10 Common MYTHS about ELECTRONICS.

[@linustechtips420]

Learnt a lot today! Thanks Ron!

[@beachdancer]

The penultimate one was beyond my knowledge.

[@MaximAnnen-j1b]

Battery capacity myth busting is only for spherical horse and some batteries, but not for all. In practice, capacity is rated at specific (and for marketing purposes very low) current, at least for typical non-rechargeable AA/AAA, for rechargeables in the car example can be busted maybe...). Connecting (at least cheapest AA) in series will decrease their capacity greatly due to higher current.

[@xrafter]

Getting a computer myth inside of an electronics myth is unexpected.

[@johnaweiss]

The title of this video is a myth.

[@ExtremalMetal]

thanks about the last one specifically

[@dreamswinger9328]

Very Nice Ron...! Keep it up....!!

[@shadowpapito]

Thank you for the presentation

[@sylwesterirla9246]

thank you

[@Amy_Price]

Yeap, I know your other channel(youtube algorithms), “Major Tom Workshop”… Good Luck=)👍

[@SergeySimeonov]

Thanks for sharing it :) Really nice channel.

[@Amy_Price]

@@SergeySimeonov My pleasure!

[@kamal92606]

👍 Thank u

[@JazzBerri]

15:56 what kind of power supply is that? I would like to know the reference pleasee!

[@waqasahmad55202]

6:45 I'm still confused about this battery capacity concept !😢

[@wisteela]

Excellent video

[@xrafter]

To be fair car igniters won't work with any 12v supply, it needs something capable of of more than 450A for a few seconds. I don't much about car ignitors other than that, but I do lnow that imdustrial igniters with 12000v transformers for boilers is common. So the voltage is still important to make a spark happen.

[@Eltctronique]

It s a frequence modulator right ?

[@MaximAnnen-j1b]

22:39 USB 3.1 not same as USB PD 3.1

[@johnaweiss]

Capacity is measured in watts?

[@YourTeddy999Roblox]

This logic is incorrect. You stated that current will go out of both paths and that the amount would be reversed to the capacity of the resistors and showed how the lower resistor passed a more amps that the bigger resistance as evidence BUT this is expected. What you have is simply A + B = C… if you have wire with 500 ma and a 50 ma then you should see 450 ma out the end of the wire. The same is true of the other wire with the bigger resistor.