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Solve : If a MB dies, is it always worth replacing the PSU as well?? |
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Answer» upon doing some research, we are both correct. the example you provided... Here you go: ...would be correct if the power supply had a current limiter set at 30 amps. the voltage would drop to compensate for the increased amperage. however, if it did not have a current limiter, the voltage would REMAIN constant and it would draw 60A, thereby either blowing the power supplies fuse or blowing the power supply itself. so it seems our argument is based on if power supplies have a current limiter or not. if they do, then yes, there will be voltage drops under load if a rated 300 watt faulty PSU can only supply 10 watts. if they dont have current limiters, then, using the same faulty PSU scenario there will be no voltage drop even under load, the PSU will simply blow a fuse. personally, i would rather have a dead power supply then a dead mothboard, graphics card, RAM, etc, caused by an undervoltage. Quote Pick a voltage. 12v. your formula is INCORRECT. I=E/R, not E x R I= E/R I= 12/0 I= ∞ anyways i believe, if you WISH to continue this discussion, we should do so via PM to avoid cluttering up the thread. i wish to continue the discussion of the 12V BATTERY example later, but as i have stated before, i am not really in the best health and it does have an effect on my ability to concentrate.Quote from: giorgio652 on March 06, 2008, 03:21:53 AM Theory being, most MB's die due to the PSU sending unstable voltage so it is fairly likely to happen again, so for the price of an extra £20 you and the customer will have piece of mind. I don't know where your pal got that theory from. Anyway, rather than giving customers peace of mind, you're more likely to have them coming back in 6-12 MONTHS to give you a piece of their mind, if you only spend £20 on the power supply. Unless you're getting them in bulk, I suppose. Good ones cost £40 - £60 I have found. By the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery. Quote from: Dias de verano on March 08, 2008, 07:52:26 AM By the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery. ...we know.Quote from: homer on March 08, 2008, 11:20:41 AM Quote from: Dias de verano on March 08, 2008, 07:52:26 AMBy the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery. Homer, I know that you know. Quote from: homer on March 07, 2008, 09:35:39 PM ... Right. Probably a typo as I re-composed. Quote anyways i believe, if you wish to continue this discussion, No. I already told you that you appear to want to simply argue with me, and I'm not interested in that. You've lost the point. All these things are examples - trying to explain the original point to you. And it still stands: When testing a power supply, simply checking for proper voltage in a no-load condition is certainly better than nothing, it is not conclusive. To see if a source of power is good, it should be checked while operating under a load. I've tried to explain it to you. Maybe I'm not doing a good job of that. Quote from: Dias de verano on March 08, 2008, 11:23:34 AM Quote from: homer on March 08, 2008, 11:20:41 AMQuote from: Dias de verano on March 08, 2008, 07:52:26 AMBy the way, when you short a battery, the current that flows is determined by the (non-zero) resistance of the short and the internal resistance of the battery. Do you have a point, or anything worthwhile to add? Quote I already told you that you appear to want to simply argue with me i am not arguing for the sake of arguing, i am arguing because alot of your examples you used go against what i have been taught. this is why i asked what your education is in this. i would like to know if i am having a discussion with someone who has been professionally taught at a university or college or someone who simply looked it up on the internet. Quote simply checking for proper voltage in a no-load condition is certainly better than nothing, it is not conclusive. as i stated before, it depends if the supply has a current limiter or not. if it does, then the voltage will drop once a load is placed on it that exceeds the current limit. if it does not, then the voltage will remain constant regardless if a load is placed on it or not. anyways, im feeling a little healthier today and i would like to bring up the battery example that you provided. you stated that voltage would fall to zero in the event of a short. i stated that was incorrect because then it would be impossible to destroy a battery. you rebuttled by saying in theory i was correct, so i assume you mean in practicality i was incorrect. however, these formulas work in theory and in practicality. you have a 12v battery with a dead short. in theory a dead short is 0 ohms, however 0 ohms will never be achieved due to the resistance of the battery and the resistance of the material used to short the terminals. now lets use your assumption that voltage falls to zero in the case of a short circuit. im going to use 0.001 ohms as the cumulative resistance of the battery and the material shorting the terminals V=IR I=V/R I=0/0.001 I=0 if there is no current, a battery will not catch fire or explode if the terminals are shorted. lets try it again with 12 volts being maintained. V=IR I=V/R I=12/0.001 I=12000 12000A will destroy a battery, 12000A will melt leads, 12000A will start the battery on fire and that is exactly what happens when you short a battery in reality. this proves that ohms law works both in theory and in practicality. to sum it up...no voltage=no current. no current=no short. your battery example is incorrect. Quote from: WillyW on March 08, 2008, 01:06:21 PM Do you have a point, or anything worthwhile to add? Well, now! If we're getting snippy, what point, or additional information did your post add, WillyW? |
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