9 Amazing Electrical Tricks That Will Make You Level 100 Master

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The coax cable repair shown will introduce an impedance mismatch at the location where it is spliced. Probably not an issue on receive but could be an issue for a transmitter. To keep the impedance you'd have to use BNC or N type connectors instead of the splice shown.

4: No need to cut a pencil. You can use one old battery instead.

@ number 2, just buy a proper plug with tension relief. It’s default on any non-garbage plug.

Nunca vi un control remoto que pudiera funcionar con la mitad de la tension de diseño

Putting uncrimped stranded wire into a screw terminal is enough to ban you from any kind of master title for all eternity.

Semua teknik di konten video ini super kereeeeen

Ad 1: mindblowing Ad 2: the twist and extra stress can damage the metal wires. However the risk would be very small, no problem. Ad 3: nice idea! Ad 4: this might work if the remote can operate with just single battery. Which is sometimes the case, there are step up circuits. But even then a wire would be better than a pencil. Graphite has high resistance. Ad 5: another nice idea Ad 6: the last option is not better than the first one. The second option works better because there is longer path of the wire which would need to slide. Friction is therefore much higher and so the force needed to pull it out is higher. The third option does not extend the path much. Only the small part on right needs to move. And once it is pulled out then only the small part at left has to move. They don't get stressed at once. The force therefore does not go up much. And having two pieces of wire next to each other could actually decrease the friction if the metal piece is not reasonably flat. Option 3 is not an improvement. Use options 1 or 2. Ad 7: this one often works. What you do here is replacing a LED by a wire. LEDs cause a voltage drop. Wire does not. In order to provide a light there must be a current flowing through the LED. Too high current means it burns out. Now let's imagine you have power source of 48V and you use 12 LEDs where each one has 3V drop. Combined they do 36V drop. Each needs 20mA of current. To limit the current you use a resistor which would be 12V / 20mA = 600 ohm. The resistor would dissipate 0.24W of heat. Now, to the point. You replace one LED by a wire. So the voltage drop of the remaining 11 LEDs is only 33V. The voltage on the resistor is up to 15V. It will give current of 25mA and power dissipation would be 0.375W. Will the LEDs survive that current increase? And won't the resistor overheat? Real situation is even worse as the manufacturers want to decrease the power lost on the resistors and so use LEDs which have voltage drop lower than the source, but still as closest to it as possible. Replacing even a single LED by a wire could then easily mean the resistors would overheat and bad things could happen. Do not do this unless you have the schema and skills to evaluate the risk involved. Ad 8: works on non-critical wiring. Which means 99.9% of wiring around us. No problem with this either. Ad 9: another perfect idea!

مبدع والله وننتظر منك المزيد وبارك الله فيك

brilliant mate. Australia

Excelente trabajo

Trik dan tips yg sangat menakjubkan

I am spaniard, from Cartagena Spain. ... and I am very pleased for your teaching videos ... thanks a lot

Thanks

電線不要纏繞會產生電感效應，固定電線要用壓接端子

ماشاءالله،عالی بود، آفرین

❤ very useful ❤

Good job! 👍💯👍💯👍💯

very good

Step 4.....JUST USE THAT EMPTY BATTERY !!