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CIRCUITS and SYSTEMS – part I Prof. dr hab. Stanisław Osowski Electrical Engineering (B.Sc.) Projekt współfinansowany przez Unię Europejską w ramach Europejskiego Funduszu Społecznego. Publikacja dystrybuowana jest bezpłatnie Lecture 3 Power in electric circuits Instantaneous power p(t) Definition p(t ) u (t )i(t ) At sinusoidal signals: u(t) = Um sin(ωt), i(t) =Im sin(ωt-φ) p (t ) u (t )i (t ) U m I m sin( t ) sin( t ) UmIm cos cos( 2t ) 2 U I [cos cos( 2t )] Active (average) power P Definition 1 P T t 0 T p(t )dt t0 For sinusidal signals P U I cos cos - power coefficient Unit of active power: wat [W] Active power (cont.) • Resistor (φ=0) P U I cos R I 2 GU 2 • Inductor and capacitor / 2 PL PC 0 Active power on reactive elements (L & C) is always zero. Reactive power Q • Definition Q U I sin • Unit of reactive power: war [var] • Resistor 0 QR 0 • Inductor QL U I sin X L I • Capacitor 2 1 U XL 2 1 QC U I sin X C I U XC 2 2 Reactive power dissipated in resistor is always zero Apparent complex power S Definition S UI * P jQ Unit of apparent power: [VA] a) b) Phasor diagram for circuit of the a) induktive charakter, b) capacitive charakter Apparent power (cont.) Module = apparent power S U I P2 Q2 Power coefficient P cos S Principle of conservation of power Power flow from sources (Ss) in a network equals the power flow (Sl) into the other elements of the network Ss=Sl In this principle the directions of currents and voltages are the same for sources and opposite for passive RLC elements . If we assume the unified directions of currents and voltages (currents and voltages directed opposite irrespective of the type of element) the principle of conservation of power may be written in the form: Ss+Sl=0 Example Determine the currents and powers of elements in the circuit at sinusoidal excitation e(t)=141 sin(ωt+45o), at ω=1. Assume: R=Ω, C=0.5F, L=1H. Solution Complex description of circuit elements: j 45o E 100e Z L jL j1 Z C j1 / C j 2 Equivalent impedances: Z RL RZ L j 45o 0.707e R ZL Z Z C Z RL 0,5 j 0,5 j 2 1,58e j 71, 6o Solution (cont.) The succeding steps of solution j 45o E 100e j116, 6o IC 63,3e o j 71 , 6 Z 1,58e U C Z C I C 126,6e j 26, 6o U RL Z RL I C 44,72e j161, 6o U RL j 71, 6o IL 44,72e ZL U RL j161.6o IR 44,72e R Phasor diagram circuit Powers of elements Apparent power of the source S E I C (2000 j 6000)V A * Powers of passive elements 2 PR I R R 2000W QL Im(U RL I L* ) 2000 var QC Im(U C I C* ) 8000 var Balance of powers: – Receivers S o PR QL QC 2000 6000 – Source S E I C (2000 j 6000)V A * Energy stored in capacitor General expression t W (t0 , t ) p( )d t0 Energy of capacitor du ( ) W (t0 , t ) u ( )i( )d u ( )C d C udu d t0 t0 u (t 0 ) t t u (t ) At zero initial condition of capacitor u (t ) 1 2 1 u(t) U W (t 0 , t ) C udu Cu (t ) W CU 2 2 2 0 Energy stored in inductor General expression t W (t0 , t ) p( )d t0 Energy of inductor i (t ) t t di( ) W (t 0 , t ) u ( )i ( )d i ( ) L d L idi d t t i (t ) 0 0 i (t ) At zero initial condition of 1 inductor 2 W (t 0 , t ) L idi 0 2 0 I Li (t ) i(t) W 1 2 LI 2