Every parallel branch has the full source voltage across it. That was the last lesson's rule. The rule raises a question: why does each branch get the full 6 V, when a series circuit makes its components share?
The answer starts with the battery's job. First, recall what a voltage states:
Each coulomb moving between two points of a circuit transfers 3 J. What is the voltage between the points?
A 6 V battery gives each coulomb of charge that passes through it 6 J of energy.
Now follow one coulomb around the two-lamp circuit. It leaves the battery, reaches the junction, and takes one branch. Say it takes the branch holding lamp L1.
The coulomb passes through L1 only. It never enters L2's branch on this trip.
On this trip the coulomb passes through just one component: L1. So it transfers all 6 J in L1. Transferring 6 J per coulomb is what a voltage of 6 V means. So the voltage across L1 is the full 6 V.
A coulomb that takes L2's branch does the same thing there. It transfers all 6 J in L2. So L2 also has the full 6 V across it.
A battery gives each coulomb 9.0 J. A coulomb takes the left branch, through one lamp. How much energy does it transfer in that lamp, in J?
A 12 V battery drives a motor and a resistor on parallel branches. A coulomb takes the motor's branch. How much energy does it transfer in the motor?
The 12 V battery gives each coulomb 12 J.
The coulomb passes through only the motor on its trip, so it transfers all 12 J there.
The coulomb transfers 12 J in the motor.
Why does each parallel branch get the full source voltage?
A third branch holding a buzzer is added to the same 12 V circuit. How much energy does a coulomb taking the buzzer's branch transfer in the buzzer?
The battery still gives each coulomb 12 J.
A coulomb on the buzzer's branch passes through only the buzzer, so it transfers all 12 J there.
The coulomb transfers all 12 J in the buzzer, so the voltage across the buzzer is the full 12 V.
Compare the two circuit types. In a series circuit, each coulomb passes through every component in turn, so its 6 J is shared between them. In a parallel circuit, each coulomb passes through only the one branch it takes, so that branch gets all 6 J.
Elena says two parallel lamps on a 10 V battery get 5 V each, because the 10 J each coulomb carries must be shared between the lamps. Which fact shows the claim is wrong?
A 24 V supply gives each coulomb 24 J. A coulomb passes through the one fan on its parallel branch. How much energy does it transfer in the fan?
In a series circuit, a coulomb's 8 J is shared between two lamps. Why is a coulomb's energy not shared between the lamps of a parallel circuit?
A third lamp branch is added to a 15 V parallel circuit. How much energy does a coulomb now transfer in whichever lamp it passes through?