Chapter 3: Electricity: Circuits and their Components

Step 1 – Recall Nihal’s original two lists

The textbook shows Nihal writing two columns:

Our task is to enrich these lists with more everyday examples.

Step 2 – Add more uses of electricity

Step 3 – Suggest other logical ways to group the uses

1. According to the main effect produced
   
   • Heating effect  –  electric iron, geyser, toaster, soldering iron
   • Lighting effect  –  bulbs, tube-lights, decorative LEDs, search-lights
   • Magnetic effect  –  loudspeakers, electric bell, cranes in junk-yards
   • Chemical effect  –  electroplating, charging batteries, metal refining
2. According to the sector of use
   
   • Domestic
   • Commercial (shops, malls, offices, hotels)
   • Industrial (factories, refineries, mines)
   • Agricultural (water-pumps, threshers, milk-chillers)
   • Transportation (railways, trams, electric cars, airports)
   • Health & research (hospitals, laboratories, observatories)
3. According to the source and scale
   
   • Battery-powered appliances (torch, remote, wrist-watch)
   • Low-voltage d.c. gadgets (mobile phone, Bluetooth speaker)
   • Mains-powered a.c. equipment (refrigerator, washing-machine)
   • High-voltage grid applications (inter-city railways, smelting furnaces)

These alternative classifications reveal that the same device can appear in different groups if we change the grouping principle. For example, an electric iron belongs to the “heating effect” group, to the “domestic sector” group, and to the “mains-powered appliances” group simultaneously.

Step 1 – Recall the condition for a lamp to glow
Current can pass through the filament of a bulb only when the electric circuit that joins the two terminals of the cell to the two terminals of the bulb is closed (complete). If the circuit is open (broken) no current flows and the filament does not heat up, so the lamp stays dark.

Step 2 – What the torch-switch does internally
Inside a torch there is a sliding (or push-button) metal strip that acts as a switch. The strip can occupy two stable positions:

• ON position: the strip touches both contact points, so the path from the positive terminal of the cell, through the filament, and back to the negative terminal is continuous. The circuit is therefore closed.
• OFF position: the strip moves away from (at least) one of the contact points. A gap appears in the conducting path, breaking the circuit.

Step 3 – Consequences of the two positions

• In the ON position the circuit is complete; electric current flows, the filament becomes white-hot, and the torch lamp glows.
• In the OFF position the circuit is open; no current flows, so the filament remains cool and the lamp does not glow.

Step 4 – Answering the question
Thus the lamp glows only in that particular position of its switch because only in that position does the switch connect all the conductors and make a closed circuit for current to flow.

Step 1 — Why more than one cell?
One dry cell supplies about $$1.5\text{ V}$$, but a torch bulb needs roughly $$3\text{ V}$$ or more to glow brightly. Two (or three) cells therefore have to be used.

Step 2 — How to obtain a higher voltage?
Voltages add when cells are connected in series. If each cell gives an emf $$V_c$$, then with n cells

For two 1.5 V cells: $$V_\text{total}=1.5+1.5=3.0\text{ V}$$.

Step 3 — Series connection inside a torch
Series connection is obtained only when the positive terminal ($$+$$) of one cell touches the negative terminal ($$-$$) of the next. Inside the torch

• a spring or metal plate presses against the free $$-$$ end of the bottom cell,
• a metal strip joins the $$-$$ of the first cell to the $$+$$ of the second,
• the free $$+$$ end of the top cell connects to the switch and bulb.

Thus the cells lie head-to-tail (+ to − to + to −), not randomly.

Conclusion
Yes, the cells in a torch must be placed in a definite order—positive of one touching negative of the next—so that their voltages add up and the bulb receives the required higher voltage.

Step 1 - Recall why several dry cells are used
To make a torch, radio, remote-control, etc., work for a long time or at a higher voltage, we do not rely on a single 1.5 V dry cell. Instead we put two, three or even more cells in the battery holder. To add their voltages, they have to be connected in series.

Step 2 - Meaning of series connection
When cells are in series, the positive terminal (marked ‘+’) of one cell must touch the negative terminal (marked ‘–’) of the next cell, so that current has only one path through all the cells one after another.

Step 3 - What happens when cells lie side by side
Many gadget designers arrange the cells side by side instead of one behind another because it saves space. A metal strip (or a spring) is fixed in such a way that it joins:

• $$+$$ of the first cell to $$-$$ of the second,
• $$+$$ of the second to $$-$$ of the third, and so on.

