Electrical Energy Part 2

What if we change it so that the two bulbs are not one after the other, but rather one or the other

This makes two parallel paths for the electrons to travel down

Each path has just one bulb

So each path has only 10Ω  of resistance

The electrons split at the junction. Those that go down the split off at the first junction will only encounter one bulb, then return to the battery

Those that carry on past the first junction will go to the other bulb, that is the only bulb they will encounter, and then they will also return

Half of all the electrons will go down the first path to the first bulb

Half of all the electrons will go down the second path to the second bulb

Notice that these bulbs are each as bright as the single bulb circuit. Why?

Well, first we need to calculate the current for the first bulb using V=IR

V=IR 

I = V/R

9V/10Ω = 0.9A

And for the second bulb

V=IR 

I = V/R

9V/10Ω = 0.9A

Now lets calculate power for each bulb using P=IV

Bulb 1

P=IV

9V x 0.9A = 8.1W

Bulb 2

P=IV

9V x 0.9A = 8.1W

So each bulb has the same power as the single-bulb circuit

In parallel circuits, the voltage is the same down each path, and the current is split

Well, the current looks the same as the original 1-bulb circuit, but look at the current leaving and returning to the battery. It is the sum of the individual currents

0.9A + 0.9A = 1.8A

Thus the power of the whole circuit is

IV = P

1.8A x 9V = 16.2W 

If the battery is transforming 16.2 joules of energy per second, in this parallel setup, then the battery will go flat faster with the parallel circuit than it will in the series circuit. 

Notice also that a large returning current. As we keep adding bulbs, they all stay the same 9V/10Ω = 0.9A, but the returning current keeps increasing.

Now, if we add 10Ω batteries in parallel, they continue to follow the pattern of 9V, 10Ω, 0.9A. 

Notice that the current to the battery always equals the sum of the currents in each of the parallel branches.  

When we have 10 bulbs, the current across each path remains 0.9A; they continue to follow the pattern of 9V, 10Ω, and 0.9A per path. 

But because there are now 10 paths, the current going into and out of the battery is the sum of 10 x 0.9A = 9Amps

If we double the Voltage to 18V, then the current will also double 

The current across each path is now 1.8A. And with 10 paths, the total current is 1.8A x 10 = 18A

Notice that the bulbs are brighter. This is because by doubling the voltage across each bulb, we have quadrupled the energy transformed at each bulb. What that is is the joules of electrical energy transformed into heat and light energy at each bulb in each second. Watt

It was:

P=IV

VI=P

9V x 0.9A = 8.1W

It is now

18V x 1.8A = 32.4W

So four times as bright

The total power for the circuit

It was: 

8.1W x 10 bulbs = 81W 

or VI=P

9Vx9A = 81W

Now with double the voltage, it is

VI=P

18V x 18A = 324W

So by doubling the voltage, the total power of the circuit has quadrupled

You might also notice something interesting in the screenshot

The battery is on fire

High voltage will not cause a fire, but high Current will. The current is high enough in this example to melt the wires connected to the battery and start a fire

This is why there is a limit to how many multiadaptors you can plug into a circuit - too many and they will fuse OR catch on fire. 

You should always buy a multiadaptor that has a fuse

Modern fuses are just a bar of metal that bends as it gets hotter, as it bends it disconnects the circuit and breaks the circuit - It'll cool back down quickly - this allows you to unplug a few devices then continue using the multiadaptor - cheap multiadaptors do not have this - so pay more and save your house from a fire (fuses in multiadaptors are not a legal requirement - hence the 52 house fires in NZ caused by multiadaptors)

In the oldern days and in old houses, the fuse was a piece of wire - I used to have this type in my house

The video below is a demonstration of how they worked