Series and Parallel Circuit Simulator

How current and voltage behave in series and parallel circuits

In a series circuit the same current flows through every component and the battery’s voltage is shared between them, so the resistances add: R = R₁ + R₂ + R₃. In a parallel circuit every branch gets the full voltage and the branch currents add, so the reciprocals add: 1/R = 1/R₁ + 1/R₂ + 1/R₃. A bulb’s brightness follows the power it converts, P = I²R, which is why the same bulb glows differently in the two.

This simulator builds four circuits from a battery, bulbs or resistors and switches, and solves each one exactly with Kirchhoff’s circuit laws (1845): the current into every junction equals the current out, and the voltages round every loop add up to the battery’s. It writes them as one set of simultaneous equations by modified nodal analysis, the method of Ho, Ruehli and Brennan (1975) that circuit simulation programs use, so it also handles the bridge circuit that the series and parallel rules cannot reduce.

Using the simulator

Choose a circuit: series, parallel, series and parallel combined, or a Wheatstone bridge. Set the battery’s EMF, its internal resistance and each load’s resistance, and open or close the switches with their controls, by tapping them on the drawing, or with the keys 1, 2 and 3 once the drawing has focus. The voltmeter hangs across one load and the ammeter sits in line with it; choose which under Meters on, tap a load, or move them with the arrow keys.

The moving dots are the current, drawn in the conventional direction from the battery’s positive terminal round to its negative one, moving faster where more current flows. Bulbs are drawn brighter as their power rises and are fully lit at 3 W. The table under the drawing lists the voltage, current and power of every load, and the lines under it state Kirchhoff’s laws with the circuit’s own numbers.

Worked example: one bulb in series with two in parallel

The simulator opens with three identical 12 Ω bulbs on a 6 V battery: bulb 1 in series with bulbs 2 and 3, which are in parallel with each other. The parallel pair is 1/(1/12 + 1/12) = 6 Ω, so the whole circuit is 12 + 6 = 18 Ω and the battery supplies I = 6/18 = 0.333 A. Bulb 1 carries all of it and takes 0.333 × 12 = 4 V, which leaves 2 V across the pair. Each of bulbs 2 and 3 then carries 0.167 A, half the current, and the powers are 1.33 W for bulb 1 and 0.333 W for each of the others, so bulb 1 is by far the brightest.

Now open S3. Bulb 3 goes out, the circuit becomes two 12 Ω bulbs in series, 24 Ω, and the current falls to 6/24 = 0.25 A. Bulb 1 dims from 1.33 W to 0.75 W, but bulb 2 brightens from 0.333 W to 0.75 W, because it now has 3 V across it instead of 2 V. Switching a bulb off made another one brighter: removing a parallel branch raised the circuit’s resistance, which moved voltage off bulb 1 and onto bulb 2.

Why bulbs in series are dimmer than bulbs in parallel

Three identical 12 Ω bulbs in series on 6 V share it equally, 2 V each, and carry 0.167 A, so each converts 0.333 W, a ninth of the 3 W one bulb alone would. In parallel each bulb has the whole 6 V across it, carries 0.5 A and converts 3 W, as bright as a single bulb, while the battery has to supply 1.5 A, nine times the series circuit’s 0.167 A. Parallel wiring is why the lights and sockets in a house each get the full supply voltage and switch on and off without affecting one another, as the parallel circuit’s S2 and S3 show.

Kirchhoff’s laws and the Wheatstone bridge

The bridge is two voltage dividers side by side, loads 1 and 2 on the left and 3 and 4 on the right, with load 5 joining their midpoints. When R₁/R₂ = R₃/R₄ the two midpoints sit at the same voltage and no current flows through load 5 whatever its resistance, which is how a bridge measures an unknown resistance against three known ones. Try 10, 20, 15 and 30 Ω: 10 × 30 = 20 × 15, and bulb 5 stays dark.

Unbalance it with 10, 20, 30, 40 and 50 Ω on 6 V and load 5 carries 7.74 mA from the left midpoint to the right. No two of those resistors are in series or in parallel, so there is no way to add them up. Kirchhoff’s current law at each midpoint gives two equations in the two unknown midpoint voltages, and solving them gives the battery current, 0.286 A, which the series and parallel rules cannot.

What internal resistance does

A real cell has resistance inside it, r, so the current it drives is I = E/(R + r) and the voltage at its terminals is V = E − Ir, less than its EMF whenever current flows. Set 1 Ω on the opening circuit and the current falls to 6/19 = 0.316 A with 5.68 V at the terminals. The more current a circuit draws, the more voltage is lost inside the cell, so adding bulbs in parallel no longer leaves the others untouched.

What this does not cover

Every load is a fixed resistance. A real filament bulb is not ohmic: tungsten’s resistance rises steeply as it heats, so a bulb on a low voltage draws more current than a fixed resistance would, and brightness here stands in for power rather than modelling light. The meters are ideal, the voltmeter drawing no current and the ammeter adding no resistance, and the wires and closed switches have none either. Nothing changes with time: a DC circuit settles the moment a switch moves, so for what happens during that moment, with a capacitor in the circuit, use the RC charge and discharge simulator.

Common mistakes

• Thinking current is used up by the bulbs. The ammeter reads the same on either side of a series bulb. What the bulb converts is energy, carried by the current, not the current itself.
• Adding parallel resistances directly. Two 12 Ω bulbs in parallel make 6 Ω, less than either one, because the second branch gives the current another path.
• Connecting a voltmeter in series or an ammeter in parallel. The voltmeter goes across a component and the ammeter in line with it, as the drawing shows.
• Expecting a bigger resistance to be brighter in parallel. With the same voltage across every branch, P = V²/R, so the smallest resistance converts the most power. In series it is the other way round, P = I²R with the same current.
• Assuming an open switch has no voltage across it. With S1 open in the series circuit no current flows, the bulbs have 0 V across them, and the whole 6 V sits across the gap in the switch.

To total a network of resistors quickly, use the resistor network calculator; for one component at a time, the Ohm’s law calculator; and to practise, the electricity practice problems.