[{"data":1,"prerenderedAt":1100},["ShallowReactive",2],{"layer:electricity:investigate":3},{"layer":4,"contentHash":1081,"dependencyHashes":1082,"approval":1094,"releaseId":1099},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":43,"sourceIds":1076,"reviewStatus":1077,"authoring":1078},1,"electricity","en","investigate","Circuits you can test","Fair tests, meters, series and parallel, Ohm's law, fuses and fruit batteries","Design fair circuit tests, place ammeters and voltmeters correctly, compare series and parallel bulbs, test Ohm's law and see a filament bulb break it, work out when an MCB trips, and build a safe lemon battery.",[13,14,15,16,17],"Plan a fair circuit investigation with independent, dependent and control variables, repeat readings and an uncertainty estimate.","Connect ammeters in series and voltmeters across components, and read meters and scales safely and honestly.","Predict and test current, voltage and brightness in series and parallel circuits, and explain why homes are wired in parallel.","Use a V–I table to decide whether a component is ohmic, and explain why a filament bulb is not.","Add appliance currents to predict overloads and explain what fuses, MCBs and short circuits do.",40,{"title":20,"rows":21},"Rating plate",[22,25,28,31,34,37,40],{"label":23,"value":24},"Depth","Investigate",{"label":26,"value":27},"Reading time","About 40 minutes",{"label":29,"value":30},"Prior knowledge","Voltage, current, resistance; V = I × R",{"label":32,"value":33},"Chapters","11",{"label":35,"value":36},"Labs","3 circuit labs, 2 Ohm's-law labs",{"label":38,"value":39},"Hands-on voltage","Batteries of 9 V or less only",{"label":41,"value":42},"Units used","V, A, mA, Ω, W",[44,48,51,57,63,69,96,99,104,119,124,132,153,157,186,190,193,208,213,218,221,235,252,292,297,302,307,310,321,330,350,364,377,381,386,389,429,434,464,469,472,488,501,529,560,575,579,584,587,620,626,653,657,661,666,671,674,707,725,729,748,753,756,769,779,783,788,791,822,842,853,857,862,865,894,904,914,918,1045,1064],{"id":45,"type":46,"markdown":47},"intro-bench","prose","Up to now electricity may have felt like something to *believe*: invisible charge, invisible pushes, invisible resistance. This layer is where you stop believing and start **testing**. Every idea here can be checked with a battery, a few bulbs, some wire and a cheap multimeter — or with the labs on this page when you do not have the kit to hand.\n\nYou will learn to set up a fair test, to put meters in the right places, to read them honestly, and to decide whether your results really support a claim. Then you will use those skills on the big practical questions: why are the lights in your home wired in parallel? Does a bulb obey Ohm's law? What exactly makes an MCB trip when someone plugs a heater into an already-busy extension board? And can a lemon really power anything?",{"id":49,"type":46,"markdown":50},"intro-rules","Two rules run through every chapter.\n\n1. **Predict before you test.** Write down what you expect *and why*. A surprise is only useful if you had an expectation to be surprised against.\n2. **Low voltage only.** Every hands-on activity here uses cells or batteries of **9 V or less**. Mains electricity (230 V in India) appears only as calculations and simulations. The physics is the same; the danger is not.",{"id":52,"type":53,"variant":54,"title":55,"markdown":56},"careful-golden-rule","callout","careful","The one rule that is never bent","Never open, probe, measure or experiment on anything connected to mains sockets, switchboards, extension boards or appliance cables. A multimeter on a school bench is not a mains tester. If something at home seems faulty, switch it off at the MCB and call a qualified electrician. Everything in this layer that involves 230 V is a *calculation*, not an experiment.",{"id":58,"type":59,"title":60,"eyebrow":61,"navLabel":62},"ch-fair-test","chapter","Designing a fair test with circuits","Chapter 01","1 Fair tests",{"id":64,"type":46,"markdown":65,"help":66},"fair-test-idea","Suppose a friend claims: *\"A longer wire makes the bulb dimmer.\"* How would you find out?\n\nThe tempting method is to grab a long wire and a short wire, connect each to a bulb, and compare. But if the long wire is thinner, or the battery was fresher for the first test, or you used a different bulb, you cannot tell *which* change caused the difference. A **fair test** changes exactly one thing on purpose and keeps everything else the same.\n\nScientists name the three kinds of variable:\n\n- **Independent variable** — the one thing you deliberately change (wire length).\n- **Dependent variable** — the thing you measure to see the effect (current through the bulb, or its brightness).\n- **Control variables** — everything else that could affect the result, which you hold steady (battery, bulb, wire thickness, wire material, temperature).",{"simplerExplanation":67,"anotherExample":68},"Change one thing, measure one thing, keep everything else the same. That is all a fair test is.","Testing whether the type of fruit affects a fruit cell's voltage: change the fruit (independent), measure voltage (dependent), keep the same two metal strips, the same gap between them and the same depth (controls).",{"id":70,"type":71,"caption":72,"columns":73,"rows":77},"tbl-variables","table","Planning table: does wire length affect the current?",[74,75,76],"Kind of variable","In this test","How you handle it",[78,82,86,90,93],[79,80,81],"Independent","Length of nichrome wire (10, 20, 30, 40, 50 cm)","Move a crocodile clip along a wire taped to a metre rule",[83,84,85],"Dependent","Current through the circuit","Read an ammeter connected in series",[87,88,89],"Control","Battery voltage","Use the same 4.5 V battery pack; check it with a voltmeter before and after",[87,91,92],"Wire thickness and material","Use one single piece of the same wire for every length",[87,94,95],"Temperature of the wire","Switch off between readings so the wire does not heat up and change its resistance",{"id":97,"type":46,"markdown":98},"fair-test-repeats","**Why repeat readings?** Any single reading can be thrown off: a loose crocodile clip, a meter that has not settled, a finger pressing a contact. Take each reading at least three times. If the repeats agree closely, you can trust them. If one is far from the others, it is an **anomaly** — check the circuit, find the cause if you can, and repeat that reading rather than quietly including it.\n\n**Why several values of the independent variable?** Two lengths only tell you *whether* something changes. Five or six lengths, evenly spread, show you the **pattern**: does the current halve when the length doubles? Is the graph a straight line or a curve? Patterns are what turn a result into an explanation.",{"id":100,"type":53,"variant":101,"title":102,"markdown":103},"aha-heating","aha","Your circuit is also a heater","Current warms wires, and warm metal has more resistance. If you leave a thin wire connected for a minute while you write things down, the later readings will creep downward. That is why good circuit investigations say *switch off between readings*: temperature is a sneaky extra variable.",{"id":105,"type":106,"prompt":107,"options":108,"explanation":118},"pred-wire-length","prediction","In the wire-length investigation, the current is 0.60 A with 20 cm of nichrome wire in the circuit. The rest of the circuit has very little resistance. What do you predict at 40 cm?",[109,112,115],{"id":110,"label":111},"a","About 0.60 A — length does not matter",{"id":113,"label":114},"b","About 0.30 A — twice the length, about half the current",{"id":116,"label":117},"c","About 1.20 A — longer wire carries more","**About 0.30 A.** Resistance of a uniform wire is proportional to its length: double the length, double the resistance. With the same voltage, I = V ÷ R, so doubling R halves the current. In a real test the answer comes out slightly above 0.30 A, because the leads, clips and battery also have a little resistance that does not double. Spotting that small difference is exactly the kind of thing a careful investigator notices.",{"id":120,"type":59,"title":121,"eyebrow":122,"navLabel":123},"ch-measuring","Measuring: meters, scales and uncertainty","Chapter 02","2 Measuring",{"id":125,"type":46,"markdown":126,"help":127},"meters-placement","Two meters do almost all the work in circuit investigations, and they are connected in opposite ways.