DELMAR’S Standard Textbook of Electricity - 5th Edition Unit 25 – Resistive-Inductive-Capacitive Parallel Circuits 1) RLC Parallel Circuits a) Voltage is the same across all branches b) Currents flowing through each branch are out of phase with each other i) Resistive element – voltage in phase with current (1) True power ii) Inductive element – voltage out of phase with current by 90° – current lagging (ELI) (1) Reactive power – VARsL iii) Capacitive element – voltage out of phase with current by 90 – current leading (ICE) (1) Reactive power – VARsC c) If VARsL > VARsC then circuit current lags the applied voltage and the power factor is a lagging power factor d) If VARsC > VARsL then circuit current leads the applied voltage and the power factor is a leading power factor e) Work examples: 25-1, 25-2 from the text 2) Calculating Circuit Values a) Total Impedance/Z i) = b) Total Current c) = d) Resistive Current i) = e) Voltage Drop Across the Resistor f) = × Notes g) True Power/Watts i) = × h) Inductive Current i) = i) Inductive VARs i) = × j) Inductance i) = !" k) Voltage Drop Across the Inductor i) = × # l) Inductive VARs (1) VARsL = EL x I m) Capacitive Current i) IC = EC/XC n) Capacitance i) $ = !" o) Voltage Drop Across the Capacitor i) EC = IC x XC p) Capacitive VARs i) VARsC = EC x IC q) Total Circuit Current i) = %& ' + & − * ' r) Apparent Power/VA i) VA = ET x IT ii) = %&' + & − * ' s) Power Factor , i) + = -. × 100 t) Angle Theta i) Cos ∠θ = PF ii) Example: Cos ∠θ = .8 (1) ∠θ = 36.87° 3) Parallel Resonant Circuits (aka “tank” circuits) a) Resonance i) The one specific frequency at which XC = XL (1) 1 = !√* ii) For a parallel circuit at resonance, IC and IL are equal and 180˚ out of phase with each other and cancel each other out (1) Minimal line current at resonance (2) Maximum impedance at resonance (a) Limited by the quality (Q) of the inductor (b) Q = XL/R b) Very high currents circulate within the “tank” i) Useful when high load currents are required with minimal line current (1) Induction heating 4) Power Factor Correction – (Step-by-step) a) Consider the following scenario i) An electric motor is connected to a 240 VAC line at a frequency of 60 Hz ii) An ammeter indicates 10 amps of current flow iii) A wattmeter indicates a true power of 1630 watts at full load b) How much capacitance should be connected in parallel to achieve a PF correction to 95 percent? i) VA = E x I VA = 240 x 10 VA = 2400 ii) PF = P/VA PA = 1630/2400 PF = .67917 (1) PF = 0.67917 X 100 (a) PF = 67.9 percent iii) 3 = √ − (1) 3 = √2400 − 1630 (a) 3 = 1761.562 iv) VA = P/PF (1) VA = 1630/.95 (a) VA = 1715.789 v) 3 = √ − (1) 3 = √1715.789 − 1630 (a) 3 = 535.754 vi) 3* = 3&=>?@?AB' - 3&A??C?C' (1) 3* = 1761.562 − 535.754 (a) 3* = 1225.808 vii) #* = -.@ DE (1) #* = F.GEG (a) #* = 46.989 Ω viii) $ = !" (1) $ = !&HE'×&DH.IGI' (a) $ = 56.45J+