(a)(i) Three batteries E₁, E₂ and E₃ of emfs and internal resistances (4mathrmV,2Ω), (2mathrmV,4Ω) and (6mathr

Question

Figure — Figure — 55/5/1 Q31 — FigureFigure — 55/5/1 Q31

E_{1}

(ii) The ends of six wires, each of resistance

R\,(=10\ \Omega)

, are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the effective resistance offered by it to the circuit.

(OR)

(b)(i) A current

I\,(=1\ \mathrm{A})

is passing through a copper rod (

n=8.5\times10^{28}\ \mathrm{m^{-3}}

) of varying cross-section as shown in the figure. The areas of cross-section at points A and B are

1.0\times10^{-7}\ \mathrm{m^{2}}

and

2.0\times10^{-7}\ \mathrm{m^{2}}

respectively. Calculate: (I) the ratio of the electric fields at points A and B; (II) the drift velocity of free electrons at point B.

(ii) Two point charges

q_{1}\,(=16\ \mu\mathrm{C})

and

q_{2}\,(=1\ \mu\mathrm{C})

are placed at points

\vec{r_{1}}=(3\ \mathrm{m})\hat{i}

and

\vec{r_{2}}=(4\ \mathrm{m})\hat{j}

. Find the net electric field

\vec{E}

at the point

\vec{r}=(3\ \mathrm{m})\hat{i}+(4\ \mathrm{m})\hat{j}

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Part (a)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. …

Part (b)Concept understanding — Drift Velocity

Drift Velocity: The Slow March of Electrons

Electrons in a metal are always moving — but randomly. At room temperature they zip around at roughly $10^6$ m/s, colliding with the lattice ions every few trillionths of a second. Without an electric field this motion cancels out: for every electron heading left another heads right, so the net velocity is zero.

Apply a battery and the field gives every electron a tiny, steady push in one direction. Between collisions the electron accelerates only briefly before smashing into an ion and losing its directed motion. What survives is a very small average velocity along the field — the drift velocity.

Note

The random thermal speed is about $10^5$ m/s, but the drift velocity is only about $10^{-4}$ m/s — about a billion times slower. An electron drifts slower than a snail, yet a lamp lights instantly, because the electric field (not the electrons) propagates at nearly the speed of light and starts every electron drifting almost at once.

The precise definition

Drift velocity ( $v_d$ ) is the average velocity acquired by the charge carriers in a conductor under an applied electric field:

v_d = \frac{eE\tau}{m}

where:

$e$ = electron charge ( $1.6 \times 10^{-19}$ C)
$E$ = electric field inside the conductor (V/m)
$\tau$ = average relaxation time — the mean time between collisions (s)
$m$ = electron mass ( $9.1 \times 10^{-31}$ kg)

v_d = \frac{eE\tau}{m}

Linking to current

Drift velocity connects the microscopic motion of electrons to the current an ammeter reads:

I = n e A v_d

where $n$ is the free-electron number density and $A$ the cross-sectional area. A larger $v_d$ means more current, but $v_d$ stays tiny because $\tau$ is tiny (about $10^{-14}$ s in copper).

Tip

For a copper wire carrying $1$ A with area $1$ mm² and $n \approx 8.5 \times 10^{28}$ m $^{-3}$ :

$v_d = \frac{I}{neA} \approx \frac{1}{(8.5\times10^{28})(1.6\times10^{-19})(10^{-6})} \approx 7 \times 10^{-5}\ \text{m/s}$ …

Why this formula?

Drift Velocity: Why the Formula Holds

Let's build this from first principles — understanding why electrons drift the way they do, not just memorizing the formula.

1. The Core Idea: What is Drift Velocity?

In a conductor, free electrons are constantly moving randomly (thermal motion, speeds ~ $10^5$ m/s). Without an electric field, their net displacement is zero — they're like a swarm of bees buzzing in all directions.

When we apply an electric field $\vec{E}$ , it gently nudges each electron in the opposite direction (since electrons are negatively charged). This small, steady net velocity superimposed on the random motion is drift velocity ( $v_d$ ).

Key insight: Drift velocity is not the speed of individual electrons — it's the average velocity of the entire electron cloud.

2. The Derivation: Step by Step

Step 1: Force on a single electron

An electron of charge $-e$ in an electric field $\vec{E}$ experiences:

\vec{F} = -e \vec{E}

The magnitude of acceleration (opposite to $\vec{E}$ ) is:

a = \frac{F}{m} = \frac{eE}{m}

where $m$ is the electron's mass.

Step 2: What happens between collisions?

Electrons don't accelerate forever — they keep colliding with atoms/ions in the metal lattice. Let the average time between collisions be $\tau$ (relaxation time). Just after a collision an electron's velocity is essentially random (zero average in the field direction); it then accelerates for time $\tau$ before the next collision.

Step 3: Drift velocity

Averaging the field-driven velocity over the relaxation time $\tau$ gives the net drift:

\boxed{v_d = \frac{eE\tau}{m}}

Here $\tau$ is the average time since the last collision, so this expression already averages over electrons at every stage between collisions — it is the standard result used in the NCERT treatment.

