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TWO ELECTRIC BULBS WHOSE RESISTANCES ARE IN THE RATIO OF 1:2ARE CONNECTED IN PAALLEL TO ACONSTANT VOLTAGE SOURCE . THE POWER DISSIPATED IN THEM HAVE THE RATIO
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|dw:1367217537922:dw|
2:1, I guess
Since \[P=\frac{ V ^{2} }{ R }\]
|dw:1367217925507:dw|
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\[I_1=2I_2\]
\[P=IV\]
You could sum the resistors first, to find... whatever it is you need to find, you didn't really say...?\[\Large \frac{ 1 }{ R _{T} }= \frac{ 1 }{ R _{1}}+\frac{ 1 }{ R _{2} }\] Where R1 and R2 are the R and 2R.
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