Every card in Electricity and magnetism, Year 9

Resistance

HintThe same word describes people pushing back against something.

WhyEvery component holds the current back to some extent. For the same push from the battery, a component that opposes the current more lets less of it flow. A bulb filament does this far more than the copper wires leading to it.

It gets smaller

HintThe charge now has two obstacles to get through, and no extra push to help.

WhyEach bulb opposes the current, and in a single loop those effects add together. The cell pushes exactly as hard as before, so with more in its way less current flows. That is why both bulbs glow more dimly than one did alone.

Bulb A

HintThe push is identical both times, so ask where the flow was held back more.

WhyWith the same potential difference each time, the only thing that can change the current is the component. The bulb that lets less current through is the one opposing it more, so the smaller ammeter reading goes with the larger resistance.

The ohm

HintIt is named after a German physicist, and its symbol is the Greek letter omega.

WhyOne of these is the resistance of a component that lets 1 amp flow when there is 1 volt across it. A torch bulb has a few of them; the plastic coating on a wire has many millions.

It is extremely high

HintCompare how strongly plastic and copper hold back the flow of charge.

WhyConductors and insulators are not two different kinds of thing so much as two ends of one scale. A metal opposes the current very little; plastic opposes it so much that the cell cannot push a measurable current through.

18

HintDividing by a half does not make a number smaller.

WhyResistance = potential difference ÷ current, so 9 ÷ 0.5. Dividing by 0.5 is the same as asking how many halves fit into 9. A small current for a given potential difference always means a large resistance.

5 ohms; 4 A

HintFind the ratio of the two readings first, then keep that ratio fixed when the push changes.

WhyResistance is 10 ÷ 2. The resistor itself has not changed, so its resistance is the same at 20 V, and the current is 20 divided by that resistance. Doubling the push across a fixed resistor doubles the current.

12 V

HintRearrange the resistance equation so the quantity you want stands alone.

WhyR = V ÷ I rearranges to V = I × R, so 3 × 4. Check it by putting the answer back: 12 ÷ 3 gives the 4 ohms you started with.

current

HintIt is the reading you would take with the meter placed in the loop itself.

WhyResistance is a ratio: how many volts of push it takes for each amp that flows. Measure the push across a component with a voltmeter and the flow through it with an ammeter, divide, and you have its resistance in ohms.

Count how many paperclips it lifts

HintYou need lots of small, identical steel objects.

WhyStrength cannot be seen, so you measure something it does. A stronger field holds more steel, so the number picked up gives a figure you can compare from one trial to the next. Repeating each trial and averaging makes the result more trustworthy.

The current

HintThink of the three things that affect the strength. One is being changed and one has been dealt with.

WhyA fair test changes one factor and holds the others steady. Winding on more wire can change how much flows, so she should check the ammeter and adjust the supply each time. Otherwise she cannot tell whether the turns or the flow made the difference.

Adding the core

HintCompare each of the last two results with the first one.

WhyDoubling the turns doubled the number lifted, from 2 to 4. Adding the iron core, with the turns back at 10, took it from 2 to 30. Iron concentrates the field so strongly that it usually matters far more than a few extra turns.

Iron loses its magnetism when the current stops

HintThink about what the device is supposed to do the instant you switch it off.

WhyBoth metals are pulled into the field and strengthen it. Steel, though, stays magnetised afterwards, which would leave a scrapyard crane unable to let go of its load. Iron gives the boost without the memory.

complete

HintCharge needs an unbroken route all the way round and back to the cell.

WhyA series circuit is a single loop, so a broken filament anywhere is a gap in the only path there is. This is why finding one dead bulb in an old string of fairy lights meant testing them one by one.

A broken bulb

HintAsk which part you have now shown is not to blame.

WhyGood fault-finding is a process of elimination. Swapping in a part you trust tests one suspect at a time. Here the fault must lie elsewhere: the cell, a wire, a connection or the switch.

In the switch

HintAsk which part the charge no longer has to pass through.

WhyThe spare wire gives the current a route around one component. If the circuit comes to life, everything else must have been working all along, and the part you bypassed is the one that was blocking the loop. Bypassing each part in turn finds a break without guesswork.

Ammeter

HintIts name is built from the unit that current is measured in.

WhyIt goes in the loop itself, so that everything flowing passes through it. A reading of zero anywhere in a series loop tells you the loop is broken somewhere. A voltmeter is the other meter, and it is connected across a component instead.

Nearly all the current takes the wire instead

HintCharge now has two routes past that point. Compare how easy each one is.

WhyThe wire has almost no resistance and the filament has a lot, so nearly all the current goes through the wire and too little is left to make the filament glow. This is a short circuit. It can make the wires and the cell dangerously hot, because the current is now much larger.