Chapter 3 · Current Electricity

3.3 Cells, EMF, Internal Resistance and Combinations of Cells

A cell pushes charge round a circuit with its emf $\varepsilon$, but some of that push is used up inside the cell itself: the terminal voltage is $V=\varepsilon-Ir$. Cells can be joined in series for a bigger emf or in parallel for a smaller internal resistance.

Turn the key of a car on a cold morning with the headlights on, and the lights dim sharply while the starter motor turns. The battery has not suddenly lost its chemistry; it is being asked for $150$ A or more, and part of its push is used up driving that current through the battery itself. An old torch cell shows the same thing in miniature: a digital voltmeter reads almost $1.5$ V across it, yet it can no longer light the bulb. Both puzzles are explained by two numbers that describe any source of electrical energy — its emf and its internal resistance.

So far a "cell" has simply maintained a potential difference across a circuit. This section looks inside it. We define the electromotive force $\varepsilon$ and the internal resistance $r$, derive the current $I=\varepsilon/(R+r)$ and the terminal voltage $V=\varepsilon-Ir$, see what happens when a battery is being charged, and find the equivalent emf and internal resistance of cells joined in series and in parallel.

EMF and Internal Resistance

Inside a conductor, positive charge flows from high potential to low potential — "downhill". Round a complete circuit, though, charge must come back to where it started. Somewhere it has to be pushed "uphill", from the low-potential terminal to the high-potential terminal, against the electric field. In a cell this push comes from chemical reactions at the two electrodes; in a generator it comes from magnetic forces (Chapter 6); in a solar cell from absorbed light. These are not electrostatic forces, and they are what keeps a current going.

📘 Definition 3.8 — Electromotive force (emf) and internal resistance
The electromotive force (emf) $\varepsilon$ of a source is the work done by its non-electrostatic agency (chemical, magnetic, …) per unit positive charge carried from its negative terminal to its positive terminal inside it: $$\varepsilon=\frac{W}{q}.$$ Its unit is the volt. Despite the name, emf is not a force. It equals the potential difference between the terminals when no current is drawn (an open circuit). The electrolyte and electrodes of a real cell also resist the current through it; this resistance $r$ is the cell's internal resistance.

Model a real cell as an ideal emf $\varepsilon$ in series with a resistor $r$, and connect it to an external resistor $R$. In one trip round the circuit, a charge $q$ gains energy $q\varepsilon$ from the cell and loses energy $qIR$ in the external resistor and $qIr$ inside the cell. Energy conservation gives $\varepsilon=IR+Ir$. The potential difference between the terminals is the part of the emf that is not used up inside the cell.

📐 Result 3.5 — Current and terminal voltage of a cell

A cell of emf $\varepsilon$ and internal resistance $r$ connected to an external resistance $R$ drives the current $$I=\frac{\varepsilon}{R+r}.$$ The terminal voltage (the potential difference across the terminals, which also equals $IR$) is $$V=\varepsilon-Ir\qquad\text{(cell delivering current)}.$$ If an external source drives current into the positive terminal (the cell is being charged), $$V=\varepsilon+Ir\qquad\text{(cell being charged)}.$$ The graph of $V$ against $I$ for a discharging cell is a straight line with intercept $\varepsilon$ on the $V$-axis and slope $-r$; it meets the $I$-axis at the short-circuit current $\varepsilon/r$.

The term $Ir$ is sometimes called the "lost volts": it is the potential drop inside the cell, turned into heat there. When $I=0$, $V=\varepsilon$ — this is why a high-resistance voltmeter, which draws almost no current, reads very nearly the emf. When the terminals are joined by a thick wire ($R\approx0$), the current is limited only by $r$: $I_\text{sc}=\varepsilon/r$. A car battery has $r$ of a few hundredths of an ohm and can give hundreds of amperes; a dry cell has a few tenths of an ohm, and an old one far more.

