basic part deux - RF Cafe Forums
Because of the high maintenance needed to monitor and filter spammers from the RF Cafe Forums, I decided that it would be best to just archive the pages to make all the good information posted in the past available for review. It is unfortunate that the scumbags of the world ruin an otherwise useful venue for people wanting to exchanged useful ideas and views. It seems that the more formal social media like Facebook pretty much dominate this kind of venue anymore anyway, so if you would like to post something on RF Cafe's Facebook page, please do.
Below are all of the forum threads, including all the responses to the original posts.
Post subject: basic part deux
Unread postPosted: Wed Sep 15, 2004 3:56 am
ok next question
10v supply going to 2 series resistors the first one is a 10ohm, but the second one is blank . what will the value be for max power transfer and what wil that power be? I know P=(I*I)(R) and that is about all i know
thanks for any help
Unread postPosted: Wed Sep 15, 2004 12:37 pm
If the power transfer is to the second resistor (ie no 3rd resistor exists) it's a well known fact that maximum power transfer occurs for matched impedences (R2 = R1 = 10ohm)
The power dissipated in that resistor is then simply ((V/2)^2)/R due to the voltage dividing effect, where V is the input voltage.