A Fresh Cell Marked Two Volts Never Actually Delivers Two Volts

Measure the voltage across a fresh cell with nothing connected to it and you get the figure printed on the side — say $2$ V. Connect it to a bulb and measure again, and the reading has fallen. The cell has not gone flat in a second.

The reason is that the cell itself has resistance. The electrolyte and electrodes inside it oppose the current just as the external circuit does, so when a current flows, part of the cell's energy is spent inside the cell and never reaches the circuit at all.

Which gives two different voltages that must be kept apart.

- The electromotive force, or emf, is the voltage the cell would supply if no current were drawn. It is the energy the cell gives to each unit of charge round the whole circuit, inside and outside
- The terminal potential difference is what the circuit actually receives — the energy per unit charge spent in the external part only

The difference between them is the energy wasted inside the cell, and it grows as the current grows. That is why a torch dims when its cell is old and why a car's headlights flicker while the engine turns over: the internal resistance has risen or the current has surged.

But before any of that can be quantified, four quantities have to be defined precisely, and this part of the chapter does so.

- Charge, the quantity of electricity
- Current, the charge passing per second
- Potential difference, the energy spent per unit charge
- Resistance, the opposition to the current

And then Ohm's law ties three of them together in the single most used equation in the subject, with a graph of $V$ against $I$ as its experimental verification and the slope of that graph giving the resistance.

The part closes with the two ways resistors combine — in series, where resistances add, and in parallel, where their reciprocals do — and with networks that mix the two.

This page covers the first part of the ICSE Class 10 Physics chapter on electricity and magnetism: the four basic quantities, Ohm's law, ohmic and non-ohmic conductors, emf and internal resistance, and resistors in series and parallel.

Four definitions and one relation between three of them.

**Charge ($Q$). The quantity of electricity. Its SI unit is the coulomb (C)**, and one coulomb is the charge carried by about $6.25\times 10^{18}$ electrons, since each electron carries $1.6\times 10^{-19}$ C.

**Current ($I$).** The rate of flow of charge:

$$I=\frac{Q}{t}$$

The SI unit is the ampere (A), and one ampere is one coulomb per second. Conventional current flows from the positive terminal to the negative terminal outside the cell, while the electrons drift the opposite way.

**Potential difference ($V$).** The work done in moving a unit positive charge from one point to the other:

$$V=\frac{W}{Q}$$

The SI unit is the volt (V), and one volt is one joule per coulomb.

**Resistance ($R$).** The opposition offered by a conductor to the flow of current through it:

$$R=\frac{V}{I}$$

The SI unit is the ohm, and one ohm is one volt per ampere.

Ohm's law. The current flowing through a conductor is directly proportional to the potential difference across its ends, provided the physical conditions — especially the temperature — remain unchanged.

$$V\propto I⟹V=IR$$

The condition matters as much as the statement. Ohm's law is not a universal truth about all conductors; it is a statement that holds for a metallic conductor at constant temperature, and a question asking for the law without its condition is incompletely answered.

How the law is verified experimentally. Connect the conductor in series with a cell, a rheostat, an ammeter and a key, with a voltmeter across the conductor.

- The rheostat is adjusted to give a series of different currents, and the ammeter and voltmeter readings are noted for each
- **The ratio $\frac{V}{I}$ is computed for each pair, and is found to be constant
- A graph of $V$ against $I$ is plotted, and comes out as a straight line passing through the origin
- The slope of that line gives the resistance**, since $\text{slope}=\frac{V}{I}=R$

Why the line must pass through the origin. With no potential difference there is no current, so the point $(0,0)$ lies on the graph. A straight line that misses the origin is not obeying Ohm's law, and that is one way a question tests the law rather than the arithmetic.

Worked example — reading a resistance off a graph. An experiment gives the following readings: $V=2$ V with $I=0.5$ A, $V=4$ V with $I=1.0$ A, and $V=6$ V with $I=1.5$ A. Does the conductor obey Ohm's law, and what is its resistance?

$$\frac{2}{0.5}=4,\frac{4}{1.0}=4,\frac{6}{1.5}=4$$

**The ratio is constant at $4$, so the conductor obeys Ohm's law and its resistance is $4$ ohm. The graph of $V$ against $I$ would be a straight line of slope $4$ through the origin.