3. Connecting to Current: The Big Picture

Drift velocity directly gives us current density $J$ :

J = n e v_d

where $n$ = number of free electrons per unit volume. …

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Part (a): three parallel batteries give a common terminal voltage $V=4.4$ V, so $I_1=-0.2$ A, $I_2=-0.6$ A (charging) and $I_3=+0.8$ A; the six-wire tetrahedron has $R_{AB}=R/2=5\ \Omega$ . Part (b): in the tapered rod $E_A/E_B=2$ and $v_d(B)\approx3.7\times10^{-4}$ m/s; the two point charges give a net field $(1\hat i+9\hat j)\times10^{3}$ N/C.

Part (a)

(i) Three batteries in parallel

As the figure shows, the three batteries are connected in parallel, so they all share one terminal voltage $V$ . The current delivered by battery $k$ (emf $\mathcal{E}_k$ , internal resistance $r_k$ ) is $I_k=(\mathcal{E}_k-V)/r_k$ , negative if the battery is being charged.

Write each current for terminal voltage $V$ :

I_1=\frac{4-V}{2},\qquad I_2=\frac{2-V}{4},\qquad I_3=\frac{6-V}{2}.

With no external load, Kirchhoff's junction rule gives $I_1+I_2+I_3=0$ . Multiplying through by 4:

2(4-V)+(2-V)+2(6-V)=0\ \Rightarrow\ 22-5V=0\ \Rightarrow\ V=4.4\ \text{V}.

Substitute back:

I_1=-0.2\ \text{A},\qquad I_2=-0.6\ \text{A},\qquad I_3=+0.8\ \text{A}.

Watch out

The negative signs are not errors — they show that the strong battery $E_3$ drives current into $E_1$ and $E_2$ , charging them.

Tip

Shortcut: $V=\dfrac{\sum \mathcal{E}_k/r_k}{\sum 1/r_k}=\dfrac{2+0.5+3}{0.5+0.25+0.5}=\dfrac{5.5}{1.25}=4.4$ V.

(ii) Six equal wires (tetrahedron)

The six wires in the figure connect four nodes $A$ , $B$ , top apex $T$ and centre $O$ , with a resistor $R$ on every one of the six pairs — i.e. a complete graph $K_4$ (tetrahedron).

Treat $A$ and $B$ as the terminals. The arms $A\text{-}T$ , $T\text{-}B$ , $A\text{-}O$ , $O\text{-}B$ form a Wheatstone bridge with $T\text{-}O$ as the bridge arm. Because all arms equal $R$ , $\dfrac{A\text{-}T}{T\text{-}B}=\dfrac{A\text{-}O}{O\text{-}B}=1$ , so the bridge is balanced and $T\text{-}O$ carries no current.
Removing $T\text{-}O$ $T - O$ leaves three parallel paths between $A$ $A$ and $B$ $B$ :
- direct $A\text{-}B$ : $R$
- $A\text{-}T\text{-}B$ : $2R$
- $A\text{-}O\text{-}B$ : $2R$
Combine in parallel: $\frac{1}{R_{AB}}=\frac{1}{R}+\frac{1}{2R}+\frac{1}{2R}=\frac{2}{R}\ \Rightarrow\ R_{AB}=\frac{R}{2}=\frac{10}{2}=5\ \Omega.$ …

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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 54 on this concept.

CBSE 2026Set A1 markMCQ
$Q. A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if (A) R = r/2 (B) R = r (C) R > r (D) R < r$
›Reveal solutionSolution

I = ε/(R+r); smaller R ⇒ larger current, so current is maximum for R < r (ideally R → 0).

The circuit current is $I = \dfrac{\varepsilon}{R+r}$ .

For a fixed emf ε and fixed internal resistance r, the current increases as the external resistance R decreases. Hence the current through R is maximum when R is as small as possible. Among the given choices, this corresponds t …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The electromotive force of an accumulator battery is 10 V and internal resistance 0.5Ω. The maximum electric current obtained from the battery will be$
(a) 5 A
(b) 10 A
(c) 20 A
(d) 0.05 A
›Reveal solutionSolution

The maximum current a cell can deliver is its short-circuit current, I = EMF / internal resistance.

A real battery has EMF (epsilon) and internal resistance r. When connected to an external circuit of resistance R, the current is I = epsilon/(R+r), which is largest when R = 0 (sho …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Drift velocity Vd varies with the intensity of electric field E as per the relation$
(a) Vd is proportional to E^2
(b) Vd is proportional to 1/E
(c) Vd is proportional to sqrt(E)
(d) Vd is proportional to E
›Reveal solutionSolution

Drift velocity is the (small) average velocity electrons gain between collisions due to the electric field, and it comes out directly proportional to E.

When an electric field E is applied to a conductor, each free electron experiences a force F = eE, giving it an acceleration a = eE/m between collisions with the lattice ions. If tau is the average time between collisions (relaxation time), the average extra velocity gained (the drift velocity) is

vd = a * tau = (eE/m) * tau = (e*tau/m) * E

…

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The quantity measured across a cell without drawing any current from it, is ..... .$
›Reveal solutionSolution

The potential difference measured across a cell's terminals when no current is drawn from it equals the cell's EMF.