✏️ Example 1 — A cell and a resistor
A cell of emf $1.5$ V and internal resistance $0.50\ \Omega$ is connected to a $2.5\ \Omega$ resistor. Find (a) the current, (b) the terminal voltage and the lost volts, (c) the current if the terminals were short-circuited.
(a) Current $I=\dfrac{\varepsilon}{R+r}=\dfrac{1.5}{2.5+0.50}=0.50$ A.
(b) Terminal voltage $V=\varepsilon-Ir=1.5-(0.50)(0.50)=1.25$ V; the lost volts are $Ir=0.25$ V. Check: $IR=(0.50)(2.5)=1.25$ V ✓.
(c) Short circuit $I_\text{sc}=\dfrac{\varepsilon}{r}=\dfrac{1.5}{0.50}=3.0$ A — all the energy now becomes heat inside the cell.
Check Interactive 1 opens on this circuit: $I=0.500$ A, $V=1.25$ V.
✏️ Example 2 — Finding emf and internal resistance from two readings
A voltmeter across a cell reads $1.40$ V when the cell delivers $0.20$ A, and $1.20$ V when it delivers $0.60$ A. Find the emf and internal resistance of the cell.
Equations $V=\varepsilon-Ir$: $\ 1.40=\varepsilon-0.20r$ and $1.20=\varepsilon-0.60r$.
Subtract $0.20=0.40r$, so $r=0.50\ \Omega$; then $\varepsilon=1.40+0.20(0.50)=1.50$ V.
Graph view The two points lie on the line $V=\varepsilon-Ir$; its slope is $-\dfrac{0.20\ \text{V}}{0.40\ \text{A}}=-0.50\ \Omega$ and its $V$-intercept is $1.50$ V. This is exactly how the emf and internal resistance are measured in the laboratory.
✏️ Example 3 — Charging a battery
A $12.0$ V battery of internal resistance $0.10\ \Omega$ is charged with a current of $5.0$ A. Find (a) the potential difference the charger must apply, (b) the power supplied by the charger, (c) the rate at which energy is stored chemically and the rate of heating inside the battery.
Concept The charger pushes current into the positive terminal, against the emf, so $V=\varepsilon+Ir$.
(a) Voltage $V=12.0+(5.0)(0.10)=12.5$ V — more than the emf.
(b), (c) Powers Supplied: $VI=(12.5)(5.0)=62.5$ W. Stored: $\varepsilon I=(12.0)(5.0)=60$ W. Heat: $I^2r=(5.0)^2(0.10)=2.5$ W. And $60+2.5=62.5$ ✓.
Lesson A terminal voltage larger than the emf means the cell is being charged (or, in a circuit with several cells, is being driven backwards by a stronger one).
Interactive 1 — A Cell with Internal Resistance

The dashed box is the real cell: an ideal emf $\varepsilon$ in series with its internal resistance $r$. Change the load $R$ (or press Play to sweep it) and watch the operating point slide down the line $V=\varepsilon-Ir$. Switch the graph to see the power delivered to the load as $R$ changes. The defaults are Example 1.

1.50.502.5
current $I$ (A)—
terminal voltage $V$ (V)—
lost volts $Ir$ (V)—
power to load $I^2R$ (W)—
efficiency $V/\varepsilon$—
✏️ Example 4 — Why the headlights dim
A car battery has emf $12.6$ V and internal resistance $0.020\ \Omega$. (a) Find its terminal voltage while the starter motor draws $150$ A. (b) The headlights are connected across the terminals. Treating them as fixed resistors, by what fraction does their power fall while the starter runs (compared with the battery at its full emf)?
(a) Terminal voltage $V=\varepsilon-Ir=12.6-(150)(0.020)=12.6-3.0=9.6$ V.
(b) Lamps For a fixed resistance $P=V^2/R$, so $\dfrac{P}{P_0}=\left(\dfrac{9.6}{12.6}\right)^2\approx0.58$: the lamps drop to about $58\%$ of their power — a clearly visible dimming.
Lesson The battery's emf has not changed; the large current makes the "lost volts" $Ir$ large. On a cold morning $r$ is larger still, which is why cars are harder to start in winter.
⚠️ Common mistakes with emf and terminal voltage

1. Thinking emf is a force. It is energy per unit charge, measured in volts.

2. Writing $V=\varepsilon$ when a current flows. Only an ideal cell ($r=0$) or an open circuit ($I=0$) has $V=\varepsilon$.

3. Always subtracting $Ir$. For a cell being charged — current entering at its positive terminal — the terminal voltage is $\varepsilon+Ir$.

4. Forgetting $r$ when finding the current. $I=\varepsilon/(R+r)$, not $\varepsilon/R$. The internal resistance is in series with everything else.

Cells in Series and in Parallel

A torch takes two or three cells end to end; a car battery is six $2.1$ V cells in series; a power bank contains cells side by side. To analyse a circuit with several cells, we replace the group by one equivalent cell with an emf $\varepsilon_\text{eq}$ and an internal resistance $r_\text{eq}$. (Resistors combine as in Class 10: in series $R=R_1+R_2+\cdots$; in parallel $\dfrac1R=\dfrac1{R_1}+\dfrac1{R_2}+\cdots$.)