\n\n**An ammeter measures current**, the flow of charge *through* something. To count what flows through a bulb, the charge must flow through the meter too, so the ammeter goes **in series**: you break the circuit and let the meter become part of the loop. A good ammeter has a *very low* resistance so it barely changes the current it is measuring.\n\n**A voltmeter measures voltage** (potential difference), the energy each coulomb of charge gains or loses *between two points*. So it connects **across** a component, in parallel, with one lead on each side. A good voltmeter has a *very high* resistance so almost no current is diverted through it.\n\nIn the water picture: an ammeter is a flow meter spliced into the pipe; a voltmeter is a pressure gauge with one tube before the pump or narrow pipe and one tube after it.",{"simplerExplanation":128,"hints":129},"Ammeter: in the loop, counts the flow. Voltmeter: beside the part, compares the push on either side.",[130,131],"Ask: am I measuring something that flows *through* (current) or a difference *between* two points (voltage)?","To add an ammeter you must open the circuit. To add a voltmeter you never need to.",{"id":133,"type":134,"tone":135,"items":136},"spec-meters","spec","blue",[137,141,145,149],{"label":138,"big":139,"value":140},"Ammeter","in series","Measures current in amperes (A). Very low resistance. Break the loop and insert it.",{"label":142,"big":143,"value":144},"Voltmeter","across","Measures potential difference in volts (V). Very high resistance. Clip it on either side of the part.",{"label":146,"big":147,"value":148},"Multimeter","both","One box, a dial and three or four sockets. The dial *and* the red lead socket must match what you measure.",{"label":150,"big":151,"value":152},"Typical bench","≤ 9 V","Cells and batteries only. Currents usually well under 1 A.",{"id":154,"type":53,"variant":54,"title":155,"markdown":156},"careful-ammeter-across","The classic multimeter mistake","An ammeter has almost no resistance. If you connect it **across** a battery instead of in series with a bulb, you have built a short circuit: a large current rushes through the meter. At best its internal fuse blows; at worst the battery and leads get hot. Before switching on, trace the path with your finger: does all the current have to go through the ammeter *and* through a bulb or resistor?",{"id":158,"type":159,"title":160,"items":161},"steps-multimeter","steps","Using a multimeter safely on a battery circuit",[162,166,170,174,178,182],{"title":163,"tag":164,"text":165},"Choose the quantity","dial","Turn the dial to V⎓ (DC volts) for voltage or A⎓ \u002F mA⎓ for current. Never measure with the dial on Ω while the circuit is powered.",{"title":167,"tag":168,"text":169},"Choose the sockets","leads","Black lead in COM. Red lead in the V socket for voltage, the mA or A socket for current. The 10 A socket is often unfused — avoid it on small circuits.",{"title":171,"tag":172,"text":173},"Start on a high range","range","If your meter is not auto-ranging, start on a higher range (20 V, 200 mA) and step down for more digits. Too low a range shows 1 or OL (over limit).",{"title":175,"tag":176,"text":177},"Connect, then switch on","circuit","Build the circuit with the switch open, check the meter position, then close the switch and read.",{"title":179,"tag":180,"text":181},"Read and record","data","Wait for the digits to settle. Write the value with its unit and the number of digits the meter shows.",{"title":183,"tag":184,"text":185},"Put it back to volts","habit","Move the red lead back to V and turn the meter off. The next person who measures voltage with the lead in the A socket will short the battery.",{"id":187,"type":53,"variant":54,"title":188,"markdown":189},"careful-meters-mains","Cheap meters and mains do not mix","Even though many multimeters show 600 V or 750 V on the dial, the cheap ones are not built to protect you from mains faults. Measuring mains is a job for trained electricians with properly rated (CAT-rated) meters and probes. In this layer, a meter only ever touches circuits powered by cells of 9 V or less.",{"id":191,"type":46,"markdown":192},"reading-scales","**Reading an analogue scale.** Older school meters have a needle. First work out what each small division is worth: if the scale goes from 0 to 1 A with 50 divisions, each is 0.02 A. Read with your eye directly above the needle — looking from the side makes the needle appear to sit over a different mark (this is called **parallax**). If the needle sits between two marks, estimate to about half a division.\n\n**Reading a digital display.** Digital meters feel exact, but the last digit can flicker. A reading of 0.46 A usually means \"somewhere around 0.455 to 0.465 A\", and the meter's own accuracy (printed in the manual, often something like ±1% plus a digit or two) adds more. The **resolution** is the smallest change the display can show: 0.01 A here.\n\n**Uncertainty** is an honest statement of how far off your answer might be. A simple, widely used estimate is **half the range of your repeat readings**.",{"id":194,"type":195,"title":196,"problem":197,"steps":198,"help":204},"we-uncertainty","worked_example","Mean and uncertainty from repeat readings","You measure the current through a resistor four times and get 0.46 A, 0.44 A, 0.62 A and 0.47 A. What should you record?",[199,200,201,202,203],"Look for anomalies. 0.62 A is far from the other three, which agree within 0.03 A. You check the circuit and find a crocodile clip had slipped onto a second wire. So 0.62 A is an anomaly with a known cause; take a fresh reading instead. Say it gives 0.45 A.","Readings to use: 0.46, 0.44, 0.47 and 0.45 A.","Mean = (0.46 + 0.44 + 0.47 + 0.45) ÷ 4 = 1.82 ÷ 4 = 0.455 A. Round to match the meter's resolution: **0.46 A** (or keep 0.455 A while you calculate further).","Range = biggest − smallest = 0.47 − 0.44 = 0.03 A. Uncertainty ≈ half the range = 0.015 A, which rounds to about ±0.02 A.","Record: **current = 0.46 ± 0.02 A**. Anyone reading this knows both your best value and how much to trust it.",{"hints":205},[206,207],"An anomaly is not just a number you dislike. Remove it only if it is clearly out of line and, ideally, you know why.","Half the range is a quick estimate, good enough for school investigations.",{"id":209,"type":53,"variant":210,"title":211,"markdown":212},"nuance-accuracy","nuance","Precise is not the same as accurate","If your three repeats are 0.453, 0.454 and 0.453 A, they are **precise** (close together). But if the meter itself reads 5% high, they are all wrong in the same direction — precise but not **accurate**. Repeats reveal random scatter; they cannot reveal a meter that is off in the same way every time. Checking against a second meter, or a known resistor, catches that kind of error.",{"id":214,"type":59,"title":215,"eyebrow":216,"navLabel":217},"ch-series","Series circuits on the bench","Chapter 03","3 Series bulbs",{"id":219,"type":46,"markdown":220},"series-setup","A **series circuit** is a single loop. Charge leaves the battery, goes through the first bulb, then the second, then back — one path, no choices.\n\nThat single path has two consequences you can test. First, the current is the **same everywhere** in the loop: an ammeter placed before the first bulb, between the bulbs or after the last one shows the same reading. Charge is not used up by bulbs; it is *energy* that is transferred. Second, the battery's voltage is **shared** between the bulbs. With identical bulbs it is shared equally.",{"id":222,"type":106,"prompt":223,"options":224,"explanation":234},"pred-series-add","A 6 V battery lights one bulb brightly. You add a second identical bulb **in series**. What happens?",[225,227,229,231],{"id":110,"label":226},"Both bulbs are as bright as the single bulb was",{"id":113,"label":228},"Both bulbs are dimmer than the single bulb was",{"id":116,"label":230},"The first bulb stays bright and the second is dim",{"id":232,"label":233},"d","The first bulb is dim and the second is bright","**Both are dimmer, and equally so.