Thus, although the cells look parallel, the internal metal strips still link them head-to-tail, exactly the same as if they were standing in a straight line.

Step 4 - Conclusion
Therefore, even in a side-by-side holder, the cells are connected positive of one to negative of the next; all the cells form a single series path for the current.

Step 1 — The parts of a torch

• Two or three dry cells (source of potential difference).
• A miniature bulb with a filament.
• A switch that can slide or push to two positions.
• Metallic strips/wires joining these parts to make one complete loop.

Step 2 — Closed vs. open circuit

• Closed circuit: conducting path is continuous, so current $$I$$ flows and the bulb glows.
• Open circuit: the path is broken, so $$I = 0$$ and the bulb remains dark.

Step 3 — Action of the switch when turned ON

The switch contains a movable metallic strip. Pressing it pushes the strip so that it touches two fixed metal contacts—one coming from the positive terminal of the cells and the other leading to the bulb. This contact completes the conducting loop:

negative terminal → metallic body → filament of the bulb → switch strip (now touching) → positive terminal.

Because the loop is closed, electrons can move, so $$I \neq 0$$ and the filament heats up and emits light.

Step 4 — Action of the switch when turned OFF

Sliding the switch back removes the strip from one of the contacts, leaving a small air gap. Air has an extremely large resistance, $$R_{\text{air}} \gtrapprox 10^{12}\,\Omega$$. By Ohm’s law $$I = \dfrac{V}{R_{\text{air}}} \approx 0$$, so practically no current flows and the bulb goes out.

Step 5 — Summary

• ON: strip bridges the gap → circuit closed → current flows → torch shines.
• OFF: gap present → circuit open → no current → torch dark.

Diagram to draw (description): Draw a cell–bulb circuit with a break labelled “switch”. Show two insets: (a) gap open, bulb dark; (b) strip touching both contacts, bulb glowing.

Concept recap
The part of a circuit that actually supplies energy to push charges round the circuit is the cell or battery. A switch cannot supply current; it only opens or closes the conducting path.

How a switch works

• ON position: the metal terminals touch; the circuit is complete → current flows.
• OFF position: the terminals separate leaving an air gap; the circuit is broken → current stops.
• This property lets us start or stop the flow of electricity whenever we want.

Testing each statement

Conclusion
Only statement (A) is incorrect.

Step 1 : Recognise what makes the lamp glow
In an electric circuit the lamp glows only when an unbroken conducting path exists from one terminal of the cell, through the lamp-filament, and back to the other terminal. In symbols we say that an electric current $$I$$ will flow only if the circuit is closed by a conductor.

Step 2 : Recall the property of the four test materials shown in Fig. 3.16

Step 3 : Explain why the cardboard fails to complete the circuit
Cardboard is mainly made of cellulose fibres. These fibres hold their electrons tightly and therefore do not allow free movement of charge carriers. In other words, the electrical resistance $$R$$ of cardboard is extremely high (millions of ohms). Ohm’s law $$I = \dfrac{V}{R}$$ then gives a current so small that the filament of the bulb does not get hot enough to glow.

Conclusion
When the gap AB is bridged with cardboard (or any similar insulating material such as plastic or rubber) the circuit remains open and the lamp stays dark.

Step 1 : Observe how the two lamps are wired in Fig. 3.17.

• The free end of each filament is joined to the same junction J₁, which goes to the positive terminal of the cell.
• The other end of each filament meets at another common junction J₂, which returns to the negative terminal.

Because both ends of every lamp are connected to the same pair of junctions, the lamps are in a parallel circuit.

Step 2 : What happens when the filament of one lamp breaks?

• If the filament of lamp 1 breaks, that branch becomes an open circuit; its resistance shoots up to an extremely large value (practically $$\infty$$), so no current flows through that branch.
• The second branch, which contains lamp 2, is still a complete path between J₁ and J₂. Its filament remains intact, so current can continue to flow through it.

Step 3 : Consequence for the second lamp

Because current can still pass through the unbroken branch, lamp 2 continues to receive energy from the cell and therefore keeps glowing.

Conclusion

In the parallel connection shown in Fig. 3.17, breaking the filament of one lamp does not interrupt the circuit of the other. Hence the other lamp will still glow.

