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Unformatted text preview: 75 ) I Re ( C 1 L R I 2 2 O O = & & Â¡ Â¢ Â£ Â£ Â¤ Â¥ + = & & Â¡ and & & & & Â¡ Â¢ Â£ Â£ Â£ Â£ Â¤ Â¥ = R C 1 L Tan 1 & & Â¢ which clearly reproduces eqs (4) & (5) . The physical current in the circuit is, of course, the real part of the phasor I in eq (16). I.4. Parallel LCR circuit : Fig 3 Consider now the parallel LCR circuit shown in fig 3. The current through the resistor can be found by calculating the equivalent impedance of the circuit. C 1 L C L j R Z Z Z Z R Z 1 Z 1 1 Z Z L C C L L C R & & = + + = + & & Â¡ Â¢ Â£ Â£ Â¤ Â¥ + =(17) Thus ( ) Â£ & & & & Â¡ Â¡ + = & & & & Â¡ Â¢ Â£ Â£ Â£ Â£ Â¤ Â¥ = = t j O t j O e I C 1 L C / L j R e Z I(18) The magnitude of current I O is given by 2 2 O O C 1 L C / L R I & & & & Â¡ Â¢ Â£ Â£ Â£ Â£ Â¤ Â¥ + = & & Â¡(19) Viewed as a function of & , it is clear that I O is now a minimum (the impedance in the denominator is maximum ) when & L = 1 / ( & L) , or ,where 76...
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This note was uploaded on 11/26/2011 for the course PHY 2053 taught by Professor Lind during the Fall '09 term at FSU.
 Fall '09
 LIND

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