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Sun 9 Oct, 2005 04:28 pm
http://img383.imageshack.us/my.php?image=physics27tn.jpg
Use Kirchoff's rules to find:
A. The current going through each resistor.
B. The power dissipated in each resistor.
Answer:
R1--- I=1.143A P=9.145W
R2--- I=1.143A P=3.919W
R3--- I=1.143A P=5.226W
I don't know what's wrong with me as in like why i cant figure out these power dissipated things or things with a bunch of batteries. but i just get so confused. so if you could..please help.
p.s. i know the answers, thats not what i need. i need help on how to do it.
thank you very very much.
Simple circuit rules
You have a nice, simple circuit with everything in series. I can tell you the rules that apply to this diagram, but the application of these rules gets much more challenging in a harder circuit and you will need more rules.
Resistances in series add together.
Voltage sources in series add together.
The total resistance for this diagram is the sum of all the resistances or 14 ohm. For the voltages, you have to pay attention to their orientations. These are actually fighting against each other, like if you put a battery in backwards. The total is 18 volts. Look at the long and short lines in the picture to see that the 24 is oriented differently from the 2 and 4 volt supplies, so those two are negative. From this point, it should be straight forward. I=V/R. Since the current is going around the loop, each resistor sees the same current. The power dissipated by each resistor is P=R*I^2.
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