Science Lab · Physics

Circuit Builder

A breadboard where nothing can get hot. Wire up batteries, bulbs, switches, resistors, an LED, a buzzer and two meters. The circuit is solved again after every change, the way engineers’ simulators do it, so bulbs glow by the power they get and the meters read what real ones would.

Scenes

Selected part

The circuit

Show

Missions

    How to build. Drag from any hole to draw a wire; everything that meets at a hole is connected. Tap a part in the tray, then tap the board (or drag the part straight onto the board). A part lies along the grid line nearest your finger. Tap a part to select it: the little buttons turn it (↻), flip it (⇄) or delete it (✕); a switch flips when you tap it. Keyboard: click the board or tab to it, then use the arrow keys to move the yellow cursor, W to start or stop drawing a wire, 1 to 9 to place a part (1 cell, 2 9 V, 3 bulb, 4 resistor, 5 LED, 6 switch, 7 buzzer, 8 ammeter, 9 voltmeter), Enter to select or flip a switch, R to turn, F to flip, + and − to change a resistor, Delete to remove, Ctrl+Z to undo.

    What you’re seeing

    The board is a grid of holes. A wire joins two neighboring holes, and every part is two squares long, with a brass ring at each end. Anything that meets at the same hole is connected; a pale dot marks a junction where three or more things meet.

    The gold dots show the current. They move the way conventional current flows: out of the battery’s + end, around the circuit, and back into its − end. Their speed is in step with the current, so a 0.3 A bulb circuit races and a 15 mA LED circuit crawls.

    In a metal wire, the charges that really move are electrons. They are negative, so they drift the other way, from − to +. Tick “Show electrons instead” to watch them in blue. Both pictures describe the same current; scientists chose the + direction before anyone knew about electrons, and it stuck.

    A bulb’s label is its brightness: the power it gets compared with its rated 0.9 W (3 V at 0.3 A), so 100% is how bright it is meant to be. An LED shows its current. The ammeter (A) reads the current flowing through it, and the voltmeter (V) reads the voltage between its two leads. On both, the red lead is the + side, so a reading can come out negative.

    The science

    Current, voltage and resistance

    Current is how much charge flows past a point each second. One ampere (A) is one coulomb per second, and one coulomb is the charge of about 6.24 × 10¹⁸ electrons (each carries 1.602176634 × 10⁻¹⁹ C, an exact value).

    Voltage is energy per charge. A 1.5 V cell gives every coulomb that passes through it 1.5 joules, and the parts of the circuit turn that energy into light and heat.

    Resistance tells how hard it is to push current through something. For a resistor, current and voltage are in proportion. That is Ohm’s law:

    V = I · R     and the power is     P = V · I = I² · R

    Example: 9 V across 220 Ω gives I = 9 ÷ 220 ≈ 41 mA. The ammeter in the simulation shows 40.6 mA, a little less, because the battery’s own resistance (1.5 Ω) is in the loop too.

    Series and parallel

    In series, parts are in one line, so the same current goes through each one and their resistances add. In parallel, each part has its own path between the same two points, so each gets the same voltage, and the currents add.

    Series: R = R₁ + R₂     Parallel: 1/R = 1/R₁ + 1/R₂
    Two 1.5 V cells and…Battery currentEach bulb
    one bulb0.288 A0.83 W, 92%
    two bulbs in series0.147 A0.22 W, 24%
    two bulbs in parallel0.556 A0.77 W, 86%

    Each bulb is 10 Ω and the two cells have 0.4 Ω inside them. Series: 3 V ÷ 20.4 Ω. Parallel: 3 V ÷ 5.4 Ω, shared by two bulbs. These are the simulation’s own numbers, and they match the hand calculation.

    Batteries and short circuits

    A battery has an EMF ε, its voltage when nothing is connected, and a small internal resistance r. When current flows, a little voltage is lost inside it:

    Vends = ε − I · r

    Join a battery’s ends with a bare wire and the only thing limiting the current is r. For a 1.5 V cell with r = 0.2 Ω, that is 1.5 ÷ 0.2 = 7.5 A, and all 11 W of power turns into heat inside the battery. That is why a short circuit makes a battery hot. In the simulation it only lights up red; with a real battery it can be dangerous, so never short a real battery.

    LEDs: one-way and steep

    An LED (light-emitting diode) lets current through only one way, from its + leg (the anode) to its − leg. Below its turn-on voltage, about 2 V for red, almost nothing flows. Just above it, the current shoots up very steeply:

    I = Is · (eV ⁄ (n·VT) − 1),    VT = kT ⁄ q ≈ 25.9 mV at 300 K

    So an LED needs a resistor in series to set its current. Small indicator LEDs usually run at about 20 mA:

    R = (Vbattery − VLED) ÷ I = (9 − 2.0) V ÷ 0.020 A = 350 Ω

    The next common size up is 390 Ω or 470 Ω. With 470 Ω the simulation gives 15.0 mA. The color sets the turn-on voltage: light is made in packets of energy E = hc ⁄ λ, about 1.97 eV for red light (630 nm) and 2.64 eV for blue (470 nm). Shorter wavelengths need a higher voltage, and blue LEDs need about 3 V.

