Elect principles 2 ac circuits parallel
- 1. AC Circuits – The Parallel R, L Circuit In parallel circuits the voltage is common to each branch of the network and is therefore taken as the reference phasor. For the two branch network shown the current flowing in the resistance, IR, is in- phase with the supply voltage, V. The current flowing in the inductance, IL, lags the supply by 90º. The supply current, I, is the phasor sum of IR and IL and lags the applied voltage by an angle between 0º and 90º. Ø = tan-1 IL IR I = IL 2 + IR 2 IR = V R V θ IL = V XL Z = V I R L IL I IR V
- 2. AC Circuits – The Parallel R, C Circuit Ø = tan-1 IC IR I = IC 2 + IR 2 IR = V R V θ IC = V XC For the two branch network shown the current flowing in the resistance, IR, is in- phase with the supply voltage, V. The current flowing in the capacitance, IC, leads the supply by 90º. The supply current, I, is the phasor sum of IR and IC and leads the applied voltage by an angle between 0º and 90º. Z = V I R I IR V C IC
- 3. AC Circuits – Parallel Networks Activity 1. A 20Ω resistor is connected in parallel with a 2.387mH inductance across a 60V, 1kHz supply. Calculate a) the current in each branch, b) the supply current, c) the circuit phase angle, d) the circuit impedance and d) the power taken from the supply. 2. A 30µF capacitor is connected in parallel with an 80Ω resistor across a 240V, 50Hz supply. Calculate a) the current in each branch, b) the supply current, c) the circuit phase angle, d) the circuit impedance and d) the true power and the power taken from the supply (apparent power). 3. Sketch the phasor diagram for each of the above.
- 4. AC Circuits – The Parallel R, L and C Network For the three branch network shown the current flowing in the resistance, IR, is in- phase with the supply voltage, V. The current flowing in the inductance, IL, lags the supply by 90º. The current flowing in the capacitor, IC, leads the supply by 90º The supply current, I, is the phasor sum of IR and IL and lags the applied voltage by an angle between 0º and 90º. R L IL I IR V C IC Ø = tan-1 ( IC – IL ) IR Z = V I I = IR 2 + ( IC 2 – IL 2 ) IR = V R V θ IL = V XL IC = V XC IC - IL
- 5. AC Circuits – Parallel Networks Activity 1. A 50Ω resistor is connected in parallel with a 150mH inductance and a 100µF capacitance across a 100V, 50Hz supply. Calculate a) the current in each branch, b) the supply current, c) the circuit phase angle, d) the circuit impedance and d) the power taken from the supply. 2. A 35µF capacitor is connected in parallel with a 200mH inductor and a 20Ω resistor across a 240V, 60Hz supply. Calculate a) the current in each branch, b) the supply current, c) the circuit phase angle, d) the circuit impedance and d) the true power and the power taken from the supply (apparent power). 3. Sketch the phasor diagram for each of the above.
- 6. AC Circuits – The True Parallel Circuit The true parallel circuit represents a more common practical system with the coil resistance as a series element to the inductive branch and a parallel connected capacitance. Current IC leads the applied voltage by 90º and is shown in phasor IC. The current IL through the inductance lags the applied voltage by an angle dependant on the value of its resistance, R. R L IL I V C IC V IC = V/XC V is the reference for the parallel capacitance branch IL VR = ILR VL = ILXL V Ø IL is the reference for the series R/ L branch Since V is a common phasor to both diagrams we can combine them to provide a diagram for the complete circuit V IC IL VR I Ø The supply current is now the phasor sum of IC and IL with the phase angle Ø to the applied voltage.
- 7. IL is the reference for the series R/ L branch AC Circuits – The True Parallel Circuit In this arrangement there are two series circuits connected in parallel. We can analyse this through two separate phasor diagrams each with the current as the reference phasor. Then superimpose them making the supply voltage the reference phasor. R1 L IL I V C IC Since V is a common phasor to both diagrams we can combine them to provide a diagram for the complete circuit The supply current is now the phasor sum of IC and IL with the phase angle Ø to the applied voltage. R2 IL VR1 = ILR1 VL = ILXL V Ø1 IC VR2 = IC R2 VC = IC XC V Ø2 IC is the reference for the series R/ C branch I V IC Ø2 IL Ø1 Ø
- 8. Parallel AC Circuits – Summary • Each branch is determined individually as a series circuit. • Individual phasor diagrams are superimposed. • At resonance the current is at its minimum. • Parallel resonant circuits are known as ‘rejector’ circuits. • Rejector circuits provide a high impedance at the resonant frequency. • The impedance at resonance is called the ‘dynamic’ impedance.