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Unformatted text preview: = ( I rms E rms cos φ ) T = µ 1 2 T ¶ E m I cos φ where we substitute I rms = I/ √ 2 and E rms = E m / √ 2. (d) The energy dissipated by the resistor is P avg , resistor T = ( I rms V R ) T = I rms ( I rms R ) T = µ 1 2 T ¶ I 2 R . (e) Since E m I cos φ = E m I ( V R / E m ) = E m I ( IR/ E m ) = I 2 R , the two quantities are indeed the same....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy

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