    Slow electrons, fast light

    The dots are a picture, not a speedometer. In a copper wire 2 mm thick carrying 20 A, the electrons drift at only about 0.45 mm per second. A bulb still lights the moment you close the switch, because the push travels through the wire at close to the speed of light, and the electrons all around the loop start moving almost at once.

    Try this

    The open switch. Load “One bulb” and switch it off. Wire a voltmeter across the switch: it reads 3 V, the whole battery. Across the bulb it reads 0 V. Switch on and the two readings swap. An open gap gets all the voltage.
    One bulb, three cells. Put three cells end to end (4.5 V) and light one bulb. It runs at about 200%: brighter than it is made for. Real bulbs pushed past their rated voltage burn out much sooner. If the bulb gets dimmer instead, flip the new cell (⇄): each cell’s + end must touch the next one’s − end, or they push against each other.
    A crowd of bulbs. Start with “Parallel” and add a third and fourth bulb across the same wires. Each new bulb draws more current from the battery, and every bulb dims a little, because more voltage is lost inside the cells.
    Color and voltage. Load “LED + resistor”, tap the LED and change its color to blue. It still glows on 9 V. Now build an LED circuit with two cells and a 100 Ω resistor: red glows, blue barely does.
    The ammeter mistake. Wire an ammeter straight across a battery and watch the warning. An ammeter has almost no resistance, so it always goes in series, in the loop. A voltmeter has a huge resistance, so it goes across a part.

    A common mistake: “the bulb uses up the current”

    It feels as if current must run out as it goes through a bulb. It doesn’t. Put an ammeter on each side of a bulb and they read the same. Charge isn’t used up anywhere in a circuit; every coulomb that leaves the battery comes back to it. What the bulb uses is energy: each coulomb gives up some of the energy the battery gave it, and that becomes light and heat. Current measures how much charge flows each second. Voltage measures how much energy each coulomb carries.

    The same idea answers a second mix-up: a battery does not push out a fixed current. It sets a voltage, and the circuit decides the current. That is why two bulbs in parallel draw twice the current of one.

    How the simulation works

    Every change is solved with modified nodal analysis, the method inside circuit simulators. Everything joined by wires, closed switches or ammeters is one node at one voltage. For every node, the currents flowing in must add up to zero (Kirchhoff’s junction rule). That gives one equation per node, solved together as a matrix.

    Batteries are an EMF with internal resistance: 0.2 Ω for a 1.5 V cell (a fresh AA alkaline cell is 0.15 to 0.3 Ω) and 1.5 Ω for the 9 V battery, which is six small cells inside. A bulb is a fixed 10 Ω. The voltmeter is 10 MΩ, like a typical digital meter. The buzzer is a 200 Ω load that hums.

    The LED follows the diode equation above with n = 2 and a 10 Ω inner resistance, tuned so that red, yellow, green and blue need 2.0, 2.1, 2.2 and 3.2 V at 20 mA. Because the equation is so steep, it is solved by Newton’s method with the same step limit used in the SPICE simulator; it settles in about 12 to 15 steps.

    The current in each wire comes from sharing a node’s current among its wires as if every wire segment had the same tiny resistance. Burnout rules: an LED pops above 40 mA, a bulb above three times its rated power (2.7 W), each after about half a second. A resistor is marked hot above ¼ W, a common size for small resistors.

    Simplified on purpose: real bulb filaments have a much lower resistance when cold; batteries here never run down; wires have no resistance; the LED ignores reverse breakdown (real LEDs can be damaged by a few volts backwards); and everything is steady direct current, with no capacitors or time.

    How it was checked

    TestExpectedSimulation
    One bulb on two cells3 ÷ 10.4 = 0.288462 A0.288462 A, exact to 15 digits
    Two bulbs in series / in parallel0.147059 A / 0.555556 Asame to 15 digits
    100 Ω + (220 Ω ∥ 470 Ω) on 9 V35.8059 mA35.8059 mA
    Unbalanced bridge of five resistorsmesh analysis, done separatelyagrees to 1 part in 10¹⁰ or better
    Red LED + 470 Ω on 9 Vbisection on the diode equation14.98 mA, agrees to 1 part in 10¹²
    Red and blue LEDs side by side, 100 Ω, 9 Vnested bisection64.0 mA red, 1.3 µA blue, both agree
    Wire across one cell1.5 ÷ 0.2 = 7.5 A7.5 A, flagged as a short
    V ÷ I from the two meters, 220 Ω220 Ω219.995 Ω (the voltmeter draws a trickle)
    Current in = current out, at every hole0within 10⁻¹³ A
    Battery power = power in parts + inside batteries0 differencewithin 10⁻¹² W
    400 random boardsall solvedall converged; worst junction error 4 × 10⁻¹⁴ of the largest current

    Sources

    @teddyrookbook GitHub