The warning about which way round the axes are.** If $V$ is on the vertical axis the slope is $R$; **if $I$ is plotted vertically instead, the slope is $\frac{1}{R}$. Reading a slope without checking the axes is a standard trap, and labelling your own axes before reading anything off is the guard.

Worked example 2 — combining two definitions.** A charge of $30$ C flows through a resistor of $5$ ohm in $10$ s. Find the current and the potential difference across it.

$$I=\frac{Q}{t}=\frac{30}{10}=3\text{ A}$$
$$V=IR=3\times 5=15\text{ V}$$

And the energy supplied, using $V=\frac{W}{Q}$:

$$W=VQ=15\times 30=450\text{ J}$$

Three definitions used in one question, which is exactly how this material is examined.

**Ohmic conductors give a straight line through the origin on a $V$ against $I$ graph; non-ohmic conductors do not.

Ohmic conductors** obey Ohm's law, so $\frac{V}{I}$ is constant.

- All metallic conductors at constant temperature — copper, silver, aluminium
- Alloys such as nichrome, manganin and constantan, which are especially good because their resistance changes very little with temperature
- The graph is a straight line through the origin

Non-ohmic conductors do not obey it, and $\frac{V}{I}$ changes with the current.

- A filament lamp, because the filament gets hotter as the current rises and its resistance increases with temperature. The graph curves away from the straight line
- A semiconductor diode, which conducts easily one way and hardly at all the other. Its graph is not even symmetric about the origin
- A diode valve and a triode valve
- An electrolyte, in many cases

The filament lamp is the example to be able to explain. Its resistance when cold is much lower than when it is glowing, so at switch-on the current surges briefly and then settles. That is why a bulb usually fails at the moment it is switched on rather than in the middle of use.

The four factors affecting resistance.

- Length. Resistance is directly proportional to the length, so a longer wire resists more
- Area of cross-section. Resistance is inversely proportional to the area, so a thicker wire resists less
- Material. Different substances resist differently, even at the same length and thickness
- Temperature. For a metal the resistance increases with temperature; for a semiconductor it decreases

Combining the first three gives the formula:

$$R=\frac{\rho l}{A}$$

where $\rho$ is the specific resistance or resistivity of the material.

Specific resistance is the resistance of a conductor of unit length and unit area of cross-section, at a given temperature. Rearranging,

$$\rho =\frac{RA}{l}$$

so its unit is the ohm metre. It is a property of the material alone and does not depend on the shape of the piece — which is exactly what makes it useful for comparing substances.

Worked example 1 — resistance from resistivity. A copper wire $10$ m long has a cross-sectional area of $1\times 10^{-6}$ m${}^2$. Taking the specific resistance of copper as $1.6\times 10^{-8}$ ohm metre, find its resistance.

$$R=\frac{\rho l}{A}=\frac{1.6\times 10^{-8}\times 10}{1\times 10^{-6}}=0.16\text{ ohm}$$

A very small resistance for a long wire, which is why copper is used for connecting wires and for transmission.

Worked example 2 — finding the specific resistance. A wire $2$ m long with an area of $0.5\times 10^{-6}$ m${}^2$ has a resistance of $4$ ohm. Find its specific resistance.

$$\rho =\frac{RA}{l}=\frac{4\times 0.5\times 10^{-6}}{2}=1\times 10^{-6}\text{ ohm metre}$$

That is far larger than copper's, so the wire is an alloy rather than a pure good conductor.

Worked example 3 — stretching a wire. A wire of resistance $4$ ohm is stretched until its length is doubled. Find its new resistance.

The volume of material is unchanged, so doubling the length halves the area:

$$R_{\text{new}}\propto \frac{2l}{A/2}=4\times \frac{l}{A}⟹R_{\text{new}}=4\times 4=16\text{ ohm}$$

Four times, not twice — because the length doubling and the area halving each double the resistance, and the two effects multiply.

Why alloys are used for heating elements. Nichrome has a high specific resistance, so a short coil gives a large resistance and a great deal of heat in a small space, and it does not oxidise readily when red hot. Manganin and constantan have a very small change of resistance with temperature, which is why they are used for standard resistance coils.

Superconductors. Certain substances lose their electrical resistance completely below a particular low temperature, called the transition temperature, and are then said to be superconducting.