Series. Join the positive terminal of cell 1 (emf $\varepsilon_1$, internal resistance $r_1$) to the negative terminal of cell 2. The same current $I$ flows through both. Going from the free negative terminal to the free positive terminal, the potential rises by $\varepsilon_1-Ir_1$ across the first cell and by $\varepsilon_2-Ir_2$ across the second, so the combination has terminal voltage $(\varepsilon_1+\varepsilon_2)-I(r_1+r_2)$. If cell 2 is turned round (its negative terminal joined to the positive terminal of cell 1), its emf opposes the first and enters with a minus sign.

📐 Result 3.6 — Cells in series

Cells in series are equivalent to one cell with $$\varepsilon_\text{eq}=\varepsilon_1\pm\varepsilon_2\pm\cdots,\qquad r_\text{eq}=r_1+r_2+\cdots,$$ with the plus sign for cells joined positive-to-negative (aiding) and the minus sign for a cell connected the other way round (opposing). For $n$ identical aiding cells: $\varepsilon_\text{eq}=n\varepsilon$, $r_\text{eq}=nr$, and with an external resistor $R$, $I=\dfrac{n\varepsilon}{R+nr}$.

Parallel. Join the two positive terminals together and the two negative terminals together, so both cells share the same terminal voltage $V$, and connect the pair to an external circuit that takes a current $I$. If the cells deliver currents $I_1$ and $I_2$, then $I=I_1+I_2$ (charge does not pile up at the junction), and $$V=\varepsilon_1-I_1r_1=\varepsilon_2-I_2r_2.$$ Solving for $I_1$ and $I_2$ and adding: $$I=\frac{\varepsilon_1-V}{r_1}+\frac{\varepsilon_2-V}{r_2}\quad\Longrightarrow\quad V=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}-I\,\frac{r_1r_2}{r_1+r_2}.$$ This has exactly the form $V=\varepsilon_\text{eq}-Ir_\text{eq}$ of a single cell.

📐 Result 3.7 — Cells in parallel

Two cells joined in parallel (like terminals together) are equivalent to one cell with $$\varepsilon_\text{eq}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2},\qquad \frac{1}{r_\text{eq}}=\frac{1}{r_1}+\frac{1}{r_2}.$$ Equivalently $\dfrac{\varepsilon_\text{eq}}{r_\text{eq}}=\dfrac{\varepsilon_1}{r_1}+\dfrac{\varepsilon_2}{r_2}$, which extends to any number of cells. For $m$ identical cells in parallel: $\varepsilon_\text{eq}=\varepsilon$, $r_\text{eq}=r/m$, and $I=\dfrac{\varepsilon}{R+r/m}$.

When is each arrangement better? Series adds up the emfs but also the internal resistances, so it helps when the external resistance is large compared with $r$. Parallel keeps the emf the same but divides the internal resistance, so it helps when the external resistance is small — and it lets each cell supply only part of the current, so the battery lasts longer. Interactive 2 compares every arrangement of a set of identical cells.

✏️ Example 5 — Four cells in a torch, one put in backwards
Four cells, each of emf $1.5$ V and internal resistance $0.25\ \Omega$, are joined in series to a $5.0\ \Omega$ lamp. (a) Find the current and the voltage across the lamp. (b) One cell is accidentally put in the wrong way round. Find the current and the lamp voltage now.
(a) All aiding $\varepsilon_\text{eq}=4(1.5)=6.0$ V, $r_\text{eq}=4(0.25)=1.0\ \Omega$; $I=\dfrac{6.0}{5.0+1.0}=1.0$ A; lamp voltage $IR=5.0$ V.
(b) One reversed The reversed cell cancels one of the others: $\varepsilon_\text{eq}=1.5+1.5+1.5-1.5=3.0$ V. Its internal resistance is still in the circuit: $r_\text{eq}=1.0\ \Omega$. So $I=\dfrac{3.0}{6.0}=0.50$ A and the lamp gets $2.5$ V — a quarter of its proper power.
Compare Interactive 2 uses older cells of $0.50\ \Omega$ each. With $N=4$ all in series and $R=5.0\ \Omega$ it reads $I=\dfrac{6.0}{5.0+2.0}\approx0.857$ A — doubling the internal resistance of each cell costs about $14\%$ of the current.
✏️ Example 6 — Two different cells in parallel
A cell of emf $2.0$ V and internal resistance $0.20\ \Omega$ is joined in parallel with a cell of emf $1.8$ V and internal resistance $0.30\ \Omega$, and the pair supplies a $1.88\ \Omega$ resistor. Find (a) the equivalent emf and internal resistance, (b) the current in the resistor, (c) the current in each cell. Comment on the result.
(a) Equivalent cell $\varepsilon_\text{eq}=\dfrac{(2.0)(0.30)+(1.8)(0.20)}{0.20+0.30}=\dfrac{0.96}{0.50}=1.92$ V; $r_\text{eq}=\dfrac{(0.20)(0.30)}{0.50}=0.12\ \Omega$.
(b) Load current $I=\dfrac{1.92}{1.88+0.12}=0.96$ A, and the terminal voltage is $V=(0.96)(1.88)\approx1.805$ V.
(c) Each cell $I_1=\dfrac{2.0-1.805}{0.20}\approx0.976$ A; $I_2=\dfrac{1.8-1.805}{0.30}\approx-0.016$ A. Check: $0.976-0.016=0.96$ A ✓.
Comment $I_2$ is negative: the weaker cell is not helping at all — the stronger cell is pushing a small current backwards through it, charging it. Mixing cells of different emf in parallel wastes energy, which is why batteries tell you not to mix old and new cells. (Section 3.4 solves such circuits systematically with Kirchhoff's rules.)
Interactive 2 — Arranging Identical Cells

