
#1
Apr2905, 07:19 PM

P: 5

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 



#2
Apr2905, 08:43 PM

Sci Advisor
HW Helper
P: 3,033





#3
Apr2905, 08:57 PM

P: 2,223

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. 



#4
Apr2905, 10:12 PM

P: 5

circuitsthat 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... 



#5
Apr2905, 10:16 PM

Emeritus
Sci Advisor
PF Gold
P: 2,977





#6
Apr2905, 10:21 PM

P: 5

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?




#7
Apr2905, 10:24 PM

P: 5

so...
R=10 V=10 I=V/R hence I=10/10? meanning P=IV P=(1)10...? 



#8
Apr2905, 10:30 PM

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HW Helper
P: 3,033





#9
Apr2905, 10:52 PM

P: 5

how do i calcuate voltyage drop arcross the 6ohm? do i use P=I^2R?




#10
Apr2905, 11:10 PM

Sci Advisor
HW Helper
P: 3,033

You used V = IR solved for I to find the current of 1amp. Now you know the current through the 6 ohm resistor. Use V = IR to find the voltage across that one resistor.



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