First a set of $n$ equal resistors of $R$ each are connected in series to a battery of emf $E$ and internal resistance $R$. A current $I$ is observed to flow. Then the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is '$n$'?

First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistan

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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 key idea is to apply Ohm’s law to both series and parallel configurations, using the battery’s internal resistance $R$ in each case. The condition that the parallel current is 10 times the series current leads to a quadratic in $n$ , whose positive solution is $n = 10$ .

Why this approach works

The problem gives you two circuits built from the same battery (emf $E$ , internal resistance $R$ ) and the same $n$ identical resistors (each of value $R$ ). In the series case, the total external resistance is $nR$ ; in the parallel case, it is $R/n$ . The battery’s internal resistance $R$ is always in series with the external load. So the total circuit resistance in each case is just the sum of the external resistance and the internal resistance. Then Ohm’s law gives the current. The only unknown is $n$ , and the ratio of the two currents is given as 10. That gives an equation you can solve.

Step-by-step solution

Series connection When $n$ resistors, each $R$ , are connected in series, the total external resistance is

R_{\text{ext, series}} = nR.

The battery has internal resistance $R$ , so the total circuit resistance is

R_{\text{total, series}} = nR + R = (n+1)R.

The current $I$ is therefore

I = \frac{E}{(n+1)R}.

Parallel connection When the same $n$ resistors are connected in parallel, the equivalent external resistance is

R_{\text{ext, parallel}} = \frac{R}{n}.

Adding the internal resistance $R$ gives

R_{\text{total, parallel}} = \frac{R}{n} + R = R\left(1 + \frac{1}{n}\right) = R\left(\frac{n+1}{n}\right).

The current in this case, call it $I'$ , is

I' = \frac{E}{R\left(\frac{n+1}{n}\right)} = \frac{nE}{(n+1)R}.

Using the given ratio The problem states that the parallel current is 10 times the series current:

I' = 10 I.

Substitute the expressions from steps 1 and 2: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$\Omega$
›Reveal solutionSolution

Equating the EMF for both cases gives internal resistance $r=3\ \Omega$ .

The EMF of the battery equals $I(R+r)$ in each case:
$E=0.6(3+r)\qquad\text{and}\qquad E=0.4(6+r)$
Setting them equal:

$0.6(3+r)=0.4(6+r)$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$0.1\ \Omega$
›Reveal solutionSolution
Identical cells in parallel keep the same emf and give internal resistance $r/n$ .

For $n$ identical cells (emf $\varepsilon$ , internal resistance $r$ ) connected in parallel:

Equivalent emf $=\varepsilon$ (unchanged).

Equivalent internal resistance $=\dfrac{r}{n}$ . …
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KEAM 2026Set eng-2026-04204 marksMCQ
$emfs$
›Reveal solutionSolution

Total emf $8$ V across total resistance $10\ \Omega$ gives $I=0.8$ A.

Circuit. Two cells (2 V and 6 V) in series, aiding, so net emf $=2+6=8$ V. Total resistance $=$ external $8\ \Omega$ $+$ two internal $1\ \Omega$ eac …
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KEAM 2026Set eng-2026-04214 marksMCQ
$3\,V$
›Reveal solutionSolution

Series loop: net driving EMF is $3\,$ V and total resistance $6\,\Omega$ , giving $0.5\,$ A.

The internal resistances add to the external resistor in series:
$R_{total} = 0.5 + 0.75 + 0.75 + 4 = 6\,\Omega.$
The only combination of the three cell EMFs ( $3, 4, 4\,$ V) that gives a current matching the options is when the $3\,$ V cell opposes the two $4\,$ V cells (or two cells oppose), leaving a net EMF of $3\,$ V: …
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KEAM 2026Set eng-2026-04224 marksMCQ
$\Omega$
›Reveal solutionSolution

The two side-by-side cells are in parallel (2 V, $0.5\,\Omega$ ); this combines in series with the third cell, then apply Ohm's law across the $4.5\,\Omega$ resistor.

Two identical 2 V, $1\,\Omega$ cells in parallel give an emf of $2$ V and internal resistance $\tfrac12(1)=0.5\,\Omega$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$'n'$
›Reveal solutionSolution
Connecting identical cells in parallel keeps the voltage the same but increases the current the battery can deliver.

Concept and Intuition

Cells in parallel all share the same EMF, so the net terminal voltage stays equal to that of a single cell. However, the effective internal resistance drops to r/n, allowing a larger current to be supplied to the load.

Step-by-Step Solution

EMF of parallel combination = EMF of one cell (voltage unchanged).

Internal resistance = r/n (decreases). …
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KEAM 2025Set eng-2025-04254 marksMCQ
$0.5\ \Omega$
›Reveal solutionSolution

The largest current a source can deliver is limited only by its own internal resistance (external $R=0$ ), so $I_{max}=E/r$ .

The terminal current of a cell is $I=\dfrac{E}{R+r}$ . This is maximised when the external resistance $R\to 0$ (a short circuit), giving …
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KEAM 2025Set eng-2025-04274 marksMCQ
$E$
›Reveal solutionSolution

Cells in parallel keep the same EMF ( $E$ ) but their internal resistances combine like parallel resistors, so the net internal resistance is $\dfrac{r}{n}$ .

When $n$ identical cells, each of EMF $E$ and internal resistance $r$ , are joined in parallel, all their positive terminals are tied together and all negatives together. Since every cell drives the same potential difference, the combined EMF is just $E$ (not $nE$ — that would be a series stack).

The $n$ internal resistances $r$ are now in parallel: …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$Q. An electric cell does 10 J of work in carrying a charge of 5 C around a simple closed circuit. The electromotive force of the cell is (A) 0.5 V (B) 1.5 V (C) 1 V (D) 6 V (E) 2 V$
›Reveal solutionSolution

EMF is work done per unit charge: $\varepsilon=W/q=10/5=2$ V.

The electromotive force is defined as the work done by the cell per unit charge carried around the circuit: …
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KEAM 2024Set eng-2024-06064 marksMCQ
$\Omega$
›Reveal solutionSolution

The emf drives current through external plus internal resistance: $\varepsilon=I(R+r)$ .

Applying $\varepsilon=I(R+r)$ :

$12=2(5.8+r)\Rightarrow 6=5.8+r\Rightarrow r=0.2\ \Omega.$ …
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KEAM 2024Set eng-2024-06084 marksMCQ
$10\,\Omega$
›Reveal solutionSolution

Open-circuit voltage = EMF = 2 V; loaded current $0.15$ A; $r=(E-V)/I=0.5/0.15=10/3\,\Omega$ .

In the open circuit the terminal voltage equals the EMF: $E = 2$ V.

When connected across $R = 10\ \Omega$ , terminal voltage $V = 1.5$ V, so the current is

$I = \dfrac{V}{R} = \dfrac{1.5}{10} = 0.15\ \text{A}.$ …
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KEAM 2024Set eng-2024-06094 marksMCQ
$\varepsilon$
›Reveal solutionSolution

The straight-line terminal-voltage plot meets the V-axis at the cell's emf, so the y-intercept $= \varepsilon$ .

For a cell of emf $\varepsilon$ and internal resistance $r$ driving a load $R$ ,

$V = \varepsilon - I r, \qquad I = \dfrac{\varepsilon}{R+r}.$ …
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