** The loop now has twice the resistance, so the current is halved. Each bulb also gets only half the voltage. Power = V × I, so each bulb gets a quarter of the power the single bulb had. There is no \"first\" bulb that grabs the energy: the current is the same all the way round, so identical bulbs behave identically.",{"id":236,"type":237,"component":238,"componentVersion":5,"config":239,"objective":245,"textAlternative":246,"help":247},"lab-series","interactive","circuit-lab",{"voltage":240,"bulbResistance":241,"bulbs":242,"mode":243,"allowModeChange":244},6,12,2,"series",false,"Measure how current, bulb voltage and brightness change as identical bulbs are added in series, and what happens when one breaks.","The lab shows a 6 V battery and identical 12 Ω bulbs in a single loop. Results for 1 to 4 bulbs:\n\n- **1 bulb:** total resistance 12 Ω, current 6 ÷ 12 = 0.50 A, bulb voltage 6 V, bulb power 3.0 W.\n- **2 bulbs:** 24 Ω, current 0.25 A, each bulb 3 V and 0.75 W.\n- **3 bulbs:** 36 Ω, current ≈ 0.17 A, each bulb 2 V and ≈ 0.33 W.\n- **4 bulbs:** 48 Ω, current 0.125 A, each bulb 1.5 V and ≈ 0.19 W.\n\nThe current is the same at every point in the loop, the bulb voltages always add up to 6 V, and every added bulb makes all of them dimmer. If you break any one bulb, the only path is cut: the current everywhere becomes 0 A and every bulb goes out.",{"simplerExplanation":248,"hints":249},"In series, adding bulbs is like adding more narrow sections to one pipe: less flow everywhere, and every bulb gets less.",[250,251],"Check that the bulb voltages always add to the battery voltage.","Compare the power per bulb, not just the current: brightness follows power.",{"id":253,"type":71,"caption":254,"columns":255,"rows":261},"tbl-series-results","Series results (6 V battery, identical 12 Ω bulbs, model values)",[256,257,258,259,260],"Bulbs","Total resistance","Current","Voltage per bulb","Power per bulb",[262,268,274,280,286],[263,264,265,266,267],"1","12 Ω","0.50 A","6.0 V","3.0 W",[269,270,271,272,273],"2","24 Ω","0.25 A","3.0 V","0.75 W",[275,276,277,278,279],"3","36 Ω","0.17 A","2.0 V","0.33 W",[281,282,283,284,285],"4","48 Ω","0.125 A","1.5 V","0.19 W",[287,288,289,290,291],"4, one broken","open circuit","0 A","0 V","0 W",{"id":293,"type":53,"variant":294,"title":295,"markdown":296},"model-limit-series","model_limit","Real bulbs are kinder than the model","The lab treats each bulb as a fixed 12 Ω. A real filament has lower resistance when it runs cooler (Chapter 7), so dim bulbs in a long series string draw a little more current than the model predicts and glow slightly more than the table suggests. The pattern — dimmer with every bulb added, all out if one breaks — is exactly right.",{"id":298,"type":53,"variant":299,"title":300,"markdown":301},"try-series-meter","try_it","Move the ammeter","With two torch bulbs in series on a 3 V or 4.5 V battery pack, put an ammeter first before bulb 1, then between the bulbs, then after bulb 2. Record all three readings. If they differ by more than your meter's last digit, look for a loose connection. Then use a voltmeter across each bulb and across the battery: the bulb voltages should add up to (almost exactly) the battery voltage.",{"id":303,"type":59,"title":304,"eyebrow":305,"navLabel":306},"ch-parallel","Parallel circuits on the bench","Chapter 04","4 Parallel bulbs",{"id":308,"type":46,"markdown":309},"parallel-setup","In a **parallel circuit** each bulb has its own branch, connected directly across the battery. Charge reaching a junction can go one way or the other, and the branches join again before returning to the battery.\n\nNow the rules flip. Every branch gets the **full battery voltage**, so each bulb glows as brightly as a single bulb would. The branch currents **add up** to give the total current drawn from the battery. And because each branch is a separate path, breaking one bulb leaves the others working.",{"id":311,"type":106,"prompt":312,"options":313,"explanation":320},"pred-parallel-add","A 6 V battery lights one bulb. You add a second identical bulb **in parallel**. What happens to the current drawn from the battery?",[314,316,318],{"id":110,"label":315},"It halves, because the current is shared between two bulbs",{"id":113,"label":317},"It stays the same",{"id":116,"label":319},"It doubles, because each bulb draws the same current as before","**It doubles.** Each bulb is connected straight across 6 V, so each draws the same current it would alone (0.50 A for a 12 Ω bulb). Two branches mean 0.50 + 0.50 = 1.0 A from the battery. Adding a branch gives the charge an *extra path*, so the total resistance goes down, not up — 12 Ω alone, 6 Ω for two in parallel. The cost: the battery works twice as hard and runs down twice as fast.",{"id":322,"type":237,"component":238,"componentVersion":5,"config":323,"objective":325,"textAlternative":326,"help":327},"lab-parallel",{"voltage":240,"bulbResistance":241,"bulbs":242,"mode":324,"allowModeChange":244},"parallel","Measure branch current, total current and brightness as identical bulbs are added in parallel, and test what happens when one breaks.","The lab shows a 6 V battery with identical 12 Ω bulbs, each on its own branch. Every bulb has 6 V across it, carries 6 ÷ 12 = 0.50 A and uses 3.0 W, however many bulbs there are. The total current is 0.50 A × the number of working bulbs: 0.50 A for one, 1.0 A for two, 1.5 A for three, 2.0 A for four. The total resistance falls: 12 Ω, 6 Ω, 4 Ω and 3 Ω. If you break one bulb, only its branch stops; the others stay exactly as bright and the total current drops by 0.50 A.",{"simplerExplanation":328,"anotherExample":329},"Parallel is like adding more taps to a full water tank: each tap flows as strongly as before, but the tank empties faster.","Two phone chargers plugged into two sockets both charge at full speed, but the house meter records both.",{"id":331,"type":71,"caption":332,"columns":333,"rows":336},"tbl-parallel-results","Parallel results (6 V battery, identical 12 Ω bulbs, model values)",[256,257,334,335,260],"Current per bulb","Total current",[337,338,341,344,347],[263,264,265,265,267],[269,339,265,340,267],"6 Ω","1.0 A",[275,342,265,343,267],"4 Ω","1.5 A",[281,345,265,346,267],"3 Ω","2.0 A",[287,342,348,343,349],"0.50 A (3 working)","3.0 W (0 W in the broken one)",{"id":351,"type":106,"prompt":352,"options":353,"explanation":363},"pred-compare-modes","Same battery, same three bulbs. Arrangement X has them in series; arrangement Y in parallel. Which battery goes flat first?",[354,357,360],{"id":355,"label":356},"x","X, the series circuit",{"id":358,"label":359},"y","Y, the parallel circuit",{"id":361,"label":362},"same","Both last the same time","**Y, the parallel circuit.** In parallel each bulb draws its full current, so the battery supplies three times the single-bulb current and transfers energy nine times faster than the series version (where the current is a third and each bulb gets a third of the voltage). Bright light costs energy; the series string is dim but thrifty.",{"id":365,"type":237,"component":238,"componentVersion":5,"config":366,"objective":371,"textAlternative":372,"help":373},"lab-compare",{"voltage":367,"bulbResistance":368,"bulbs":369,"mode":243,"allowModeChange":370},9,18,3,true,"Switch the same three bulbs between series and parallel and compare brightness, battery current and the effect of a broken bulb.","This lab uses a 9 V battery and three identical 18 Ω bulbs, and lets you switch arrangements.\n\n- **Series:** total resistance 54 Ω, current 9 ÷ 54 ≈ 0.17 A, each bulb 3 V and 0.5 W; total power 1.5 W. One broken bulb puts all three out.\n- **Parallel:** each bulb 9 V, 0.50 A and 4.5 W; total current 1.5 A and total power 13.5 W, nine times the series total. One broken bulb leaves the other two fully bright and the total current falls to 1.0 A.