Choose how many identical cells you have ($\varepsilon=1.5$ V, $r=0.50\ \Omega$ each) and how to arrange them: $n$ cells in series in each row, $m$ rows in parallel. The bars compare the current through the load for every possible arrangement; the highlighted one is drawn on the left. Slide $R$ from small to large and watch the best arrangement move from "all in parallel" to "all in series". Press Play to sweep $R$.

3.0
$\varepsilon_\text{eq}=n\varepsilon$ (V)—
$r_\text{eq}=nr/m$ (Ω)—
current in $R$ (A)—
current in each cell (A)—
best arrangement here—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — maximum power to a load, and mixed grouping of cells

Not required by the CBSE syllabus. Electrical power and the combination of cells in series and parallel are listed by JEE Main, NEET and JEE Advanced; maximum power transfer and mixed grouping are standard applications that no syllabus names.

Maximum power transfer. The power delivered to the load is $$P=I^2R=\frac{\varepsilon^2R}{(R+r)^2}.$$ It is zero for $R=0$ (short circuit) and tends to zero for very large $R$ (almost no current), so it has a maximum in between. Setting $\dfrac{dP}{dR}=\varepsilon^2\dfrac{(R+r)-2R}{(R+r)^3}=0$ gives $$R=r,\qquad P_\text{max}=\frac{\varepsilon^2}{4r}.$$ At this point the efficiency $V/\varepsilon=R/(R+r)$ is only $50\%$: half the energy heats the cell. (Interactive 1, "power to load" graph, shows the peak at $R=r$.) Power stations do not work at this point — they want high efficiency — but small signal circuits such as aerials and loudspeakers are "matched" this way.

Mixed grouping. $N=nm$ identical cells in $m$ parallel rows of $n$ series cells give $\varepsilon_\text{eq}=n\varepsilon$ and $r_\text{eq}=nr/m$, so $I=\dfrac{n\varepsilon}{R+nr/m}$. For fixed $N$ the current is greatest when $\dfrac{nr}{m}=R$ — again a matching condition, the same as in Interactive 2's best bar (when that arrangement exists). Neither result is required for the board examination.

