight: as $R$ increases, the current decreases. Less current means less voltage drop inside the cell, so more of $E$ appears across $R$ . When $R$ is very large, almost no current flows — and the terminal voltage nearly equals $E$ . When $R$ is zero (a short circuit), all the voltage drops inside the cell, and the terminal voltage is zero.

Let's turn this intuition into mathematics.

Step-by-step derivation

1. Write the circuit equation

For a cell of emf $E$ and internal resistance $r$ , connected to an external resistance $R$ , the total resistance in the circuit is $R + r$ . By Ohm's law, the current is:

I = \frac{E}{R + r}

2. Express the terminal voltage

The potential difference across $R$ (which is also the terminal voltage of the cell) is:

V = I R = \frac{E R}{R + r}

This is the function we need to plot: $V$ as a function of $R$ .

3. Examine the behaviour at the extremes

When $R = 0$ (short circuit):

V = \frac{E \cdot 0}{0 + r} = 0

The entire emf is dropped across $r$ , so nothing appears across $R$ .

When $R \to \infty$ (open circuit): Divide numerator and denominator by $R$ :

V = \frac{E}{1 + \frac{r}{R}}

As $R \to \infty$ , $\frac{r}{R} \to 0$ , so $V \to E$ .

The terminal voltage approaches the emf asymptotically.

4. Check the slope and shape

Differentiate $V$ with respect to $R$ :

\frac{dV}{dR} = E \cdot \frac{(R+r) - R}{(R+r)^2} = \frac{E r}{(R+r)^2}

This is always positive — the graph is strictly increasing. The slope is steepest near $R=0$ (where it equals $E/r$ ) and gradually flattens as $R$ grows. …

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Q.A cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$ . Plot a graph showing the variation of P.D. across $R$ , versus $R$ .

Concept understanding — Internal Resistance

Internal Resistance – The Battery That Fights Itself

The Intuition

The Precise Statement

What Happens in Different Situations?

Why Does Internal Resistance Matter?

Why this formula?

Internal Resistance: Why the Key Formulas Hold

1. The Core Idea: A Real Battery = An Ideal Source + A Resistor

2. The Terminal Voltage Formula

3. The Short-Circuit Current Formula

4. Power Delivered to External Load

Why this problem matters — and the intuition

Step-by-step derivation

Method: Sketching How Terminal Voltage Varies with External Resistance

Steps

Q.A cell of emf EEE and internal resistance rrr is connected across an external resistance RRR. Plot a graph showing the variation of P.D. across RRR, versus RRR.

Why this problem matters — and the intuition

Step-by-step derivation

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Q.A cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$ . Plot a graph showing the variation of P.D. across $R$ , versus $R$ .

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