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circuits |
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| Apr29-05, 07:19 PM | #1 |
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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 |
| Apr29-05, 08:43 PM | #2 |
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Recognitions:
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| Apr29-05, 08:57 PM | #3 |
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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. |
| Apr29-05, 10:12 PM | #4 |
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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... |
| Apr29-05, 10:16 PM | #5 |
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| Apr29-05, 10:21 PM | #6 |
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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?
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| Apr29-05, 10:24 PM | #7 |
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so...
R=10 V=10 I=V/R hence I=10/10? meanning P=IV P=(1)10...? |
| Apr29-05, 10:30 PM | #8 |
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| Apr29-05, 10:52 PM | #9 |
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how do i calcuate voltyage drop arcross the 6ohm? do i use P=I^2R?
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| Apr29-05, 11:10 PM | #10 |
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Recognitions:
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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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