Exercises

I
EMF, internal resistance and terminal voltage.
  • 1 A cell of emf $2.0$ V and internal resistance $0.40\ \Omega$ is connected to a $3.6\ \Omega$ resistor. Find the current and the terminal voltage.
  • 2 A cell reads $1.6$ V on open circuit and $1.4$ V when it delivers $0.50$ A. Find its internal resistance.
  • 3 A $1.5$ V cell gives a short-circuit current of $5.0$ A. Find its internal resistance.
  • 4 The $V$–$I$ graph of a battery meets the $V$-axis at $6.0$ V and has slope $-2.0\ \Omega$. Find its emf and internal resistance, and the current through a $4.0\ \Omega$ resistor connected to it.
  • 5 A $9.0$ V battery has internal resistance $1.0\ \Omega$. What load resistance makes its terminal voltage $8.0$ V?
  • 6 Distinguish between the emf of a cell and its terminal voltage. When are they equal? When is the terminal voltage greater than the emf?
  • 7 A $6.0$ V battery of internal resistance $0.50\ \Omega$ is charged with a current of $2.0$ A. Find the terminal voltage, the rate at which energy is stored and the rate of heating in the battery.
  • 8 A voltmeter of resistance $1000\ \Omega$ is connected across a cell of emf $1.50$ V and internal resistance $1.0\ \Omega$. What does it read? Why does a high-resistance voltmeter read nearly the emf?
1 $I=\dfrac{2.0}{3.6+0.40}=0.50$ A; $V=2.0-(0.50)(0.40)=1.8$ V.
2 $\varepsilon=1.6$ V; lost volts $0.2$ V at $0.50$ A, so $r=0.40\ \Omega$.
3 $r=\varepsilon/I_\text{sc}=1.5/5.0=0.30\ \Omega$.
4 $\varepsilon=6.0$ V, $r=2.0\ \Omega$; $I=\dfrac{6.0}{4.0+2.0}=1.0$ A.
5 Lost volts $1.0$ V $=I(1.0)$, so $I=1.0$ A and $R=V/I=8.0\ \Omega$.
6 The emf is the energy given per unit charge by the cell's chemistry; the terminal voltage is the energy per unit charge available to the external circuit, $\varepsilon\mp Ir$. They are equal when no current flows. The terminal voltage exceeds the emf when the cell is being charged (current entering at its positive terminal).
7 $V=6.0+(2.0)(0.50)=7.0$ V; stored at $\varepsilon I=12$ W; heat $I^2r=2.0$ W (total $VI=14$ W ✓).
8 $I=\dfrac{1.50}{1001}$, $V=IR_V=1.50\times\dfrac{1000}{1001}\approx1.4985$ V $\approx1.50$ V. The meter draws so little current that the lost volts $Ir$ are negligible.
II
Cells in series and in parallel.
  • 1 Three cells, each $1.5$ V and $0.20\ \Omega$, are joined in series to a $4.4\ \Omega$ resistor. Find the current.
  • 2 The same three cells are joined in parallel to the $4.4\ \Omega$ resistor. Find the current.
  • 3 Six cells, each $2.0$ V and $0.10\ \Omega$, are meant to be in series, but one is connected the wrong way round. The battery drives a $7.4\ \Omega$ resistor. Find the current.
  • 4 A cell of $1.5$ V, $0.30\ \Omega$ and a cell of $1.2$ V, $0.20\ \Omega$ are joined in parallel. Find the emf and internal resistance of the combination.
  • 5 With nothing else connected, find the current circulating between the two cells of part 4, and the potential difference across them.
  • 6 $n$ identical cells ($\varepsilon$, $r$) in series drive a resistor $R$. Show that $I=\dfrac{n\varepsilon}{R+nr}$. What does $I$ approach for large $n$ when $R\ll nr$?
  • 7 $m$ identical cells in parallel drive a resistor $R$. Show that $I=\dfrac{\varepsilon}{R+r/m}$. Why does this arrangement make the battery last longer?
  • 8 Two cells, each $1.5$ V and $1.0\ \Omega$, are available. Which arrangement gives the larger current through (a) $0.50\ \Omega$, (b) $10\ \Omega$? Give the currents.
1 $\varepsilon=4.5$ V, $r=0.60\ \Omega$; $I=\dfrac{4.5}{5.0}=0.90$ A.
2 $\varepsilon=1.5$ V, $r=0.20/3=0.067\ \Omega$; $I=\dfrac{1.5}{4.467}\approx0.34$ A.
3 The reversed cell cancels one aiding cell: $\varepsilon=(5-1)(2.0)=8.0$ V; $r=0.60\ \Omega$; $I=\dfrac{8.0}{8.0}=1.0$ A.
4 $\varepsilon_\text{eq}=\dfrac{(1.5)(0.20)+(1.2)(0.30)}{0.50}=1.32$ V; $r_\text{eq}=\dfrac{0.06}{0.50}=0.12\ \Omega$.
5 Round the loop the emfs oppose: $I=\dfrac{1.5-1.2}{0.30+0.20}=0.60$ A (charging the $1.2$ V cell). $V=1.5-(0.60)(0.30)=1.32$ V $=\varepsilon_\text{eq}$ ✓ — on open circuit the combination shows its equivalent emf, while energy is wasted inside.
6 $\varepsilon_\text{eq}=n\varepsilon$, $r_\text{eq}=nr$, so $I=\dfrac{n\varepsilon}{R+nr}$. If $R\ll nr$, $I\to\dfrac{\varepsilon}{r}$ — adding more cells does not help.
7 $\varepsilon_\text{eq}=\varepsilon$, $r_\text{eq}=r/m$, so $I=\dfrac{\varepsilon}{R+r/m}$. Each cell carries only $I/m$, so each discharges $m$ times more slowly.
