# lecture23 - 24 LRC series circuit 1 Impedance Current i(t...

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24. LRC series circuit 1) Impedance ( 29 ( 29 ( 29 + = - - = + + = φ ϖ ϖ ϖ t V t V V t V v v v t v C L R C L R cos sin cos C L L R IX VC IX V IR V = = = Current i(t) is the same in all elements of the series circuit. Note! There is no current inside the capacitor, but we can apply Kirchhoff’s rules taking into account displacement current ( 29 ( 29 ( 29 = - = - = + = = = t V t V v t V t V v t V v t I t i C C C L L L R R ϖ π ϖ ϖ π ϖ ϖ ϖ sin 2 / cos sin 2 / cos cos cos R L C ( 29 t v IZ V = ( 29 ( 29 2 2 2 2 1 C L R X X R Z C L ϖ ϖ - + = - + R C L R X X C L ϖ ϖ φ 1 tan - = - = Impedance: φ cos R Z =

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( 29 ( 29 IZ V X X R I V V V V C L C L R = - + = - + = 2 2 2 2 ( 29 2 2 C L X X R Z - + R X X C L - = φ tan - = - = IR IX IX V V V C L R C L φ tan ( 29 ( 29 ( 29 φ ϖ ϖ ϖ + = - + = + + = t V t V V t V v v v t v L C R C L R cos sin cos Calculations: ( 29 , cos sin cos 2 2 φ + + = - x B A x B x A φ φ sin cos 2 2 2 2 = + = + B A B B A A where A B = φ tan , where Z X X z R C L - = = φ φ sin cos Trigonometry:
2) Impedance and phasor I t ϖ R V L V C V C L V V - φ V ( 29 2 2 C L R V V V - + = R C L V V V - = φ tan ( 29 ( 29 ( 29 - = + = = = 2 / cos 2 / cos cos cos π ϖ π ϖ ϖ ϖ t V v t V v t V v t I t i C C L L R R

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3) Power in AC circuit ( 29 ( 29 ( 29 + = + = + = = t t t IV t t t IV t V t I iv p ϖ ϖ φ φ ϖ ϖ ϖ φ φ ϖ ϖ φ ϖ cos sin sin cos cos cos sin sin cos cos cos cos 2 φ φ cos cos 2 1 rms rms V I IV p = = Recall that z R = φ cos 0 0 = = p R ( 29 0 cos sin ; 2 / 1 cos sin sin cos cos cos 2 = = - = + x x x y x y

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