Danno, is the drawing below correct? That the main difference between the two circuits is the current that falls on the diode? In circuit (II) a negligible current and therefore it will not heat up, nor burn?
View attachment 45302
It's actually opposite. When you have parallel circuits, each one operates independently and draws its own current regardless of other. Let's back up and look at single circuit 1st, say only front winker light. First, specs on factory winker bulbs is
12v 21w
We'll use these equations
Watts=VI and
Volts=IR
21w=12v*I
I=21/12=
1.75 amps
12v=1.75*R
R=12/1.75=
6.86 ohms
So factory winker bulbs have 6.86-ohms resistance and when powered by 12v, draws 1.75-amps for 21-watts consumption. Idealised, if you actually measure resistance of bulb, it won't be 6.86-ohms when cold, but it will be when filament heats up to operating temps. Right-front winker circuit looks like this, it draws 1.75-amps for 21w consumption.
Now, what would happen if we connected
another winker light to bike, say… for right-rear? In Y-split configuration...
It would make sense that each circuit would draw 21w on its own? And “trunk” wire that feeds both circuits would have to supply
double power 42w? To simplify in schematic-drawing, we would combine both into parallel circuit, each one consuming its own 21w
independently from each other.
Independent and separate is important because you can disconnect either one and it won’t affect other in any way. If you pull rear winker-bulb out, front would still work.
Same if your front winker-bulb died, rear would still work.
It's like they have their own wires going directly to battery. Such as adding fog-lights to your bike or heated-grips. Each circuit is separate and independent of any other parallel circuit already on bike. Each additional circuit you add, results in more power-consumption...
It’s actually possible to calculate how much current is flowing in trunk wire that feeds multiple parallel circuits with:
So we know resistance of 1st circuit = 6.86 ohms. Second circuit using same bulb would also be 6.86 ohms.
R1 = 6.86 ohms, R2=6.86 ohms
1/Rtotal=1/6.86 + 1/6.86 = 2/6.86 = 1/3.43
Rtotal = 3.43 ohms
Let’s calculate current-draw and power-consumption:
V=IR, 12=I3.43
I=12/3.43 =
3.50 amps
W=VI, W=12*3.50
Watts = 12*3.50 =
42 watts
We see that effective or equivalent resistance of parallel circuit is lower than either one individually. This is resistance if we replaced
both 21w bulbs with
single 42w bulb. Battery and trunk wire won’t care if it’s 2x21w bulbs or single 42w bulb. Resistance (equivalent) is same and identical 3.50-amp current would flow.
Parallel circuit makes sense? Really important to get this before we move onto why LED bulbs needs ballast-resistors and how to wire them up properly.
