Circuits in series and parallel.

Step 1 of 6

In series

On the board are three bulbs, A, B and C, of 6 Ω each, and a 12 V battery. In a series circuit the current has only one path: it leaves the battery, passes through the bulbs one after another and comes back. Every electron has to squeeze through all three resistances, so the total resistance is their sum:

$R = R_A + R_B + R_C$⁠

Ohm’s law, applied to the total resistance, gives the current from the battery. Measure the circuit is fixed at Circuit. Work out the current.

Step 2 of 6

The current is not used up

The current leaves the battery and passes through the bulbs one after another. Predict: does it weaken on the way? Once you answer, the Scene moves to Currents and shows a tag with the current over each branch.

Step 3 of 6

The voltage is shared

Bulb C has been swapped for a 12 Ω bulb, and A and B are still 6 Ω. The current is still the same in every bulb, but the voltage across each bulb, $V = IR$⁠, depends on its resistance. Each coulomb gains 12 J in the battery and hands all of it over on the way round, so the voltages across the bulbs add up to the battery’s voltage:

$V_A + V_B + V_C = V$⁠

What is the voltage across bulb C?

Step 4 of 6

In parallel

In a parallel circuit each bulb is connected straight across the battery’s two terminals and has a path of its own. So each bulb has the full voltage, 12 V, across it, and the current in each branch is $\dfrac{12}{6} = 2\ \mathrm{A}$$\dfrac{12}{6} =$$2\ \mathrm{A}$⁠. At a junction where branches meet, the current flowing in equals the current flowing out, so the current from the battery is the sum of the branch currents.

Switch for bulb C is open now, and only A and B are lit: the battery gives 4 A. Predict what will happen to the current from the battery when you close the switch and add C.

Step 5 of 6

Resistance in parallel

Every extra branch is one more path for the current, so the total resistance of bulbs in parallel is smaller than the resistance of any one of them:

$\frac{1}{R} = \frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}$$\frac{1}{R} =$$\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C}$⁠

Bulb C is now 3 Ω, and A and B are 6 Ω. What total resistance does the battery see?

Step 6 of 6

Mixed

In Mixed, bulb A is in series with a pair: B and C, in parallel with each other. All the current passes through A, and after the junction it splits between B and C. So A shines brightest. The tags show the currents. Predict what will happen to bulb B when you open Switch for bulb C.