Given: A simple electric circuit is to be made with a cell (battery), a lamp (bulb) and connecting wires. The metal conducting core of the wires has been left covered with its plastic insulator instead of being exposed at the ends.

To decide: Will the lamp glow or not?

Key facts needed

• For an electric current to flow, every component in the circuit must form a continuous conducting path.
• The metallic core of a connecting wire is a conductor (usually copper or aluminium). The coloured plastic coating is an insulator.
• Insulators do not allow electric charge to pass through them easily.
• A lamp glows only if an electric current passes through its filament.

Step-by-step reasoning

1. At every joint in the circuit (for example, between the cell terminal and the wire, and between the wire and the lamp terminal) the metal of the wire has to touch the metal of the device. This completes the conducting path.

2. If the plastic insulation is not removed, the plastic sits between the wire and the device terminal. Plastic is an insulator, so it offers extremely high resistance (effectively an open circuit):
   $$R_\text{plastic}\;{\gg}\;10^6\,\Omega$$

3. Because of that very large resistance, the current in the circuit becomes practically zero:
   $$I = \dfrac{V}{R_\text{total}} \approx 0\;(\text{ampere})$$

4. With no current through the lamp’s filament, the filament does not get hot and therefore the lamp cannot glow.

Conclusion: Since the insulating plastic prevents the formation of a closed conducting path, the circuit remains open. Hence the lamp will not glow.

Step 1 – List the components in a torch

• One dry cell (the electric source)
• One switch (to open or close the circuit)
• One electric bulb (the load)
• Connecting conducting wires

Step 2 – Recall the standard symbols

• Cell : two parallel lines, the longer line marked $$+$$ (positive), the shorter marked $$-$$ (negative)
• Switch : a small break in the line with a movable lever; drawn open when the torch is OFF and closed when it is ON
• Bulb : a circle containing a cross (✖) or filament symbol
• Wires : straight lines joining the symbols

Step 3 – Arrange the symbols in series

1. Start at the positive terminal of the cell (long line).
2. Draw a straight wire leading to the switch symbol.
3. From the other end of the switch draw a wire to the bulb symbol.
4. From the opposite side of the bulb draw a wire back to the negative terminal (short line) of the cell, completing the loop.

Step 4 – Label the terminals

Write $$+$$ near the longer plate of the cell and $$-$$ near the shorter plate so that current direction can be traced.

What the finished diagram shows: a single closed loop consisting of cell → switch → bulb → back to cell. When the switch is closed, the bulb glows; when it is open, the circuit breaks and the bulb goes off.

Given information

Vidyut has connected a cell (battery), a switch, conducting wires and a lamp exactly as in Fig. 3.19 of the NCERT Class 7 textbook. After the switch is turned ON, the lamp still does not glow.

We must list every possible reason, then describe—step by step—how to find the real cause.

Recall of basic ideas

• An electric lamp glows only when an unbroken conducting path (closed circuit) exists and when the cell can provide a potential difference large enough to push current through the filament.
• Current $$I$$ in a simple circuit is given by Ohm’s law $$I = \dfrac{V}{R}$$, where $$V$$ is the cell voltage and $$R$$ is the total resistance. If $$V = 0$$ (a dead cell) or if $$R \to \infty$$ (open circuit, broken filament, loose connection), then $$I = 0$$ and the lamp remains dark.

Part A Possible reasons for the lamp not glowing

1. Dead cell / discharged battery. The cell may be exhausted, giving almost $$V = 0\;\text{V}$$.
2. Cell connected the wrong way or in reverse. A single bulb normally still glows, but if two or more cells are incorrectly arranged, their voltages can cancel.
3. Insulating cover on the cell terminals. Sometimes a thin plastic disc is left on a new dry cell; it prevents contact.
4. Switch left open (OFF) or internally broken.
5. Loose connection at any screw, crocodile clip or binding post.
6. Broken connecting wire. The copper core may have snapped inside the plastic cover.
7. Corrosion on terminals or clips. An oxide layer behaves like an insulator $$R \approx \infty$$.
8. Bulb fused (filament broken). An open filament also means $$R \approx \infty$$.
9. Bulb holder faulty. The central foot contact or the side contact may not touch the bulb base firmly.
10. Excessive resistance in the circuit (too thin or too long a wire). Current $$I$$ becomes too small to heat the filament to incandescence.
11. Short circuit bypassing the lamp. If the two wires touching the lamp terminals accidentally touch each other, current misses the lamp completely.