\n\nSo parallel wins on brightness and reliability; series draws far less from the battery.",{"hints":374},[375,376],"Work out the total power in each arrangement: P = V × I for the whole circuit.","Nine times the power means the battery would last about a ninth as long (in this simple model).",{"id":378,"type":53,"variant":294,"title":379,"markdown":380},"model-limit-battery","A real 9 V battery would struggle","The lab uses an *ideal* battery that keeps its voltage whatever current you draw. A real rectangular 9 V battery has internal resistance of roughly a few ohms, so asking it for 1.5 A would make its terminal voltage sag well below 9 V and the bulbs would be dimmer than predicted. It would also get warm and flatten quickly. Real parallel circuits drawing big currents need a stiffer supply — which is exactly what the mains is.",{"id":382,"type":59,"title":383,"eyebrow":384,"navLabel":385},"ch-homes","Why homes are wired in parallel","Chapter 05","5 Wiring a home",{"id":387,"type":46,"markdown":388},"homes-parallel","Every socket, fan point and light point in your home is connected **in parallel** across the live and neutral wires that come from the meter and MCB board. Put the bench results next to what a home needs and the reason is obvious:\n\n- **Every appliance gets the full 230 V.** A 1500 W geyser is designed to work at 230 V. In series with a fan and a TV it would get only a share of the voltage and barely warm the water.\n- **Each appliance can be switched independently.** Turning off the bedroom light should not turn off the fridge. In series, one open switch would break the only loop and switch everything off.\n- **One failure does not black out the house.** A burnt-out tube light breaks its own branch only.\n- **Adding an appliance does not dim the others.** Each branch draws its own current at 230 V.\n\nThe price, as the bench showed, is that the currents **add up** in the shared wires. That is the whole reason fuses and MCBs exist, and it is the subject of Chapter 8.",{"id":390,"type":391,"title":392,"prompt":393,"options":394},"explorer-wiring","explorer","What if a home were wired differently?","Pick a wiring plan for a room with a light, a fan and a phone charger.",[395,407,419],{"id":396,"label":397,"chain":398,"badge":403,"note":406},"all-series","All in series",[399,400,401,402],"One loop","Voltage shared three ways","Every device under-powered","One switch off → all off",{"text":404,"tone":405},"Useless for a home","no","With 230 V shared between a 10 W LED light, a 60 W fan and a charger, none gets the voltage it was designed for. The fan would hardly turn, and turning off the light (opening the loop) would stop everything. Series is used deliberately only where a shared current is wanted — like old strings of tiny decorative lamps.",{"id":408,"label":409,"chain":410,"badge":415,"note":418},"all-parallel","All in parallel",[411,412,413,414],"Separate branches","Each gets 230 V","Switch each alone","Currents add in main wire",{"text":416,"tone":417},"How real homes are wired","yes","Each appliance sees the full 230 V and has its own switch in its own branch. The only thing to watch is the total current in the shared cable, which is why every circuit is protected by an MCB sized for its wire.",{"id":420,"label":421,"chain":422,"badge":426,"note":428},"switch-in-series","Switch in series with load",[423,424,425],"Branch per device","Switch in that branch","Controls only that device",{"text":427,"tone":417},"Yes — this is the detail","Within a parallel home, each switch sits *in series* with the one appliance it controls, on the live wire. So real wiring uses both ideas: parallel between appliances, series between a switch and its load.",{"id":430,"type":53,"variant":431,"title":432,"markdown":433},"example-diwali","example","Series lights you have probably seen","Cheap strings of tiny decorative lamps used at Diwali or weddings were traditionally wired in series: dozens of low-voltage bulbs sharing 230 V, so each gets only a few volts. The famous annoyance — one bulb fails and the whole string goes dark, and you test bulb after bulb to find it — is the series circuit behaving exactly like the lab. Many newer LED strings use a mix of series groups wired in parallel so one failure only darkens a section.",{"id":435,"type":71,"caption":436,"columns":437,"rows":441},"tbl-series-parallel","Series versus parallel: the bench results side by side",[438,439,440],"Question","Series","Parallel",[442,446,449,453,457,460],[443,444,445],"Voltage across each bulb","Shared: adds up to the supply","Full supply voltage on every branch",[258,447,448],"Same everywhere in the loop","Branch currents add up to the total",[450,451,452],"Adding a bulb","All bulbs get dimmer; total current falls","Others unchanged; total current rises",[454,455,456],"One bulb breaks","Everything goes out","Only that branch goes out",[257,458,459],"Goes up with every bulb","Goes down with every branch",[461,462,463],"Where you see it","Switch and its load; old decorative light strings; cells in a torch","Every socket and light point in a home",{"id":465,"type":59,"title":466,"eyebrow":467,"navLabel":468},"ch-ohms","Ohm's law: the fixed-resistor experiment","Chapter 06","6 Ohm's law test",{"id":470,"type":46,"markdown":471},"ohms-setup","Ohm's law says that for some conductors, **current is proportional to voltage**: double the voltage, double the current. The ratio V ÷ I is then constant, and we call that constant the **resistance**, R.\n\nThat is a *claim about the world*, so let us test it. The classic method uses a fixed resistor (say one marked 10 Ω), a variable supply or a battery pack you can change from 1.5 V to 9 V in 1.5 V steps, an ammeter in series and a voltmeter across the resistor.\n\n- **Independent variable:** voltage across the resistor.\n- **Dependent variable:** current through it.\n- **Controls:** the same resistor, kept at room temperature (switch off between readings), same leads and meters.",{"id":473,"type":474,"items":475},"formulas-ohm","formulas",[476,479,482,485],{"expression":477,"caption":478},"V = I × R","Voltage (V) equals current (A) times resistance (Ω).",{"expression":480,"caption":481},"I = V ÷ R","Rearranged to predict the current from a known voltage and resistance.",{"expression":483,"caption":484},"R = V ÷ I","Rearranged to work out resistance from a pair of meter readings.",{"expression":486,"caption":487},"P = V × I","Power in watts: how fast energy is transferred.",{"id":489,"type":106,"prompt":490,"options":491,"explanation":500},"pred-ohm-double","With the 10 Ω resistor, 3.0 V gives a current of 0.30 A. What do you predict at 6.0 V?",[492,494,496,498],{"id":110,"label":493},"0.30 A",{"id":113,"label":495},"0.60 A",{"id":116,"label":497},"1.20 A",{"id":232,"label":499},"Impossible to say without testing","If the resistor is **ohmic**, doubling the voltage doubles the current: **0.60 A**. Option d is a fair scientific instinct — you *should* test — but a prediction is still worth making. The test is what checks it.",{"id":502,"type":237,"component":503,"componentVersion":5,"config":504,"objective":522,"textAlternative":523,"help":524},"lab-ohms","ohms-law",{"voltage":505,"resistance":507,"presets":510},{"min":506,"max":367,"initial":369,"step":506},0.5,{"min":5,"max":508,"initial":509,"step":5},100,10,[511,514,517,519],{"label":512,"voltage":513,"resistance":509},"10 Ω resistor, 1.5 V",1.5,{"label":515,"voltage":516,"resistance":509},"10 Ω resistor, 4.5 V",4.5,{"label":518,"voltage":367,"resistance":509},"10 Ω resistor, 9 V",{"label":520,"voltage":367,"resistance":521},"47 Ω resistor, 9 V",47,"Collect a table of voltage and current for a fixed resistor, check that current is proportional to voltage, and calculate R for each pair.","The lab applies a chosen voltage (0.5 to 9 V) across a chosen fixed resistance (1 to 100 Ω) and shows the current I = V ÷ R. With R fixed at 10 Ω the readings are: 1.5 V → 0.15 A, 3.0 V → 0.30 A, 4.5 V → 0.45 A, 6.0 V → 0.60 A, 7.5 V → 0.75 A and 9.0 V → 0.90 A. Every pair gives V ÷ I = 10 Ω, and a graph of current against voltage is a straight line through the origin whose steepness is 1 ÷ R. Swapping to 47 Ω at 9 V gives 9 ÷ 47 ≈ 0.19 A: a bigger resistance gives a less steep line. The lab's resistance never changes with temperature, which is why its results are perfectly proportional.",{"simplerExplanation":525,"hints":526},"Keep