Experiment 10: A Simple Electric Circuit

A complete loop, a power source and something to use the energy: that is every circuit ever built, from a torch to a phone. In this experiment you will build a real battery-and-LED circuit, discover that an LED is a one-way device, sort conductors from insulators — and in the simulation, meet the law that runs all of it: Ohm's law, I = V/R.

The physics behind the practical

A circuit works when charge can flow in a closed loop: out of the battery, through the components, and back. Break the loop anywhere and the flow stops everywhere — the switch's whole job is to make and break that loop. Inside the loop, the battery's voltage V is the push (energy per unit charge), the current I is the flow rate (charge per second), and resistance R is how hard the components make the flow. Ohm's law ties them together: I = V/R. Double the voltage at fixed resistance and the current doubles; double the resistance and the current halves. The power delivered to a bulb or LED, its brightness, is P = V × I.

Electrons actually drift from the negative terminal to the positive (the sim's moving dots show this), while the convention inherited from history draws current from + to −; both describe the same loop. An LED adds a twist: it is a diode, conducting in one direction only — long leg (anode) towards +. Connected backwards it behaves like a break in the loop, which is why the hands-on build works in one orientation and not the other. Conductors (metals, graphite) offer low resistance; insulators (rubber, plastic, dry wood) offer so much that the LED stays dark.

Variables and controls

In the virtual lab

Start at Battery voltage 6.0 V and Resistance 30 Ω with the Switch ON: the readout gives Current I = V/R = 0.20 A and Bulb power = 1.2 W, brightness "medium", and the electron dots stream around the loop. Double the voltage to 12 V — current doubles to 0.40 A and the bulb blazes; halve it to 3 V and current halves to 0.10 A. Now restore 6 V and raise resistance to 100 Ω: the dots crawl, current falls to 0.06 A, and the bulb dims — resistance chokes flow. Flip the switch OFF: no loop, no current, dark bulb, however big the voltage. Finally, plot or eyeball the Data Visualization line of current vs voltage at fixed resistance: a straight line through the origin with slope 1/R — Ohm's law as a picture. Log at least four (V, R, I) combinations and verify each against V/R before logging the next.

Hands-on: build it, then test materials

Tape two AA cells end-to-end (positive to negative) into a ~3 V pack. Wire the pack's positive end to the LED's longer leg and the LED's shorter leg back to the pack's negative end: the LED lights. Reverse the LED: darkness — polarity demonstrated. Now break one side of the loop and bridge the gap with test objects in turn — paperclip, coin, pencil graphite, rubber band, plastic spoon — recording LED on/off for each. Safety is absolute here: AA/AAA cells only, never a wall outlet; never connect the pack's two ends directly together (a short circuit heats cells fast); if anything feels warm, disconnect at once and tell an adult.

A worked investigation

From the sim, at fixed R = 30 Ω and then fixed V = 6.0 V:

Step 1 — Proportionality. Doubling V (3 → 6 → 12 V) doubled I each time (0.10 → 0.20 → 0.40 A): at fixed R, current is directly proportional to voltage.

Step 2 — Resistance's choke. At 6.0 V, raising R from 30 Ω to 100 Ω cut I from 0.20 A to 0.06 A; halving R to 15 Ω doubled I to 0.40 A: current is inversely proportional to resistance.

Step 3 — Power check. P = V × I = 6.0 × 0.20 = 1.2 W at the starting setting, matching the sim readout. Note the 12 V row: power quadrupled (4.8 W), because doubling V doubled both V and I — brightness is greedier than it looks.

Step 4 — Finding an unknown resistor. Fix V = 6.0 V; a mystery component draws I = 0.15 A. R = V/I = 6.0 ÷ 0.15 = 40 Ω. That is precisely the trick a multimeter performs in resistance mode — and the bench version of it is why a dimmer LED in your materials test means higher resistance, not "less electricity getting through" in some vague sense: the paperclip (low R) lights it brightly, the graphite (higher R) dimly, the rubber band (astronomical R) not at all.

