## circuits

ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s?

ok so what i did was that.... and i dont know where to go after... please help
Attached Thumbnails

 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Recognitions:
Homework Help
 Quote by poopoo16 ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s? ok so what i did was that.... and i dont know where to go after... please help
That's a good start. Now what about the 6 ohm resistor?
 Reduce the circuit to an equivalent resistor, find the flowing current, and from there use your power equations to find the power. You are on the right track.

## circuits

 Quote by OlderDan That's a good start. Now what about the 6 ohm resistor?

that s my question... dont know how to treat the 6 ohms... do i add it to the 4?

and then use P=IV where I=V/R?...if i do that, its give me 1...and that is not the right answer...

Recognitions:
Gold Member
Staff Emeritus
 Quote by poopoo16 that s my question... dont know how to treat the 6 ohms... do i add it to the 4?
Yes, because it's in series with the other two. Be careful, however, in the power calculation. Is the current flowing through the 20 ohm resistor the same as through the whole circuit?
 well, after u combine the 2 parallel ones, its no longer going thru the 20 per se right, its going thru a 6 and a 4?...soo... it then a series right?
 so... R=10 V=10 I=V/R hence I=10/10? meanning P=IV P=(1)10...?

Recognitions:
Homework Help