8 (a) Series: $\dfrac{3.0}{0.50+2.0}=1.2$ A; parallel: $\dfrac{1.5}{0.50+0.50}=1.5$ A — parallel wins. (b) Series: $\dfrac{3.0}{12}=0.25$ A; parallel: $\dfrac{1.5}{10.5}\approx0.14$ A — series wins.
III
Real batteries.
  • 1 Two $1.5$ V cells, each of internal resistance $0.15\ \Omega$, are in series in a torch with a $2.7\ \Omega$ bulb. Find the current, the power in the bulb, the power wasted in the cells and the efficiency.
  • 2 A phone battery of emf $3.7$ V and internal resistance $0.15\ \Omega$ delivers $2.0$ A. Find its terminal voltage and the heat produced inside it per second.
  • 3 A cell's terminal voltage is $2.80$ V at $0.50$ A and $2.40$ V at $1.50$ A. Find its emf and internal resistance.
  • 4 An old dry cell reads $1.5$ V on a digital voltmeter but will not light a torch bulb. Explain.
  • 5 Show that the efficiency of a cell supplying a resistor $R$ is $\eta=\dfrac{R}{R+r}$. Find $\eta$ when $R=4r$.
  • 6 A car battery reads $12.6$ V on open circuit and $12.0$ V when connected to a $3.0\ \Omega$ load. Find the current and the internal resistance.
  • 7 A car battery has emf $12.6$ V and internal resistance $0.020\ \Omega$. What current would make its terminal voltage fall to $10.0$ V?
1 $\varepsilon=3.0$ V, $r=0.30\ \Omega$: $I=\dfrac{3.0}{3.0}=1.0$ A. Bulb: $I^2R=2.7$ W; cells: $I^2r=0.30$ W; efficiency $\dfrac{2.7}{3.0}=90\%$.
2 $V=3.7-(2.0)(0.15)=3.4$ V; heat $I^2r=(2.0)^2(0.15)=0.60$ W.
3 Slope: $r=\dfrac{2.80-2.40}{1.50-0.50}=0.40\ \Omega$; $\varepsilon=2.80+(0.50)(0.40)=3.00$ V.
4 Its emf is still about $1.5$ V (the voltmeter draws almost no current), but its internal resistance has become very large. With the bulb connected, the current is small and almost all the emf is lost inside the cell.
5 $\eta=\dfrac{\text{power to }R}{\text{power from emf}}=\dfrac{I^2R}{\varepsilon I}=\dfrac{IR}{I(R+r)}=\dfrac{R}{R+r}$. For $R=4r$: $\eta=0.80=80\%$.
6 $I=\dfrac{12.0}{3.0}=4.0$ A; lost volts $0.6$ V, so $r=\dfrac{0.6}{4.0}=0.15\ \Omega$ (an old battery).
7 $Ir=12.6-10.0=2.6$ V, so $I=\dfrac{2.6}{0.020}=130$ A.
IV
Harder problems.
  • 1 Cells of $6.0$ V, $1.0\ \Omega$ and $4.0$ V, $1.0\ \Omega$ are joined in series opposing each other, with a $3.0\ \Omega$ resistor. Find the current and the terminal voltage of each cell.
  • 2 A cell drives $0.50$ A through a $2.0\ \Omega$ resistor and $0.25$ A through a $5.0\ \Omega$ resistor. Find its emf and internal resistance.
  • 3 For what load resistance is the terminal voltage of a cell equal to $75\%$ of its emf?
  • 4 A load $R=r$ is supplied first by one cell ($\varepsilon$, $r$) and then by two such cells in parallel. Find the terminal voltage in each case as a fraction of $\varepsilon$.
  • 5 A battery of $10$ cells, each $2.0$ V and $0.050\ \Omega$, is charged from a $25$ V d.c. supply through a series resistor. What resistor gives a charging current of $5.0$ A?
  • 6 Cells of $1.5$ V, $0.50\ \Omega$ and $1.5$ V, $1.0\ \Omega$ are joined in parallel across a $2.0\ \Omega$ resistor. Find the current in the resistor and in each cell.
1 $\varepsilon_\text{eq}=6.0-4.0=2.0$ V, $r_\text{eq}=2.0\ \Omega$: $I=\dfrac{2.0}{5.0}=0.40$ A. The $6.0$ V cell delivers current: $V_1=6.0-0.40=5.6$ V. The $4.0$ V cell is being charged: $V_2=4.0+0.40=4.4$ V.
2 $\varepsilon=0.50(2.0+r)=0.25(5.0+r)\Rightarrow0.25r=0.25\Rightarrow r=1.0\ \Omega$, $\varepsilon=1.5$ V.
3 $V/\varepsilon=\dfrac{R}{R+r}=0.75\Rightarrow R=3r$.
4 One cell: $V=\varepsilon\dfrac{r}{2r}=\dfrac{\varepsilon}{2}$. Two in parallel ($r/2$): $V=\varepsilon\dfrac{r}{r+r/2}=\dfrac{2\varepsilon}{3}$.
5 The supply drives current against the battery's emf: $25=20+5.0\,(0.50+R)$, so $R=0.50\ \Omega$.
6 $\varepsilon_\text{eq}=1.5$ V (equal emfs), $r_\text{eq}=\dfrac{0.50}{1.5}=0.333\ \Omega$; $I=\dfrac{1.5}{2.333}\approx0.64$ A; $V\approx1.286$ V. $I_1=\dfrac{1.5-1.286}{0.50}\approx0.43$ A, $I_2=\dfrac{1.5-1.286}{1.0}\approx0.21$ A — the cell with the smaller $r$ does twice the work.
🎯 Exam corner — cells, emf and internal resistance BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Two cells of emfs $\varepsilon_1$, $\varepsilon_2$ and internal resistances $r_1$, $r_2$ are connected in parallel. Derive expressions for the emf and internal resistance of the equivalent cell. Hence find them for cells of $2.0$ V, $0.40\ \Omega$ and $1.5$ V, $0.60\ \Omega$.