the resistor the same, turn up the voltage in equal steps, and watch the current go up in equal steps too.",[527,528],"Divide each voltage by its current. What do you notice?","Proportional means: a straight-line graph that passes through (0, 0).",{"id":530,"type":71,"caption":531,"columns":532,"rows":536},"tbl-ohm-data","A real set of results for a resistor marked 10 Ω (one reading per voltage, meters to 0.01)",[533,534,535],"Voltage (V)","Current (A)","R = V ÷ I (Ω)",[537,541,545,549,552,556],[538,539,540],"1.5","0.15","10.0",[542,543,544],"3.0","0.29","10.3",[546,547,548],"4.5","0.46","9.8",[550,551,540],"6.0","0.60",[553,554,555],"7.5","0.74","10.1",[557,558,559],"9.0","0.91","9.9",{"id":561,"type":195,"title":562,"problem":563,"steps":564,"help":571},"we-ohm-analysis","Is the resistor ohmic? Analysing the table","Use the six results above to decide whether current is proportional to voltage and to estimate the resistance.",[565,566,567,568,569,570],"Calculate R for each row: 1.5 ÷ 0.15 = 10.0 Ω; 3.0 ÷ 0.29 ≈ 10.3 Ω; 4.5 ÷ 0.46 ≈ 9.8 Ω; 6.0 ÷ 0.60 = 10.0 Ω; 7.5 ÷ 0.74 ≈ 10.1 Ω; 9.0 ÷ 0.91 ≈ 9.9 Ω.","Mean R = (10.0 + 10.3 + 9.8 + 10.0 + 10.1 + 9.9) ÷ 6 = 60.1 ÷ 6 ≈ **10.0 Ω**.","Spread: the values run from 9.8 to 10.3 Ω, a range of 0.5 Ω, so uncertainty ≈ ±0.25 Ω — call it **R = 10.0 ± 0.3 Ω**.","The scatter is small and random: R does not climb or fall steadily as the voltage rises. That is the signature of an **ohmic** conductor.","Why the small wobble? A current meter reading to 0.01 A cannot tell 0.285 A from 0.294 A. At 3.0 V a one-digit change in the current moves R by about 0.3 Ω, so the 10.3 Ω is well within the meter's resolution.","Conclusion: within experimental uncertainty, current is proportional to voltage, and the resistor's resistance is about 10 Ω, matching its label.",{"hints":572},[573,574],"Look for a trend in the R column, not just whether the numbers are exactly equal.","Small readings (like 0.15 A) have the biggest *percentage* uncertainty.",{"id":576,"type":53,"variant":101,"title":577,"markdown":578},"aha-graph","Read the graph, not just the numbers","Plot current (vertical) against voltage (horizontal). For an ohmic resistor the points lie close to a **straight line through the origin**. A straight line that does *not* pass through (0, 0), or a curve that bends over, tells you something else is going on — a meter zero error in the first case, a changing resistance in the second.",{"id":580,"type":59,"title":581,"eyebrow":582,"navLabel":583},"ch-filament","The bulb that breaks the rule","Chapter 07","7 Filament bulbs",{"id":585,"type":46,"markdown":586},"filament-intro","Now repeat the experiment with a small filament bulb instead of a resistor — say a 6 V, 0.5 A torch bulb. At first the numbers look familiar. Then something odd happens: doubling the voltage **does not** double the current. The current keeps rising, but more and more slowly.\n\nCalculate V ÷ I for each row and the reason jumps out: the bulb's resistance is **not constant**. It climbs as the voltage goes up.",{"id":588,"type":71,"caption":589,"columns":590,"rows":592},"tbl-bulb-iv","Typical readings for a 6 V, 0.5 A filament torch bulb (illustrative values)",[533,534,535,591],"What you see",[593,598,603,608,611,616],[594,595,596,597],"0.5","0.20","2.5","No glow; filament barely warm",[599,600,601,602],"1.0","0.26","3.8","Faint red",[604,605,606,607],"2.0","0.34","5.9","Dull orange",[542,609,553,610],"0.40","Yellowish",[612,613,614,615],"4.0","0.44","9.1","Bright",[550,617,618,619],"0.50","12.0","Full brightness",{"id":621,"type":46,"markdown":622,"help":623},"filament-why","The filament is a very thin coil of tungsten wire. As more current flows it gets hotter — at full brightness a filament runs at well over 2000 °C. In a hot metal the atoms vibrate harder, so the drifting electrons collide with them more often, and the resistance rises. Between cold and white-hot, a tungsten filament's resistance typically rises by a factor of **roughly 10 to 15**.\n\nSo a filament bulb is **non-ohmic**. Its I–V graph starts steep near the origin (low, cold resistance) and bends over as it gets hotter (high resistance). The fixed resistor's graph was straight because it stayed at roughly room temperature.",{"simplerExplanation":624,"anotherExample":625},"The bulb gets hot, and hot metal resists current more. So the more you push, the harder it pushes back.","The same thing happens to the thin wire in the Chapter 1 investigation if you leave it switched on: it warms up and the current creeps down.",{"id":627,"type":71,"caption":628,"columns":629,"rows":634},"tbl-cold-hot","Cold versus hot filament resistance (hot values from R = V² ÷ P; cold values approximate)",[630,631,632,633],"Bulb","Hot resistance (lit)","Cold resistance (approx.)","Ratio",[635,639,644,649],[636,264,637,638],"6 V, 0.5 A torch bulb","about 1 Ω","about 12×",[640,641,642,643],"60 W, 230 V household bulb","230² ÷ 60 ≈ 880 Ω","about 60 Ω","about 15×",[645,646,647,648],"100 W, 230 V household bulb","230² ÷ 100 ≈ 530 Ω","about 35–40 Ω","about 14×",[650,651,652,643],"100 W, 120 V bulb (USA)","120² ÷ 100 = 144 Ω","about 9.5 Ω",{"id":654,"type":53,"variant":101,"title":655,"markdown":656},"aha-switch-on","Why old bulbs usually died at switch-on","At the instant you flick the switch, the filament is cold and its resistance is about a fifteenth of its working value. For a moment the current is many times the normal current — a surge lasting a fraction of a second until the filament heats up. That thermal shock is why incandescent bulbs so often failed with a flash just as they were switched on, not in the middle of the evening.",{"id":658,"type":53,"variant":294,"title":659,"markdown":660},"model-limit-ohms-lab","The Ohm's-law lab keeps R constant","The simulated lab uses a resistance that never changes with temperature, so it always gives a straight-line graph. That is a good model for a fixed resistor, and a poor one for a filament bulb. To model a bulb honestly you would need a resistance that depends on how hot the filament is — which itself depends on the current. Real investigations are how you find out where a model stops working.",{"id":662,"type":53,"variant":663,"title":664,"markdown":665},"misconception-ohm","misconception","\"V = I × R is Ohm's law, so everything obeys it\"","The equation R = V ÷ I *defines* resistance at any moment, for any component. **Ohm's law** is the separate, testable claim that R stays constant as V changes. A filament bulb has a resistance at every voltage — it just is not the same resistance. Diodes and LEDs break the law even more dramatically: almost no current until a threshold voltage, then a lot.",{"id":667,"type":59,"title":668,"eyebrow":669,"navLabel":670},"ch-fuses","Fuses, MCBs and overloads","Chapter 08","8 Fuses and MCBs",{"id":672,"type":46,"markdown":673},"fuse-idea","Parallel wiring means the currents of everything switched on in a circuit **add up** in the wires they share. Those wires are sized to carry a certain current safely. Push more than that through them and, by P = I²R, they heat up — slowly at first, then enough to soften insulation and start a fire.\n\nA **fuse** is a deliberately weak link: a short, thin wire that melts when the current exceeds its rating, breaking the circuit before the house wiring overheats. A **miniature circuit breaker (MCB)** does the same job but can be reset. It has two trip mechanisms: a bimetal strip that heats and bends on a moderate overload (tripping after seconds or minutes) and an electromagnet that snaps it open almost instantly on a huge current, such as a short circuit.