Part B Step-by-step test plan to locate the fault

1. Visual inspection
   • Look for any obvious breaks, loose ends, or bare wires touching.
   • Check that the switch lever actually touches the other terminal when in the ON position.
   • Examine the bulb to see whether the thin tungsten filament inside is intact.
2. Check the cell
   • Touch its two terminals momentarily with the tongue (not recommended in school) or better
   • Connect a voltmeter; if $$V \lt 1\;\text{V}$$ for a 1.5 V cell, the cell is probably dead.
   • Alternatively, connect the cell directly to a known-good torch bulb; if it does not glow at all, replace or recharge the cell.
3. Test the bulb separately
   • Remove the bulb and connect it straight across the same cell with two short wires.
   • If it fails to light, the filament is broken or the bulb rating is too high for the cell voltage.
4. Check continuity of wires and switch
   • Use a continuity tester or multimeter set to the lowest resistance range.
   • One by one, test each wire, then the switch when ON. The meter should read almost $$0\;\Omega$$. A reading $$R \to \infty$$ shows an open circuit.
5. Clean and tighten all connections
   • Scrape oxide layers gently with sand-paper.
   • Tighten screws/clips so that shiny metal touches shiny metal. Good contact keeps $$R$$ small.
6. Reassemble and re-test the whole circuit
   • Close the switch again. If the lamp now glows, the fault has been removed.
   • If not, repeat steps 2 to 5 until the defective part is finally identified and replaced.

Summary

The lamp fails to glow whenever no current or an insufficient current flows. This may be due to a defect in the cell, bulb, switch, wires, contacts, or the way parts are connected. A systematic isolation test—checking one component at a time—quickly reveals the exact reason.

Observing the four circuits shown in Fig. 3.20

1. Case (i) – both leads coming from the same terminal of the cell.
   Current has no return path to the other terminal, therefore the circuit is open and no current can flow.
2. Case (ii) – the connection touches only the glass/plastic part of the bulb/LED at one side.
   That point is an insulator; hence the circuit again remains open and no current flows.
3. Case (iii) – both leads are joined to the same metallic terminal of the lamp/LED (either both on the cap or both on the side).
   The two terminals of the source are not linked through the filament/LED chip, so the circuit is incomplete.
4. Case (iv) – one lead goes to the cap (or longer anode leg of the LED) and the other to the side terminal (or shorter cathode leg) and each lead reaches a different cell terminal.
   There is now a continuous conducting path from one terminal of the cell, through the switch, through the filament/LED, and back to the other terminal. Current flows and the lamp/LED glows.

Thus the lamp/LED fails to glow in all the arrangements except the fourth one.

Objective: To find out which end of an unmarked battery is positive ($$+$$) and which is negative ($$-$$).

Scientific fact used: A light-emitting diode (LED) conducts electric current in only one direction. The LED glows when its anode (the longer lead) is connected to the higher potential and its cathode (the shorter lead, often next to a flat edge on the plastic body) is connected to the lower potential.

If $$V_\mathrm{anode}$$ and $$V_\mathrm{cathode}$$ denote the potentials at the two leads, the LED behaves as follows:

Here $$V_f$$ is the forward-bias voltage of the LED. For a common red LED, $$V_f \approx 1.8\text{–}2.0\,\mathrm{V}$$ (green, blue and white LEDs need even more), so a 3 V battery (or two 1.5 V cells in series) is needed to light it.

Apparatus required

• One LED (a red LED, which needs about $$2\,\mathrm{V}$$ to glow).
• One $$220\,\Omega$$–$$470\,\Omega$$ resistor used as a current limiter.
• Two short pieces of insulated copper wire with bare ends.
• The unmarked battery whose terminals are to be identified (use a 3 V battery, or two 1.5 V cells in series, so the LED's forward voltage is comfortably exceeded).

Step-by-step procedure

1. Identify the longer lead of the LED; this is its anode. The shorter lead is the cathode.
2. Twist (or solder) the resistor in series with either lead of the LED. The resistor keeps the current small enough that the LED is not damaged.
3. Connect one piece of wire to the free end of the resistor and another piece of wire to the remaining lead of the LED.
4. Touch the two free wire ends to the two unknown terminals of the battery. Note which terminal is touched by the wire coming from the anode (long-lead) side.
5. Observation
   • If the LED glows, current is flowing from anode to cathode through the LED. Therefore the terminal touched by the anode-side wire is the positive (+) terminal, and the other terminal is the negative (–) terminal.
   • If the LED does not glow, swap the two wires at the battery. The LED should now light up. The terminal that is now in contact with the anode-side wire is the positive (+) terminal, and the other one is the negative (–) terminal.