Setup Positive terminals joined at $B_1$, negative at $B_2$; currents $I_1$, $I_2$ leave the cells and $I=I_1+I_2$ flows in the external circuit; $V=V(B_1)-V(B_2)$.
Equations $V=\varepsilon_1-I_1r_1=\varepsilon_2-I_2r_2\Rightarrow I=\dfrac{\varepsilon_1-V}{r_1}+\dfrac{\varepsilon_2-V}{r_2}$, so $V=\dfrac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}-I\dfrac{r_1r_2}{r_1+r_2}$. Comparing with $V=\varepsilon_\text{eq}-Ir_\text{eq}$ gives Result 3.7.
Numbers $\varepsilon_\text{eq}=\dfrac{(2.0)(0.60)+(1.5)(0.40)}{1.0}=1.8$ V; $r_\text{eq}=\dfrac{(0.40)(0.60)}{1.0}=0.24\ \Omega$. Exam tip: a quick check — $\varepsilon_\text{eq}$ must lie between $\varepsilon_1$ and $\varepsilon_2$, and $r_\text{eq}$ must be smaller than both.

E2. Board · 2 marks The graph of terminal voltage $V$ against current $I$ for a cell is a straight line meeting the $V$-axis at $1.5$ V and the $I$-axis at $3.0$ A. Find the emf and internal resistance of the cell, and the current it drives through a $2.5\ \Omega$ resistor.

Reading the graph $V$-intercept $=\varepsilon=1.5$ V. $I$-intercept $=\varepsilon/r=3.0$ A, so $r=0.50\ \Omega$ (the magnitude of the slope).
Current $I=\dfrac{1.5}{2.5+0.50}=0.50$ A. Exam tip: the $I$-intercept is the short-circuit current.

E3. Board · 1 mark Assertion (A): The potential difference across the terminals of a battery can be greater than its emf. Reason (R): When a battery is being charged, current flows through it from its positive terminal to its negative terminal. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept During charging the current enters at the positive terminal, flowing against the emf, so $V=\varepsilon+Ir\gt\varepsilon$. Both statements are true and R is the reason.
Answer (A). Exam tip: direction of current inside the cell decides the sign of $Ir$.

E4. JEE Main A battery is made of $n$ identical cells, each of emf $\varepsilon$ and internal resistance $r$, in series, but one cell has been connected the wrong way round. The battery drives a resistor $R$. The current is (A) $\dfrac{n\varepsilon}{R+nr}$ (B) $\dfrac{(n-1)\varepsilon}{R+(n-1)r}$ (C) $\dfrac{(n-2)\varepsilon}{R+nr}$ (D) $\dfrac{(n-2)\varepsilon}{R+(n-2)r}$.

emf $n-1$ cells aid and one opposes: $\varepsilon_\text{eq}=(n-1)\varepsilon-\varepsilon=(n-2)\varepsilon$.
Resistance All $n$ internal resistances are still in series: $r_\text{eq}=nr$.
Answer (C). Exam tip: (D) is the classic slip — reversing a cell cancels its emf (and one more), never its resistance.