\n\nIn many Indian homes, lighting and fan circuits are protected by 6 A or 10 A MCBs, and power circuits for sockets, geysers and ACs by 16 A or larger ones (the exact sizes depend on the wiring and the electrician's design).",{"id":675,"type":237,"component":503,"componentVersion":5,"config":676,"objective":700,"textAlternative":701,"help":702},"lab-fuse",{"voltage":677,"resistance":680,"fuseAmps":684,"presets":685},{"min":508,"max":678,"initial":679,"step":509},240,230,{"min":681,"max":682,"initial":683,"step":506},5,500,53,16,[686,689,692,694,697],{"label":687,"voltage":679,"resistance":688},"Iron, 1000 W",52.9,{"label":690,"voltage":679,"resistance":691},"Kettle, 2000 W",26.45,{"label":693,"voltage":679,"resistance":691},"Room heater, 2000 W",{"label":695,"voltage":679,"resistance":696},"Kettle + iron",17.6,{"label":698,"voltage":679,"resistance":699},"Kettle + iron + heater",10.6,"Simulate appliances on a 230 V circuit protected by a 16 A MCB and find which combinations make it trip.","This is a simulation only: never experiment with mains. The lab applies 230 V across a load and compares the current with a 16 A protective device. Each appliance is modelled as a resistance: a 1000 W iron is 230² ÷ 1000 ≈ 52.9 Ω and draws about 4.3 A; a 2000 W kettle or heater is ≈ 26.5 Ω and draws about 8.7 A. Several appliances in parallel act like one smaller resistance. Kettle + iron ≈ 17.6 Ω draws about 13 A — under 16 A, so the MCB holds. Kettle + iron + heater ≈ 10.6 Ω draws about 21.7 A — over 16 A, so the MCB trips. Lower resistance means more current; adding appliances in parallel always lowers the total resistance.",{"simplerExplanation":703,"hints":704},"Every extra appliance is an extra branch. More branches, more total current. Too much current and the MCB switches off.",[705,706],"Current for one appliance = power ÷ voltage.","The MCB sees the total of all branch currents.",{"id":708,"type":195,"title":709,"problem":710,"steps":711,"help":720},"we-overload","One extension board, three appliances","On a winter morning someone plugs a 2000 W kettle, a 1000 W iron and a 2000 W room heater into one extension board, on a socket circuit protected by a 16 A MCB. The supply is 230 V. Will the MCB trip?",[712,713,714,715,716,717,718,719],"Each appliance draws I = P ÷ V.","Kettle: 2000 ÷ 230 ≈ **8.70 A**.","Iron: 1000 ÷ 230 ≈ **4.35 A**.","Heater: 2000 ÷ 230 ≈ **8.70 A**.","They are in parallel, so the currents add: 8.70 + 4.35 + 8.70 = **21.7 A**.","Check another way: total power 2000 + 1000 + 2000 = 5000 W, and 5000 ÷ 230 ≈ 21.7 A. Same answer.","21.7 A is well over 16 A, so the MCB will trip — after some seconds, because this is an overload rather than a short circuit. Any two of the three would draw at most 17.4 A (kettle + heater), which is only just over 16 A; kettle + iron at about 13 A is safe on this circuit.","The extension board itself matters too. Many cheap boards are rated at 6 A or 10 A (check the label). Even 13 A would badly overload a 6 A board: its thin cord and contacts would heat up long before the 16 A MCB in the distribution box noticed anything.",{"anotherExample":721,"hints":722},"A 1500 W geyser (≈ 6.5 A) and a 1500 W AC (≈ 6.5 A) on the same 16 A circuit total about 13 A: fine. Add a 2000 W heater and it jumps to about 21.7 A.",[723,724],"Use I = P ÷ V for each appliance.","Parallel branch currents add.",{"id":726,"type":53,"variant":54,"title":727,"markdown":728},"careful-extension","Extension boards are the weak point","An MCB protects the fixed wiring in the walls, not the flexible cord of an extension board. Never plug high-power heating appliances (kettle, heater, iron, geyser, induction cooktop) into a multi-way board together, never plug one board into another, and treat a warm plug, a burning smell or a discoloured socket as a warning to switch off and have it checked. And never replace a blown fuse with a thicker wire, or tape an MCB on: that removes the only thing standing between an overload and a fire.",{"id":730,"type":134,"tone":731,"items":732},"spec-protection","copper",[733,737,740,744],{"label":734,"big":735,"value":736},"Rewirable fuse","melts","Thin fuse wire in a ceramic carrier. Cheap, but easily replaced with the wrong wire.",{"label":738,"big":735,"value":739},"Cartridge fuse","Sealed element with a printed rating, e.g. inside some plugs, adaptors and appliances.",{"label":741,"big":742,"value":743},"MCB","trips","Thermal trip for overloads, magnetic trip for short circuits. Resettable with a switch.",{"label":745,"big":746,"value":747},"RCCB \u002F RCD","leak","Trips on a small leakage to earth (often 30 mA) — protects people, not wires. Covered in the safety layers.",{"id":749,"type":59,"title":750,"eyebrow":751,"navLabel":752},"ch-short","Short circuits","Chapter 09","9 Short circuits",{"id":754,"type":46,"markdown":755},"short-idea","A **short circuit** is a path with almost no resistance connected across a supply, so the current *takes the short cut* and skips the components it was meant to go through. Because I = V ÷ R and R is tiny, the current becomes enormous — limited only by the resistance of the wires and the supply itself.\n\nOn the bench, a short circuit is a bare wire accidentally touching both terminals of a battery. In a home, it is a live wire touching neutral inside a damaged cable or appliance. The mains supply can deliver hundreds or even thousands of amps for a moment, which is why the MCB's magnetic trip exists: it opens the circuit in a few thousandths of a second.",{"id":757,"type":106,"prompt":758,"options":759,"explanation":768},"pred-short-bulb","Two bulbs are in series with a battery and both glow. You clip a thick copper wire across bulb B only, from one of its terminals to the other. What happens?",[760,762,764,766],{"id":110,"label":761},"Both bulbs stay the same",{"id":113,"label":763},"Bulb B goes out and bulb A gets brighter",{"id":116,"label":765},"Both bulbs go out",{"id":232,"label":767},"Bulb B gets brighter","**Bulb B goes out and bulb A gets brighter.** The copper wire has almost no resistance, so nearly all the current flows through it instead of through B: B is *shorted out*. The loop now contains only bulb A, so the total resistance halves and the current roughly doubles, and A has the whole battery voltage across it. If you shorted *both* bulbs, the battery itself would be shorted — the dangerous case.",{"id":770,"type":195,"title":771,"problem":772,"steps":773},"we-short-cell","How big is a short-circuit current?","An AA cell (1.5 V) is accidentally shorted by a piece of wire with 0.05 Ω resistance. The cell's own internal resistance is about 0.15 Ω. Estimate the current and compare it with a torch bulb's normal current of about 0.3 A.",[774,775,776,777,778],"Total resistance in the loop = wire + internal = 0.05 + 0.15 = 0.20 Ω.","Current I = V ÷ R = 1.5 ÷ 0.20 = **7.5 A**.","That is 25 times the 0.3 A a torch bulb draws (7.5 ÷ 0.3 = 25).","Power in the loop P = V × I = 1.5 × 7.5 ≈ 11 W, all turned into heat in a small wire and inside the cell. The wire gets hot, the cell heats and drains fast, and some cells can leak or vent.","Notice that the cell's internal resistance, not the wire, sets the limit. Batteries with very low internal resistance (lithium-ion phone and laptop cells, car batteries) can deliver far bigger short-circuit currents — hundreds of amps for a car battery — which is why they are genuinely dangerous to short.",{"id":780,"type":53,"variant":54,"title":781,"markdown":782},"careful-short","Do not try this one","Do not short-circuit any battery on purpose, and keep loose cells away from coins, keys and foil in pockets and bags. Never short a lithium-ion battery or a car battery. Always include a bulb or resistor in any bench circuit so there is something to limit the current.",{"id":784,"type":59,"title":785,"eyebrow":786,"navLabel":787},"ch-fruit","Lemon and potato cells","Chapter 10","10 Fruit cells",{"id":789,"type":46,"markdown":790},"fruit-idea","A battery cell needs three things: two **different metals** and an **electrolyte** between them, a liquid or paste that can carry charge as ions. A lemon's acidic juice or a potato's watery flesh can be the electrolyte. Push in a strip of zinc (a galvanised iron nail is coated in zinc) and a piece of copper (a copper wire or strip), and a chemical reaction at the zinc releases electrons that flow through an outside wire to the copper.