Result: The terminal that allows the LED to glow when joined to its longer lead (anode) is the positive terminal of the battery; the other terminal is the negative one.

Note: Any one-way (polarity-sensitive) device—a silicon diode in series with a bulb, an analogue galvanometer, or a digital multimeter set on the d.c. voltage range—could be used instead of the LED, but the LED is the quickest and safest option for a school laboratory.

Step 1 – Requirement of the LED
An ordinary LED needs about $$V_{\min}\approx2\,\text{V}$$. Two dry cells in series give $$1.5\,\text{V}+1.5\,\text{V}=3\,\text{V}$$, which is sufficient.

Step 2 – What is wrong in Fig. 3.21?
In the figure both cells are joined positive-to-positive and negative-to-negative. That is a parallel connection, so the available potential difference is only

$$V=V_1=1.5\,\text{V}

Hence the LED cannot glow.

Step 3 – Correct connection of the cells

• Join the positive terminal of the first cell to the negative terminal of the second cell (series connection).
• Connect the free positive terminal to the longer lead (anode) of the LED.
• Take a wire from the shorter lead (cathode) of the LED to one end of the switch.
• Join the other end of the switch to the free negative terminal of the second cell.

Now the total voltage is $$V=V_1+V_2=3\,\text{V}\ge V_{\min}$$, so the LED will glow.

Diagram to draw: Two cells end-to-end (series), their touching terminals linked by a short wire. From the free positive terminal run a wire to the LED’s anode. The LED’s cathode goes to a switch, and the switch returns to the free negative terminal of the second cell, completing the circuit.

Understanding the situation

In our homes and in the locality almost every activity that feels “normal” relies, directly or indirectly, on the continuous flow of electric current. When the supply fails for two complete days, every device that needs electricity stops working. Below is a careful, point-by-point discussion of each such activity. The aim is to make you consciously connect the action with the electrical component that makes it possible.

1. Lighting and visibility
   • No tube-lights, LED bulbs, CFLs or night-lamps will glow.
   • You would have to depend on candles, kerosene lamps or battery torches for illumination at night.
2. Cooling and ventilation
   • Ceiling fans, table fans and exhaust fans stop.
   • Air-conditioners and coolers cannot run, leading to an uncomfortably hot and stagnant room, especially in summer.
3. Heating appliances
   • Electric geysers, immersion rods and room-heaters remain off, so you cannot get instant hot water for bathing or washing dishes, nor can you warm the room in winter.
4. Food-related work
   • Refrigerators turn off — ice melts and food may spoil.
   • Electric mixers, grinders, microwave ovens, induction cook-tops, toasters and rice-cookers all become useless.
   • The kitchen chimney and exhaust fan also stop.
5. Cleaning and washing
   • Electric water-pump (if your building uses one) will not lift water to overhead tanks; tap water supply inside flats may stop.
   • Washing machines, dryers and vacuum cleaners do not work.
   • Electric irons cannot be used to press clothes.
6. Entertainment and information
   • Televisions, radios (unless battery-operated), set-top boxes, DVD players and music systems remain silent.
   • Computers, laptops (after the battery discharges) and gaming consoles cannot be used.
   • Internet routers shut down, so Wi-Fi is lost even if your phone still has some battery.
7. Communication
   • Mobile-phone chargers, cordless-phone base units and land-line modems stop; once the battery of a phone is drained you cannot call or message.
8. Study and office work
   • No electric light to read under after sunset.
   • Printers, scanners and desktop PCs cannot be switched on.
9. Water and sanitation
   • Electric water purifiers (RO systems) and UV filters are off.
   • Motor-driven flush systems in some modern toilets stop working.
10. Transportation aids
    • Lifts (elevators) in multi-storey buildings and escalators in malls or metro stations do not run.
    • Electric vehicle-charging points cannot charge e-bikes or e-scooters.
11. Safety and security
    • CCTV cameras, electronic doorbells, automatic gates and alarm systems are disabled unless they have independent battery backups.
    • Street-lights in many areas go dark, reducing night-time safety.
12. Medical and health-care devices at home
    • Nebulisers, digital thermometers (rechargeable types), electric heating pads or blankets and health-monitoring devices (BP machine, glucometer chargers) cannot be used once their internal batteries discharge.
13. Miscellaneous small gadgets
    • Hair-dryers, electric shavers or trimmers, electric toothbrush chargers, mosquito repellant vapourisers and aroma diffusers all stop.

