trying to figure out how to calculate power distribution with mixed impedance. i have 3 13 ohm (weird huh)loads in parallel with a single 4 ohm load. final resistance is 2 ohms which gives them a shared 100w of power. i know it's not the best thing to mix impedance but it doesn't really matter with this project. just wanna make sure one load isn't seeing 70w and the others seeing like 10wea.

2/28/2009 4:38:33 AM

You have 3 parallel 13 ohm loads in parallel with a 4 ohm load (i.e. 4 parallel loads; 3, 13 ohm and 1, 4 ohm)?

Using in-my-head math (not known for accuracy):

13 ohm loads: ~16.7 W each
4 ohm load: ~50 W

[Edited on February 28, 2009 at 9:47 AM. Reason : ]

2/28/2009 9:46:54 AM

cool that actually works out they way i wanted. just for future refernce what's the formula u use to get those numbers?

2/28/2009 12:15:33 PM

http://en.wikipedia.org/wiki/Current_divider

2/28/2009 12:29:39 PM

^

2/28/2009 12:43:22 PM

^^

P_total = 3 * (V^2) / (R_13) + (V^2) / (R_4)

P_total = 3 * (V^2) / (R_13) + 3.25 * (V^2) / (R_13)

P_total = 6.25 * (V^2) / (R_13)

P_total = 6.25 * P_13

so...

P_13 = 16 W
P_4 = 52 W

2/28/2009 3:10:05 PM