\n\nThe **fruit is not the source of the energy** — the zinc is, as it slowly dissolves. The fruit is the electrolyte that lets the reaction happen. A typical lemon cell with zinc and copper gives **about 0.9 V**, and a current of **at most around 1 mA**. That is thousands of times too little for a torch bulb, but a red LED needs only about 1.8–2 V and a few milliamps to glow faintly — so several fruit cells in series can light one, dimly, in a dark room.",{"id":792,"type":159,"title":793,"items":794},"steps-fruit","Home investigation: build a lemon or potato battery",[795,799,803,807,810,814,818],{"title":796,"tag":797,"text":798},"Gather the kit","materials","4 lemons or potatoes, 4 galvanised (zinc-coated) nails, 4 pieces of thick copper wire or copper strip, 5 crocodile-clip leads, a multimeter, a red LED.",{"title":800,"tag":801,"text":802},"Soften and prepare","setup","Roll each lemon on the table to free the juice. Push a nail and a copper piece into each, about 2 cm apart and 2–3 cm deep, not touching.",{"title":804,"tag":805,"text":806},"Measure one cell","measure","Multimeter on DC volts (2 V or 20 V range). Red lead to copper, black to zinc. Record the voltage. Repeat three times, moving the clips off and on.",{"title":808,"tag":243,"text":809},"Build a series chain","Clip the copper of lemon 1 to the zinc of lemon 2, and so on. Measure the total voltage across the free ends after adding each lemon.",{"title":811,"tag":812,"text":813},"Light the LED","test","Connect the chain to the LED: longer LED leg to the copper end. Look in a dark room. No glow? Flip the LED round; LEDs work one way only.",{"title":815,"tag":816,"text":817},"Change one variable","fair test","Now test one factor: fruit type, electrode gap, depth or electrode size. Keep everything else the same and take repeat readings.",{"title":819,"tag":820,"text":821},"Clean up","safety","Throw the fruit away afterwards — do not eat it, because metal from the nails has dissolved into it. Wash your hands.",{"id":823,"type":71,"caption":824,"columns":825,"rows":829},"tbl-fruit-results","What to expect from lemon cells in series (typical values; yours will vary)",[826,827,828],"Cells in series","Expected voltage","Can it light a red LED?",[830,833,836,839],[263,831,832],"about 0.9 V","No — below the ~1.8 V the LED needs",[269,834,835],"about 1.8 V","Maybe, very faintly",[275,837,838],"about 2.7 V","Usually a faint glow in the dark",[281,840,841],"about 3.6 V","A clearer glow; current still only about 1 mA",{"id":843,"type":106,"prompt":844,"options":845,"explanation":852},"pred-fruit-size","You cut a lemon in half and use just one half as the cell, with the same electrodes, gap and depth. What happens to the voltage?",[846,848,850],{"id":110,"label":847},"It halves",{"id":113,"label":849},"It stays about the same",{"id":116,"label":851},"It drops to zero","**It stays about the same.** A cell's voltage is set by the *chemistry* — which two metals are used — not by the size of the fruit. What size can change is how much current the cell can deliver and how long it lasts. Bigger electrode surfaces and a shorter gap usually let more current flow. That is a good follow-up fair test.",{"id":854,"type":53,"variant":663,"title":855,"markdown":856},"misconception-fruit","\"The lemon's juice is the energy\"","The energy comes from the zinc reacting and dissolving, not from anything nutritious in the lemon. Swap the copper for another zinc nail and the voltage falls to nearly zero, because two identical metals give no difference to drive current. A glass of salty water or vinegar works as an electrolyte too.",{"id":858,"type":59,"title":859,"eyebrow":860,"navLabel":861},"ch-batteries","Batteries in series and in parallel","Chapter 11","11 Cells combined",{"id":863,"type":46,"markdown":864},"battery-combos","The fruit chain showed that **cells in series add their voltages**: each cell lifts the energy of every coulomb a little more, like pumps stacked one after another. That is why a TV remote uses two 1.5 V AA cells for 3 V, and why a rectangular 9 V battery is six small 1.5 V cells stacked inside one case.\n\n**Cells in parallel** keep the same voltage as one cell but share the job of supplying current. With two identical cells side by side, each supplies half the current, so the pair can deliver more current and lasts roughly twice as long. Power banks and electric vehicle battery packs use both tricks: groups in parallel for capacity, groups in series for voltage.",{"id":866,"type":71,"caption":867,"columns":868,"rows":873},"tbl-cells","Combining identical 1.5 V cells",[869,870,871,872],"Arrangement","Total voltage","What changes","Everyday example",[874,878,882,886,890],[875,284,876,877],"1 cell","—","Wall clock",[879,272,880,881],"2 in series","Double the voltage","TV remote, many torches",[883,266,884,885],"4 in series","Four times the voltage","Toys, some bicycle lights",[887,284,888,889],"2 in parallel","Same voltage, about twice the capacity","Some solar garden lights, battery packs",[891,284,892,893],"3 in series, 1 reversed","The reversed cell cancels one other","A torch put together carelessly",{"id":895,"type":106,"prompt":896,"options":897,"explanation":903},"pred-reversed","A torch takes four 1.5 V cells in series. Someone puts one in the wrong way round. What voltage does the bulb get?",[898,899,901,902],{"id":110,"label":266},{"id":113,"label":900},"4.5 V",{"id":116,"label":272},{"id":232,"label":290},"**3.0 V.** Three cells push one way (+4.5 V) and the reversed cell pushes against them (−1.5 V): 4.5 − 1.5 = 3.0 V. The bulb glows at about half voltage, so dimly — and the reversed cell is being forced backwards, which can make it heat up or leak.",{"id":905,"type":195,"title":906,"problem":907,"steps":908},"we-cells","Designing a pack for a 6 V motor","A school project needs a small 6 V motor to run for as long as possible. You have eight AA cells (1.5 V each). How should you arrange them?",[909,910,911,912,913],"Voltage first: 6 V ÷ 1.5 V = **4 cells in series** for each string.","Eight cells make **two strings** of four.","Connect the two 6 V strings **in parallel**. The voltage stays 6 V, and each string supplies half the motor's current.","Result: 6 V, lasting roughly twice as long as a single string of four.","Practical note: only do this with cells that are identical — same type, same brand, equally fresh. Otherwise the stronger string pushes current backwards into the weaker one.",{"id":915,"type":53,"variant":54,"title":916,"markdown":917},"careful-batteries","Battery safety for investigations","Use only cells and batteries of **9 V or less**. Do not mix old and new cells or different types in the same pack. Never connect cells of different voltages in parallel. Never try to recharge a non-rechargeable cell, cut open a battery, or heat one. Keep lithium coin cells away from small children — swallowed, they can cause serious internal burns within hours. Anything bigger than a 9 V battery, including car batteries and lithium-ion packs, is for supervised adults only.",{"id":919,"type":920,"title":921,"questions":922},"quiz-investigate","quiz","Check your investigation skills",[923,936,949,962,973,986,999,1012,1025,1035],{"itemId":924,"prompt":925,"options":926,"correct":116,"why":935},"electricity.investigate-control-variable","You test whether the number of lemons in series affects the voltage. Which of these is a **control** variable?",[927,929,931,933],{"id":110,"label":928},"The number of lemons",{"id":113,"label":930},"The voltage reading",{"id":116,"label":932},"The gap between the zinc and copper in each lemon",{"id":232,"label":934},"The glow of the LED","The number of lemons is the independent variable, and the voltage (or glow) is what you measure. The electrode gap could affect the result, so it must be kept the same: that makes it a control.",{"itemId":937,"prompt":938,"options":939,"correct":113,"why":948},"electricity.investigate-ammeter-position","How should an ammeter be connected to measure the current through a bulb?",[940,942,944,946],{"id":110,"label":941},"Across the bulb, one lead on each side",{"id":113,"label":943},"In series with the bulb, so the same current flows through both",{"id":116,"label":945},"Directly across the battery",{"id":232,"label":947},"It does not matter","An ammeter measures what flows *through* it, so it must be part of the same loop, in series. Across the battery it would make a short circuit because of its very low