user40603: I am trying to use the Adafruit Metro Mini as Arduino (can be used the same way) to program an IR Break Beam sensor and a stepper motor. I connected the beam sensor to the breadboard and plugged the Arduino into my computer. But  my MAC pops up with the notification:

"USB Device Disabled - Unplug the device using too much power to
  re-enable USB Devices."

What is the problem causing the issue? Did I short circuit something? I attached photos to my circuit to the post.


Thanks so much in advance. 

user32827: The stepper motor driver is installed with the error (shifted) to the breadboard.
The electrolytic capacitor at the output of the voltage stabilizer in reverse polarity.
Power to  Adafruit Metro Mini is put in inverse polarity ...

40603: <p>I am trying to use the <a href="https://www.adafruit.com/product/2590" rel="nofollow noreferrer">Adafruit Metro Mini</a> as Arduino (can be used the same way) to program an <a href="https://learn.adafruit.com/ir-breakbeam-sensors" rel="nofollow noreferrer">IR Break Beam sensor</a> and a stepper motor. I connected the beam sensor to the breadboard and plugged the Arduino into my computer. But  my MAC pops up with the notification:</p>

<blockquote>
  <p>"USB Device Disabled - Unplug the device using too much power to
  re-enable USB Devices."</p>
</blockquote>

<p>What is the problem causing the issue? Did I short circuit something? I attached photos to my circuit to the post.</p>

<p><a href="https://i.stack.imgur.com/GxbPf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GxbPf.jpg" alt="enter image description here"></a> 
<a href="https://i.stack.imgur.com/XOPTi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XOPTi.jpg" alt="enter image description here"></a></p>

<p>Thanks so much in advance. </p>
 32827: <p>The stepper motor driver is installed with the error (shifted) to the breadboard.</p>

<p>The electrolytic capacitor at the output of the voltage stabilizer in reverse polarity.</p>

<p>Power to  Adafruit Metro Mini is put in inverse polarity ...</p>
