Hey,
Both diodes are pointing in the same direction, right?
Maybe it will become clear to me when I get a short explanation of a diode. 🙂
LG
Hey,
Both diodes are pointing in the same direction, right?
(No Ratings Yet)
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Very good question.
If the diode were not there in front of lamp 1, both lamps would light (lamp 1 would, however, be significantly weaker, because only the voltage applied to it is applied, which also falls over the top diode. Depending on the diode, this would be only about 0.5-1V).
However, since there is also a diode in front of the lamp 2, the same voltage drops off at this diode as at the upper diode, and thus only a voltage of 0 V “usual” remains for the lamp 1, so it no longer lights.
The upper branch (red in the picture in my answer) has virtually no resistance compared to the branch in the middle (blue in my answer).
Accordingly, the entire current flows through the upper branch, instead of through the lower branch.
The branch in the middle thus becomes through the upper branch short-term, so that hardly any current flows in the middle branch (in any case not enough for lamp 1 to light), since the current prefers to take the upper path.
Hi Finn!
This is because electricity always takes the path with the least resistance. In the case at the top, because in the middle a lamp that has a lot of resistance would have to be ” overcome”.
It doesn’t fit to the explanation and is also incorrectly formulated… Electricity flows wherever possible.
This is true, but if a path has a greater resistance than another, the current always takes the path with the slightest resistance. Of course, a little power also drops off at the other, only this clamping is not enough to bring the LED to run. In principle, you’re right, but for the sake of simplicity, I put it that way. Thank you!
However, in the case of a parallel connection of two different resistors, current flows through both, and not only through the one with the lower resistor. But what he should do if the idea “takes the path with the least resistance” would be correct.
And, socket strips should not work…
About 0.7 volts drop across the upper diode. This voltage is equal to the sum of the voltages which drop together via diode and lamp underneath.
Since about 0.7 volts also drop above the diode below, about 0.7 volts minus 0.7 volt -> 0 volt are left for the lamp.
The flow voltage of about 0.7V is present above the upper diode. Also on the lower diode. There’s about 0V left for the lamp.