resistance.",{"itemId":950,"prompt":951,"options":952,"correct":116,"why":961},"electricity.investigate-series-break","Three identical bulbs are in series with a battery. One filament breaks. What happens?",[953,955,957,959],{"id":110,"label":954},"The other two get brighter",{"id":113,"label":956},"The other two stay the same",{"id":116,"label":958},"All the bulbs go out",{"id":232,"label":960},"Only the bulb after the broken one goes out","A series circuit has only one path. A broken filament opens it, so the current everywhere becomes zero and every bulb goes out.",{"itemId":963,"prompt":964,"options":965,"correct":116,"why":972},"electricity.investigate-parallel-current","Four identical bulbs are in parallel across 6 V. Each bulb draws 0.5 A. What current flows from the battery?",[966,967,969,970],{"id":110,"label":283},{"id":113,"label":968},"0.5 A",{"id":116,"label":346},{"id":232,"label":971},"24 A","In parallel the branch currents add: 4 × 0.5 A = 2.0 A. (0.125 A would be the current for four such bulbs in series.)",{"itemId":974,"prompt":975,"options":976,"correct":113,"why":985},"electricity.investigate-home-parallel","What is the main reason homes are wired in parallel?",[977,979,981,983],{"id":110,"label":978},"It uses less wire",{"id":113,"label":980},"Every appliance gets the full 230 V and can be switched on and off independently",{"id":116,"label":982},"It keeps the total current low",{"id":232,"label":984},"It makes appliances use less power","Parallel branches each get the full supply voltage and have their own switches. It actually makes the total current *higher*, which is why every circuit needs an MCB.",{"itemId":987,"prompt":988,"options":989,"correct":110,"why":998},"electricity.investigate-ohmic-test","A component gives 2.0 V → 0.20 A, 4.0 V → 0.40 A and 8.0 V → 0.80 A. What can you conclude?",[990,992,994,996],{"id":110,"label":991},"It is ohmic, with a resistance of 10 Ω",{"id":113,"label":993},"It is a filament bulb",{"id":116,"label":995},"Its resistance rises with voltage",{"id":232,"label":997},"Its resistance is 0.1 Ω","V ÷ I = 10 Ω for every pair, and doubling the voltage doubles the current. Constant resistance means it is ohmic. A filament bulb's V ÷ I would rise as the voltage rises.",{"itemId":1000,"prompt":1001,"options":1002,"correct":116,"why":1011},"electricity.investigate-filament","A 60 W, 230 V filament bulb has a resistance of about 880 Ω when lit. Measured cold with a multimeter, its resistance will be…",[1003,1005,1007,1009],{"id":110,"label":1004},"About 880 Ω, because resistance is fixed",{"id":113,"label":1006},"Much more than 880 Ω",{"id":116,"label":1008},"Much less — roughly 60 Ω",{"id":232,"label":1010},"Zero","Tungsten's resistance rises steeply with temperature. A cold filament typically has about a fifteenth of its hot resistance, which is why there is a current surge at switch-on.",{"itemId":1013,"prompt":1014,"options":1015,"correct":113,"why":1024},"electricity.investigate-overload","A 2000 W kettle and a 2000 W heater run on the same 230 V circuit. Roughly what total current flows?",[1016,1018,1020,1022],{"id":110,"label":1017},"About 8.7 A",{"id":113,"label":1019},"About 17.4 A",{"id":116,"label":1021},"About 4000 A",{"id":232,"label":1023},"About 0.06 A","Each draws 2000 ÷ 230 ≈ 8.7 A, and parallel currents add: 8.7 + 8.7 ≈ 17.4 A. That is over the rating of a 16 A MCB, so it would eventually trip.",{"itemId":1026,"prompt":1027,"options":1028,"correct":113,"why":1034},"electricity.investigate-fruit-cells","Each lemon cell gives about 0.9 V. How many in series are needed to reach at least the ~1.8 V a red LED needs?",[1029,1030,1031,1032],{"id":110,"label":263},{"id":113,"label":269},{"id":116,"label":281},{"id":232,"label":1033},"10","Voltages add in series: 2 × 0.9 = 1.8 V, just enough for a faint glow. In practice 3 or 4 cells give a more reliable glow, because each real cell loses some voltage when current flows.",{"itemId":1036,"prompt":1037,"options":1038,"correct":116,"why":1044},"electricity.investigate-reversed-cell","Two 1.5 V cells are in series but one is reversed. What is the total voltage?",[1039,1040,1041,1042],{"id":110,"label":272},{"id":113,"label":284},{"id":116,"label":290},{"id":232,"label":1043},"0.75 V","The two cells push in opposite directions: +1.5 − 1.5 = 0 V. Nothing will work.",{"id":1046,"type":1047,"title":1048,"points":1049},"cheat-sheet","summary","Cheat sheet",[1050,1051,1052,1053,1054,1055,1056,1057,1058,1059,1060,1061,1062,1063],"**Fair test:** change one independent variable, measure one dependent variable, keep every control variable the same. Take at least three repeats and several evenly spread values.","**Ammeter in series** (very low resistance); **voltmeter across** the part (very high resistance). Never put an ammeter across a battery.","**Multimeter habits:** right dial, right socket, start on a high range, and put the red lead back in V when finished. Never use it on mains.","**Uncertainty ≈ half the range** of your repeats. Remove an anomaly only when it is clearly out of line, then repeat that reading.","**Series:** one loop, same current everywhere, voltage shared, adding bulbs dims them all, one break turns everything off.","**Parallel:** full voltage on every branch, branch currents add, adding bulbs raises the total current, one break affects only its branch.","**Homes are parallel** so each appliance gets 230 V and its own switch. Each switch is in series with its own load.","**Ohm's law:** for an ohmic conductor at constant temperature, I is proportional to V and R = V ÷ I is constant. The I–V graph is a straight line through the origin.","**Filament bulbs are non-ohmic:** the hot filament has roughly 10–15 times its cold resistance, so the I–V graph bends over.","**Overload:** add the currents (I = P ÷ V). Kettle 8.7 A + iron 4.35 A + heater 8.7 A ≈ 21.7 A, which trips a 16 A MCB.","**Fuse** melts; **MCB** trips (thermal for overloads, magnetic for short circuits). Never bypass either.","**Short circuit:** a near-zero-resistance path; the current is limited only by wires and the supply's internal resistance.","**Fruit cells:** zinc + copper + electrolyte ≈ 0.9 V and up to about 1 mA per lemon. Several in series can light an LED faintly.","**Cells in series add voltages; in parallel they keep the voltage and add capacity.** A reversed cell subtracts.",{"id":1065,"type":1066,"sourceIds":1067},"sources-investigate","sources",[1068,1069,1070,1071,1072,1073,1074,1075],"elec-investigate-pc-series","elec-investigate-pc-parallel","elec-investigate-pc-ohms-law","elec-investigate-pc-symbols","elec-investigate-bbc-symbols","elec-investigate-wiki-incandescent","elec-investigate-wiki-fuse","elec-investigate-wiki-lemon",[1068,1069,1070,1073,1075],"needs_review",{"generatedBy":1079,"notes":1080},"claude-code","Draft generated locally; pending owner review.","a7fe811e99ef6aaf2502d57647bf29ed733b22270e43b199f182a98aafaa8920",{"component:circuit-lab@1":1083,"component:ohms-law@1":1084,"logic:practice":1085,"source:elec-investigate-bbc-symbols":1086,"source:elec-investigate-pc-ohms-law":1087,"source:elec-investigate-pc-parallel":1088,"source:elec-investigate-pc-series":1089,"source:elec-investigate-pc-symbols":1090,"source:elec-investigate-wiki-fuse":1091,"source:elec-investigate-wiki-incandescent":1092,"source:elec-investigate-wiki-lemon":1093},"20ac7be2c5201768a46360df1d470c51d9dd2633c0d59a7e6848a8b646ee82e4","8e5bbfb66fe919a96db1fce1b9d535d2b3c030c57cf90efb2c7172829a413105","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","b041117e288b1e6864061bb7f040cce2c7c07c3d1f51c9560ab5b30999cb922e","32adcf08c9b81f51a54968bed2a4ff184ae6b07c1facb11791bb241dd5720c67","da35190ed67eb6a8b73f842dcc44b91bc285cf9bf130bcb6ca82e7b91ad72722","1ab4e3fd49154bc2ad00d1fb46f531de46f34dec4ef069e8a0366b5772aff9cb","75539961cabfbad42a94e91a2a500956073f02a83f1deecace3a83f157fe6772","724c64c8296e9a30eef860fb3dd1f603a18b2d526566b74b7c360e176a06adc4","a58978d0f143b900915d1d47b6d4ed708200c613a0f74ef822ef116476d1639d","9831b96d45a204eb9c11eaf8b91afa57e5cb4e8b35b90f2ea345abd24d37160e",{"state":1095,"reviewer":1096,"selfReview":370,"reviewedAt":1097,"method":1098},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899599726]