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Two resistors in a Bosch AL 1115 CV battery charger 127V burned up yesterday, and I am trying to figure out the correct value to replace them.

I have found in the internet this photo where I suppose that these resistors are of the same value of that burned in my board:

enter image description here

This is a image of a similar board that I found on the internet:

https://radikal.ru/fp/b0963f27083043e185cba4facee7ad1c

This is a photo of my real board, that was burnt:

enter image description here

In that two resistors that burnt, in the similar board figure, I see the colors:

brown, white, red, red, black.

What is the value of this resistor?

I have learned that in 5 bands resistors, the first, second and third band are the value, the fourth id the multiplier and the fifth is the tolerance.

Black does not exist as a tolerance and in some five band resistors the third band is the multiplier, and not the fourth.

How do J know when take into account the third band as multiplier or the fourth band as multiplier?

What is the real value of the resistors in the figure?

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    \$\begingroup\$ Does this answer your question? How do I read these blasted five-band resistors? \$\endgroup\$
    – Elliot Alderson
    Jan 20 at 23:37
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    \$\begingroup\$ Fourth band looks to be silver rather than white. That should make things easier for you to figure out. \$\endgroup\$
    – Dwayne Reid
    Jan 21 at 0:20
  • \$\begingroup\$ In addition to the band looking silver instead of white, are those really two red bands, or is one of them orange? \$\endgroup\$
    – Justme
    Jan 21 at 5:28
  • \$\begingroup\$ I add a full image of the board where u could compare the orange color that is hard to see because the low resolution of the image. \$\endgroup\$
    – Guilherme
    Jan 21 at 13:14
  • \$\begingroup\$ Taking in consideration that the fourth band are silver, so this resistor are something around 1.22 Ohm? Of fact, is few probably that are a white multiplier because white multiplier have a value of 9 zeroes and it is impracticable. What do you think? \$\endgroup\$
    – Guilherme
    Jan 21 at 13:19

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