- Their resistance becomes practically zero, so a current once started in a closed loop continues without any source to drive it
- Mercury and lead become superconducting when cooled to within a few kelvin of absolute zero
- The practical difficulty is the cooling, which is why superconductors are not used in ordinary wiring
- They are used in very strong electromagnets, including those in medical imaging machines

One boundary case that ties this section together. A superconductor has $R=0$, so Ohm's law in the form $V=IR$ gives $V=0$ however large the current. A current can flow with no potential difference across it at all — which is impossible for any ordinary conductor and is the clearest sign that superconductivity is a genuinely different state of matter.

The emf is what the cell supplies to the whole circuit; the terminal voltage is what the external circuit receives; the difference is what is spent inside the cell.

**Electromotive force ($\epsilon$). The emf of a cell is the energy spent by the cell in carrying unit positive charge round the complete circuit, inside the cell as well as outside it. Its unit is the volt.

And its practical meaning: the emf is the potential difference between the terminals of the cell when no current is being drawn — that is, in an open circuit. So an emf is measured with a high-resistance voltmeter across a cell that is doing nothing.

Terminal potential difference ($V$). The energy spent in carrying unit positive charge through the external circuit only. This is the voltage actually available to the bulb or the resistor, and it is what a voltmeter across the cell reads while current flows.

Internal resistance ($r$). The resistance offered by the electrolyte and the electrodes of the cell itself to the flow of current through it.

The relation between the three.** Of the total energy $\epsilon$ given to each coulomb, part is spent in the external circuit and part inside the cell:

$$\epsilon =V+Ir$$

so

$$V=\epsilon -Ir$$

Which shows three things at once.

- The terminal voltage is always less than the emf when the cell is supplying current
- The drop is larger for a larger current, which is why a torch dims more with a brighter bulb
- **When $I=0$ the terminal voltage equals the emf, which is why an open-circuit reading gives the emf

And the current in a simple circuit**, obtained by applying Ohm's law to the whole circuit:

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

The total resistance is the external resistance plus the internal one, because the current has to pass through both in series.

Worked example 1 — a cell driving a resistor. A cell of emf $2$ V and internal resistance $0.5$ ohm is connected to a resistor of $3.5$ ohm. Find the current and the terminal potential difference.

$$I=\frac{\epsilon }{R+r}=\frac{2}{3.5+0.5}=\frac{2}{4}=0.5\text{ A}$$
$$V=IR=0.5\times 3.5=1.75\text{ V}$$

Check with the other relation:

$$V=\epsilon -Ir=2-(0.5\times 0.5)=2-0.25=1.75\text{ V}$$

The two routes agree, and the $0.25$ V that went missing was spent inside the cell.

Worked example 2 — a smaller external resistance. The same cell is now connected to a resistor of $0.5$ ohm. Find the current and the terminal voltage.

$$I=\frac{2}{0.5+0.5}=2\text{ A}$$
$$V=2\times 0.5=1\text{ V}$$

The terminal voltage has fallen to half the emf, because the current is four times as large and so the internal drop $Ir=2\times 0.5=1$ V is four times as large. This is why a cell cannot deliver its rated voltage into a low-resistance load, and why short-circuiting a cell wastes almost all its energy inside itself.

Worked example 3 — finding the internal resistance. A cell of emf $1.5$ V gives a current of $0.3$ A through a $4$ ohm resistor. Find its internal resistance and the terminal voltage.

$$r=\frac{\epsilon }{I}-R=\frac{1.5}{0.3}-4=5-4=1\text{ ohm}$$
$$V=IR=0.3\times 4=1.2\text{ V}$$

Check: $\epsilon -Ir=1.5-0.3=1.2$ V. Correct.

Why the internal resistance of a cell rises as it ages. The electrolyte becomes depleted and products of the chemical reaction build up on the electrodes. So an old cell has a nearly unchanged emf but a much larger internal resistance — which is why it still reads close to its rated voltage on an open-circuit test and yet cannot light a bulb. That single fact explains the most confusing thing about flat batteries, and it is a favourite examination question.

One thing to be precise about. The emf belongs to the cell and does not depend on the external circuit at all. The terminal voltage depends on the current, and therefore on what the cell is connected to. So a question asking for the emf has one answer whatever the load, while a question asking for the terminal voltage needs the circuit.