Conclusion and evaluation

Every sim reading matched I = V/R and P = VI, the LED's one-way behaviour demonstrated polarity, and the materials test split cleanly into conductors and insulators with graphite as the instructive middle case. Evaluation: the bench test's on/off judgement is crude — it cannot rank two conductors, and a nearly-flat battery pack can make a good conductor look like an insulator; checking the pack with the LED alone before and after the test series controls that drift. Foil-strip "wires" also add their own resistance and loose-contact errors, which is why real leads and firm clips matter.

Safety

Batteries only — never mains electricity. No direct wire between the pack's terminals (short circuit). Disconnect anything warm immediately, work with dry hands away from water, and treat a leaking cell as an adult's job.

Common mistakes

• Declaring the LED broken when it is simply reversed — check polarity before blaming components.
• Loose foil or tape joints masquerading as insulators; wiggle-test every connection.
• Short-circuiting the pack while fiddling — the "experiment" becomes battery heating, and cells can be damaged.
• Testing materials with different gap sizes, so distance, not material, changes.
• Assuming brightness is linear — power goes as V × I, so small voltage changes have outsized effects.

Exam Experience

Question 1 — calculation [4 marks]. A bulb is connected to a 9.0 V battery. Its resistance is 45 Ω. Calculate the current through the bulb and the power it converts. Then state the new current if the resistance is doubled at the same voltage.

Scoring guide:

• I = V/R = 9.0 ÷ 45 = 0.20 A [1]
• P = V × I = 9.0 × 0.20 = 1.8 W [1]
• Doubling R to 90 Ω halves the current [1]
• New current = 0.10 A [1]

Question 2 — evaluation and design [4 marks]. A student's LED circuit does not light. Rather than rebuilding randomly, describe a systematic fault-finding sequence, and then say how you would improve the conductor/insulator test to rank two metals against each other.

Scoring guide:

• Check the loop is closed and connections are firm, testing the LED alone across the pack first [1]
• Check LED polarity (long leg to positive) and swap in a known-good battery/LED to isolate the faulty part [1]
• Change one thing at a time so each test identifies a single fault [1]
• Ranking improvement: measure current with a meter (or compare brightness against the sim's P = VI) at fixed voltage instead of a yes/no LED test [1]

Multiple choice 1. A circuit stops working when a single wire is removed, even though the battery is healthy. This is because:

• A. Batteries need two wires to make electricity
• B. Current needs a closed loop; any break stops the flow everywhere
• C. The bulb has used up the electrons
• D. Resistance becomes zero in an open circuit

Answer: B. Charge flows only around a complete loop — electrons are never "used up"; a break anywhere stops the whole circuit.

Multiple choice 2. At fixed resistance, doubling the battery voltage will:

• A. Halve the current
• B. Leave the current unchanged
• C. Double the current and quadruple the power
• D. Double the resistance

Answer: C. I = V/R doubles; P = V × I then grows by 2 × 2 = 4 — as the sim's 3 V, 6 V, 12 V rows show.

Going deeper: why LEDs need a bodyguard resistor

The sim's bulb obeys Ohm's law politely; a real LED does not. An LED holds a nearly fixed voltage across itself (about 2 V for a red one) and lets current explode if the supply pushes harder — connect one straight across a 9 V battery and it draws far more current than it can survive, converting itself into a brief, bright obituary. The fix is the series resistor: it soaks up the leftover voltage and sets the current. Want 0.02 A through a 2 V LED on a 9 V supply? The resistor must drop 9 − 2 = 7 V at 0.02 A, so R = V/I = 7 ÷ 0.02 = 350 Ω. That single calculation — or its cousin — sits inside every LED lamp, strip and indicator on Earth.

Batteries have a quieter version of the same story: internal resistance. A real cell behaves like an ideal voltage in series with a small resistance of its own, so under heavy load its terminal voltage sags — the reason a nearly-flat pack lights an LED alone but fails the moment a motor joins the circuit, and the reason your conductor test in this lesson demands a fresh pack: the "component under test" includes the battery, whether you listed it or not. Naming hidden components is half of experimental maturity.