Timeline for Explantation for fried charger with Multimeter
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2015 at 17:11 | comment | added | George | Oops, I can't read! :P If you had a 100 ohm resistor, the current would drop significantly. From V=IR -> I = V/R, you can see 9/100 is 0.09A, which is a tiny amount of current. Remember the higher the resistance the lower the current, as the two are inversely proportional. You can buy resistors on ebay for about 99p. I bought a set of about 30 types for £2 from China, and they are accurate. They are useful. Here is an example: ebay.co.uk/itm/291399638610. I would NOT buy this though, as the lowest value is 10 ohms, which is slightly too high. Just have a look round on ebay :) | |
| Jul 1, 2015 at 17:09 | history | edited | George | CC BY-SA 3.0 |
I cannot read ;)
|
| Jul 1, 2015 at 13:31 | comment | added | NULL | "+1" thank you for the clarification! I said the voltage was 9v though, so it would be 9 ohms, correct? What would happen if I used a slightly higher value resistor(I would hate to have to buy a 9 ohm when I could use a 10ohm on hand!)? The measurement would be slightly incorrect though right? | |
| Jul 1, 2015 at 6:50 | history | answered | George | CC BY-SA 3.0 |