physics_ac_currentAlternating Current
AC Voltage applied to registor
AC Voltage and current are in same phase with each other
AC Voltage applied to Inductor
The average power supplied to an inductor over one complete cycle is zero.
AC Voltage applied to Capacitor
Capacitive reactanceXC=1πœ”C
The average power supplied to a capacitor over one complete cycle is zero.
AC Voltage applied to LCR Circuit
V=Vmsinπœ”t i=imsin(πœ”t+πœ™)
Peak Voltage VRm=imRVCm=imXCVLm=imXL
Input voltage peak value = Vm Vm2=VRm2+(VCm-VLm)2
im=VmZ i=imsin(πœ”t+πœ™)
The current phasor ( I ) is always parallel to VR πœ™ is angle between VR and V If XC>XL then πœ™ is positive and the circuit is predominantly capacitive.So in this case the current in the circuit leads the source voltage.
If XC<XL then πœ™ is negative and the circuit is predominantly inductive.So in this case the current in the circuit lags the source voltage.
Frequencies at which the response of the amplitude is maximum is known as resonant frequencies.Current amplitude is maximum at resonant frequencies.
In RLC circuitim=VmZ=VmR2+(XC-XL)2
Current amplitude is maximum when impedance (Z) is minium Z is minumim when XC=XL The frequency at which impedance (Z) is minimum is called resonant frequency.
Resonant CircuitIn AC circuit with inductor and capacitor: When frequency of AC is changed at certain frequency so that impedance becomes maximum or minimum, then that condition is called resonance and such a circuit is called resonant circuit
Series Resonance
Parallel Resonance
Sharpness of resonance
We choose πœ” where current amplitude is 12 times its maximum value. πœ”1=πœ”0+π›₯πœ”πœ”2=πœ”0-π›₯πœ”πœ”1-πœ”2=2π›₯πœ”=bandwidthofcircuit
In order to determine π›₯πœ” we consider im=immax2
At 12 times the current amplitude , power dissipated by he circuit becomes half.
Sharpness of resonance = Quality factorQ=πœ”0LR
Power in AC Circuit (RLC circuit)
Inatantenous power p suplied by sourcep=VI=(Vmsinπœ”t)Γ—(imsin(πœ”t+πœ™))p=Vmim2[cosπœ™-cos(2πœ”t+πœ™)]AveragePower=P=Vmim2cosπœ™P=Vm2im2cosπœ™P=VIcosπœ™P=I2Zcosπœ™
LC Oscillation
Capacitor is charged and connected to a conductor.
Transformer works using principal of mutual induction.
Primary Coil= It is input coil.Number of turns=Np
Secondary Coil= It is output coil.Number of turns=Ns
Induced emf in secondary coilVS=-NSdπœ™dt
Back emf in primary coilVP=-NPdπœ™dt
Transformer equationVSVP=NSNP
Assumption in transformer equationa. primary resistance and current are smallb. same flux links both primary and secondaryc. the secondary current is small.
For 100% efficiencyiPVP=iSVsiPiS=VSVP=NSNP
Energy losses happens in transformer due to followinga. Flux Leakage: There is always some flux leakage; that is, not all ofthe flux due to primary passes through the secondary due to poordesign of the core or the air gaps in the core. It can be reduced bywinding the primary and secondary coils one over the other. b. Resistance of the windings: The wire used for the windings has someresistance and so, energy is lost due to heat produced in the wire (I2R) . In high current, low voltage windings, these are minimised byusing thick wire. c. Eddy currents: The alternating magnetic flux induces eddy currentsin the iron core and causes heating. The effect is reduced by having alaminated core. d. Hysteresis: The magnetisation of the core is repeatedly reversed bythe alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss.
Rectification is the process in which an alternating current is forced to only flow in one direction.
Half-wave Rectification:a. Dalf wave rectifier is allows one half-cycle of an AC voltage waveform to pass, blocking the other half-cycle.b. half of the power is lost in half-wave rectification.
Full-wave Rectificationa. Full wave rectifier converts both halves of each cycle of an alternating wave (AC signal) into pulsating DC signal.
