My raspberry pi doesn't work after I execute "gpio -g write 1 0"

Question

Sorry for my poor English.

I just bought the raspberry PI 3B+ a few days, don't know what the gpio -g write 1 0 mean.After try this, my pi doesn't work now. the green led is also not lit.

I have an LED light, I want to be able to light it, then I connect a wire to the first pin, which is +3.3V, the other wire is connected to GND, the light is successfully lit, then I Google searched for gpio -g write to customize the output of the pin, and thought that 1 represents the first +3.3V pin, so I want to test whether it can be used by gpio -g write 1 0 The output is set to 0, so that LED should be extinguished, but when I execute it, there is no current output on both pins of +3.3V, it completely fails, and PI does not work.It will not start automatically when it is powered on, only the red light is always on and the green light is not lit.

Another small supplement, I am connected to the circuit when the Raspberry Pi is started, and plug and unplug multiple times.

It seems that it really burned completely.

Is there any way to save it? Buying a Raspberry Pi is a big expense for a student.

Answer 1

It would have been a better idea if you had found one of the perhaps hundreds of tutorials about how to light an LED and followed it to the letter. Unfortunately, because you did not, you likely have (as Milliways observes in comments) destroyed the device.

I connect a wire to the first pin, which is +3.3V

This was a bad idea. That pin is not a GPIO (general purpose input output) pin. It is a power pin connected directly to the 3.3V system on the Pi. Although I don't think this powers any of the components directly, I don't think it is fused either, meaning if you short this pin to ground, you could create a current surge which would extend from the 5V system which is connected to the SoC (which steps it down to 1.2V, but is not surge protected either).

the other wire is connected to GND

Again, you might have been saved if you had followed a tutorial and used a resistor in the circuit, which is a necessity with an LED. Since you did not, you effectively shorted the power to ground.

gpio -g write 1 0

There are two numbering systems used for the breakout. The predominant one is the "Broadcom numbering", in which the numbers correlate to the GPIO numbers from the SoC. This is the one you find on most pinout diagrams, such as the one here, and as observed there, "the numbering of the GPIO pins is not in numerical order". You will also notice there is no pin 1, and there is an explanation of why in the same comment.

The other, much less used, numbering system, is the physical one, which simply numbers the pins in order on the breakout. This one may be a bit misleading, because it could imply these are all GPIO numbers, but not all the pins are GPIOs (the power pins and ground are not).

Now I connect the line as before, the LED will not be lit. [...] But connected to the 5V output, the LED can still be lit

This means the LED is okay, which is too bad, because it also means the 3.3V system is not okay.

Regardless of which components are and are not damaged, you will not be able to repair the device. Please learn from this and next time:

1. Follow some directions if you do not know what you are doing.

2. If you are uncertain about something, ask before you do it.