When current $I$ flows, the terminal voltage is $V = \varepsilon - Ir$ (less than the EMF $\varepsilon$ due to the voltage drop across internal resistance $r$ ). When no current is drawn ( $I=0$ , open circuit, e.g. measured with a …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The internal resistance of a cell depends on:$
(a) the area of the plates
(b) the distance between the plates
(c) the concentration of the electrolyte
(d) All of the above
›Reveal solutionSolution

A cell's internal resistance behaves like the resistance of the electrolyte column between its electrodes, so it depends on every geometric and chemical factor that affects that column.

Inside a cell, current flows through the electrolyte between the two electrodes. Treating the electrolyte as a conducting medium, its resistance follows the same rule as any conductor: $r=\rho\dfrac{l}{A}$ , where $\rho$ is the electrolyte's resistivity, $l$ the distance between the plates, and $A$ the area of the plates. So (i) a larger plate area $A$ gives a lower resistance (more parallel paths for current), (ii) a larger separation $l$ between plates gives a higher resistance (lon …

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CBSE 2026Set ANNUAL1 mark
$Q. If the current flowing in a copper wire be allowed to flow in another copper wire of same length but of doubled the radius then what will be the effect on the drift velocity of the electron?$
›Reveal solutionSolution

For the same current, $v_d \propto 1/A$ , and doubling the radius quadruples the cross-sectional area.

Current is related to drift velocity by $I = nAev_d$ , so for the same current $I$ (and the same material, hence the same $n$ ), $v_d = \dfrac{I}{nAe} \propto \dfrac{1}{A}$ . If the radius is doubled, the cross-sectional area $A=\pi r^2$ becomes 4 times larger. So the drift velocity becomes

…

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CBSE 2026Set ANNUAL1 mark
$Q. State Ohm's law in terms of current density, specific conductance and electric field intensity.$
›Reveal solutionSolution

Microscopic Ohm's law: current density J = σE (σ = conductivity, E = field).

The usual Ohm's law is V = IR. In microscopic (vector) form, it relates the current density J (current per unit cross-sectional area) to the electric field E inside the conductor through the material's specific conductance (conductivity) σ:

J = σ E.

…

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CBSE 2026Set SEM31 markMCQ
$Q. Which of the following statement(s) is/are true ? A potential difference of V is applied at the two ends of a conductor of length l and area of cross-section A. Statement I : When potential difference is doubled, current density also gets doubled. Statement II : When potential difference is doubled, drift velocity gets halved. Statement III : When area of cross-section is doubled, current density decreases.$
(a) I and II are true
(b) Only I is true
(c) Only III is true
(d) II and III are true
›Reveal solutionSolution

Doubling V doubles E, so J = σE and v_d = μE both double — Statement I true, Statement II (drift velocity halved) false. J = V/(ρl) is independent of area, so Statement III (J decreases when A doubles) is also false. Only I is true → option (b).

Statement I: J = σE and E = V/l, so doubling V doubles E and hence doubles the current density J. TRUE.

Statement II: drift velocity v_d = (eE/m)τ ∝ E ∝ V. Doubling V doubles v_d, it does not halve it. FALSE.

…

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CBSE 2025Set D1 markMCQ
$Q. The relation between drift velocity v of free electrons in conductor in electric conduction and potential difference V between ends of conductor is (A) proportional to V (B) inversely proportional to V (C) proportional to V^2 (D) inversely proportional to V^2$
›Reveal solutionSolution

Drift velocity is directly proportional to the potential difference V.

In a conductor of length L across which a potential difference V is applied, the electric field is E = V/L. Free electrons acquire a drift velocity

$v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$ …

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CBSE 2025Set D1 markMCQ
$Q. If the length of a conductor is doubled while keeping the potential difference across it constant, then the drift velocity of electron will (A) remain the same (B) be double (C) be halved (D) increase fourfold$
›Reveal solutionSolution

Doubling the length at constant V halves the drift velocity.

The drift velocity is

$v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$ …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The maximum current that can be drawn from a cell is when$
(a) R = 0
(b) r = 0
(c) R > r
(d) r > R
›Reveal solutionSolution

Current from a cell $I = E/(R+r)$ is largest when the total resistance $(R+r)$ is smallest, i.e. when the external resistance R = 0.

For a cell of emf E and internal resistance r connected to an external resistance R, the current is

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

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A thick wire is stretched so that its length becomes two times. What is the ratio of change in resistance of the wire to the initial resistance of the wire?$
(i) 2 : 1
(ii) 4 : 1
(iii) 3 : 1
(iv) 1 : 4
›Reveal solutionSolution

New resistance is 4 times the old, so the change is 3 times the original: ratio 3 : 1.

Resistance $R = \rho L/A$ . Stretching keeps the volume $AL$ constant, so if length doubles ( $L \to 2L$ ) the area halves ( $A \to A/2$ ). Then $R' = \rho (2L)/(A/2) = 4\rho L/A = 4R$ . The change in …

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Part (a)

(i) Three batteries in parallel

(ii) Six equal wires (tetrahedron)

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