Conclusion

A two-day power cut affects practically every sphere of daily life — comfort, food, hygiene, communication, entertainment, safety and even health. The situation highlights our heavy dependence on electricity and the importance of having backup solutions such as batteries, inverters or generators for essential needs.

Required components

• One small solar panel (Fig. 3.22a)
• Two or three short pieces of insulated connecting wire with crocodile clips
• One simple key / switch
• One toy fan fitted with a d.c. motor (Fig. 3.22b)

1. Identify the terminals

1. The solar panel has two fixed leads marked + (red) and − (black).
2. The motor of the toy fan has two solder-lugs that act as its + and − terminals.

2. Plan the closed path for current

Solar + → Key → Motor + → Motor coil → Motor − → Solar −.

3. Actual connections

1. Clip one wire between the + terminal of the solar panel and one screw of the key.
2. Clip a second wire from the other screw of the key to the + terminal of the motor.
3. Clip a third wire from the − terminal of the motor back to the − terminal of the solar panel. The circuit is now complete but open at the key.

4. Testing

1. Keep the key open → no current, the fan stays still.
2. Place the panel in bright sunlight and press (close) the key → current $$I$$ flows; electrical energy from the panel is converted by the motor into mechanical energy, so the fan blades rotate.

5. Circuit diagram to draw

• Rectangle labelled “Solar panel” with terminals + and −.
• From +: straight line to the symbol of an open key.
• From the other side of the key: line to the symbol of a d.c. motor (circle with M or miniature fan).
• Return line from the second motor terminal to the − terminal of the panel, completing a loop.
• Arrow showing conventional current direction $$I$$ from the panel’s + back to its −.

Closing the key produces the arrangement shown in Fig. 3.22c and the toy fan runs solely on solar power.

Objective of the visit
Prepare a report for Chapter 3 (Electricity) by identifying as many different kinds of electric cells as an ordinary electrical-items shop keeps and find out at least one device in which each cell is normally used.

Planning before the visit

1. Carry a notebook, pen, and a mobile phone for photographs (taken only after obtaining the shopkeeper’s permission).
2. Prepare the following questions for the shopkeeper.
   
   • What is the name of the cell?
   • Is it primary (single-use) or secondary (rechargeable)?
   • What is its standard voltage $$V$$ (nominal)?
   • Which everyday device generally needs it?
3. Recall the symbol for a cell $$\bigl(\text{one long line and one short line}\bigr)$$ so that you can recognise packaged primary cells easily.

Observation table prepared in the shop

*Chemical type is mentioned exactly as it appeared on the packet or as told by the shopkeeper.

Key learning points explained to the shopkeeper and noted in the report

1. A primary cell is meant for single use. Its chemical reaction is not designed to be reversed. Example: the zinc–carbon AA cell.
2. A secondary cell is rechargeable because its reaction can run both forward and backward. Example: the lead–acid or Li-ion cell.
3. When several $$1.5\,\text{V}$$ cells are connected in series, the net emf becomes the sum:
   $$V_{\text{total}} = n \times 1.5\,\text{V}$$ where $$n$$ is the number of cells.
4. Different shapes (cylindrical, disc, pouch) make it easy for manufacturers to prevent wrong insertion and to fit into the allotted space inside a device.

Safety advice given by the shopkeeper

• Do not throw cells into fire; they can burst.
• Always match charger type with battery chemistry; e.g. never charge a zinc–carbon cell in a NiMH charger.
• Recycle depleted button and coin cells because they contain heavy metals.

Conclusion
The shop stocked nine clearly different types of cells. Their voltages ranged from about $$0.5\,\text{V}$$ (single solar cell) to $$12\,\text{V}$$ (sealed lead–acid unit). Each type is optimised for a particular set of devices. The variety shows that no single cell can satisfy every requirement — long life, high current, small size, rechargeability, or low cost — at the same time.