Questions and Answer
Common Questions
Q.What is rectification?A.Rectification is the process in which an alternating current is forced to only flow in one direction.
Q.A capacitor is used in the primary circuit of an induction coil. Why ?A.When the circuit is broken, then a large induced voltage (that comes from unduction coil) is used up in changing capacitor. This helps in avoiding spark.
Q.7.1A 100Ξ© resistor is connected to a 220 V, 50 Hz ac supply.(a) What is the rms value of current in the circuit?(b) What is the net power consumed over a full cycle?
Q.7.7A charged 30 ΞΌF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? A.7.7πœ”0=1LCπœ”0=127Γ—10-3Γ—30Γ—10-6=181Γ—10-8πœ”0=19Γ—10-4=1.1Γ—103rad/sec
Q.7.18 A circuit containing a 80 mH inductor and a 60 ΞΌF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.(a) Obtain the current amplitude and rms values.(b) Obtain the rms values of potential drops across each element.(c) What is the average power transferred to the inductor?(d) What is the average power transferred to the capacitor?(e) What is the total average power absorbed by the circuit? [β€˜Average’ implies β€˜averaged over one cycle’.] A.7.18(a)Capacitance=C=60Γ—10-6FInductance=L=80Γ—10-3Ffrequency=50Hzπœ”=2πœ‹πœˆ=2Γ—3.14Γ—50=314capacitivereactance=XC=1πœ”CXC=1314Γ—60Γ—10-6=53.1inductivereactance=XL=πœ”LXL=314Γ—80Γ—10-3=25.12Impedance=Z=R2+(XC-XL)2=27.98Vrms=230VVm=2302Im=VmZ=230227.98=11.623AIrms=VrmsZ=23027.98=8.22A(b)RMSPotentialdropacrosscapacitance=V1V1=IrmsΓ—XC=8.22Γ—53.1=436.482VRMSPotentialdropacrosscapacitance=V2V2=IrmsΓ—XL=8.22Γ—25.12=206.486V(c)The average power supplied to an inductor over one complete cycle is zero.Paverage=-imVm2γ€”sin2πœ”t⦘valueofγ€”sin2πœ”t⦘is0inonecycle(d)The average power supplied to an capacitor over one complete cycle is zero.Paverage=imVm2γ€”sin2πœ”t⦘valueofγ€”sin2πœ”t⦘is0inonecycle(e)AveragePower=I2Zcosπœ™πœ™=tan-1XC-XLRhereR=0,πœ™=90Β°so,cosπœ™=cos90Β°=0so,AveragePower=I2Zcosπœ™=0
Q.7.20A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ξ© is connected to a 230 V variable frequency supply.(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?(d) What is the Q-factor of the given circuit? A.7.20(a)L = 0.12 HC = 480Γ—10-9F im=VmR2+(XL-XC)2=VmR2+(πœ”L-1πœ”C)2imismaximumwhenπœ”=πœ”0=1LC πœ”0=10.12Γ—480Γ—10-9=112Γ—12Γ—4Γ—10-10πœ”0=112Γ—2Γ—10-5=4167rad/secfrequency=𝜈=πœ”2πœ‹=41672Γ—3.14=663Hzimmax=VmRVm=V2=2302immax=230223=14.14A (b)Power=P=VIcosπœ™Formaximumpower,cosπœ™=1MaximumpowerwillhavewhenXL=XCandsourcefrequencyis663HzV=230VPmax=V2R=230Γ—23023=2300W (c)At 12 times the current amplitude , power dissipated by he circuit becomes half.So πœ” at which power will be dissipated becomes half will beπœ”=πœ”0Β±π›₯πœ”π›₯πœ”=R2L=232Γ—0.12=96rad/secπœ”1=4167+96=4263Hzπœ”2=4167-96=4071Hz𝜈1=πœ”12πœ‹=678Hz𝜈2=πœ”22πœ‹=647Hz (d)Qualityfactor=Q=πœ”0LRQ=4167Γ—0.1223=21.74
Q.7.23A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?A.7.23VSVP=NSNP⟹2302300=NS4000⟹NS=400turns