E5. NEET A cell drives a current of $1.0$ A through a $4.0\ \Omega$ resistor and $0.50$ A through a $9.0\ \Omega$ resistor. Its internal resistance is (A) $0.50\ \Omega$ (B) $1.0\ \Omega$ (C) $2.0\ \Omega$ (D) $4.0\ \Omega$.

Equation $\varepsilon=1.0(4.0+r)=0.50(9.0+r)\Rightarrow4.0+r=4.5+0.5r\Rightarrow r=1.0\ \Omega$ (and $\varepsilon=5.0$ V).
Answer (B). Exam tip: the general result is $r=\dfrac{I_2R_2-I_1R_1}{I_1-I_2}$.

E6. JEE Advanced (Integer answer.) Twenty-four identical cells, each of emf $1.5$ V and internal resistance $1.0\ \Omega$, are arranged in $m$ identical parallel rows, each row containing $n$ cells in series ($nm=24$), to send the largest possible current through an external resistance of $1.5\ \Omega$. Find that current in amperes.

Concept Mixed grouping is beyond the board syllabus (see the 🚀 box): $I=\dfrac{n\varepsilon}{R+nr/m}$, greatest when $\dfrac{nr}{m}=R$.
Condition $\dfrac{n(1.0)}{m}=1.5$ with $nm=24$: $n^2=36$, so $n=6$, $m=4$.
Current $\varepsilon_\text{eq}=9.0$ V, $r_\text{eq}=1.5\ \Omega$: $I=\dfrac{9.0}{3.0}=3$ A. Answer: 3. Exam tip: if $nr/m=R$ has no whole-number solution, compare the two neighbouring arrangements directly.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How does the internal resistance of a cell change as it is used up and as it gets cold — and why do electric vehicles lose range in winter?

Why it matters. Battery engineers in Norway, China, India and Canada study cold-weather performance intensively: at low temperature the chemical reactions and the movement of ions in the electrolyte slow down, the internal resistance rises, more energy is lost as heat inside the battery and the voltage sags under load. Battery management systems in phones and cars estimate $r$ continuously to judge the "state of health" of a battery. You can measure the same quantity with a cell, two resistors and a multimeter.

What to do.

  1. Use an AA alkaline cell, a multimeter and a resistor of about $4.7\ \Omega$ (with a switch). Measure the open-circuit voltage $V_0$ (≈ emf), then the terminal voltage $V_L$ with the resistor connected (read it within a few seconds). Compute $I=V_L/R$ and $r=(V_0-V_L)/I$.
  2. Repeat for a fresh cell, a half-used cell and a nearly flat cell.
  3. Place a fresh cell in a sealed bag in a refrigerator for an hour, then measure again quickly. (Do not freeze, heat or short-circuit cells.)
  4. Tabulate $\varepsilon$ and $r$ for each case. Using Result 3.5, predict the current each cell would give a $2.0\ \Omega$ torch bulb.

Tools: Multimeter, $4.7\ \Omega$ resistor, switch, AA cells in different states, refrigerator, spreadsheet.

Hand in: Your table of $\varepsilon$ and $r$, the predicted bulb currents, and a short explanation of what changes inside the cell.

Going further. Read how a car's battery management system estimates internal resistance from small changes in current and voltage while driving. Which of the two quantities, $\varepsilon$ or $r$, is the better warning sign of an ageing battery?

Instructor note: sample reading $V_0=1.52$ V, $V_L=1.40$ V across $4.7\ \Omega$ gives $I\approx0.30$ A and $r\approx0.40\ \Omega$. Fresh alkaline AA cells typically show $0.1$–$0.4\ \Omega$; flat cells several ohms while $V_0$ stays near $1.2$–$1.4$ V; a refrigerated cell shows a noticeably larger $r$. The emf changes little; the rise in $r$ is the better indicator, which is why battery management systems track it.

🧪 Self-check quiz

0/5 correct

Q1. The terminal voltage of a cell equals its emf when

Q2. A cell of emf $1.5$ V and internal resistance $0.50\ \Omega$ is short-circuited. The current is

Q3. While a battery is being charged, its terminal voltage is

Q4. Two identical cells, each of emf $\varepsilon$ and internal resistance $r$, are joined in parallel. The combination has

Q5. On a graph of terminal voltage $V$ (vertical) against current $I$ for a discharging cell, the slope is

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