Tiêu đề 07. ALTERNATING CURRENT.pdf ✅

As we have seen earlier a rotating coil in a magnetic field, induces an alternating emf and hence an alternating current. Since the emf induced in the coil varies in magnitude and direction periodically, it is called an alternating emf. The significance of an alternating emf is that it can be changed to lower or higher voltages conveniently and efficiently using a transformer. Also the frequency of the induced emf can be altered by changing the speed of the coil. This enables us to utilize the whole range of electromagnetic spectrum for one purpose or the other. For example domestic power in India is supplied at a frequency of 50 Hz. For transmission of audio and video signals, the required frequency range of radio waves is between 100 KHz and 100 MHz. Thus owing to its wide applicability most of the countries in the world use alternating current. AC circuit containing pure resistance Let at any instant t, the current in the circuit = I. Potential difference across the resistance = IR with the help of Kirchhoff’s circuital law E – I R = 0  E0 sin t = IR  I = E0 R sin ω t = I0sint (I0 = peak or maximum value of current) Alternating current developed in a pure resistance is also of the sinusoidal nature. In a.c. circuits containing pure resistance, the voltage and current are in the same phase. The vector or phasor diagram which represents the phase relationship between alternating current and alternating e.m.f. are as shown in figure. a.c. circuit having R only, as current and voltage are in the same phase, hence in fig. both phasors E0 and I0 are in the same direction, making an angle t with OX. Their projection on Y–axis represent the instantaneous values of alternating current and voltage. CHAPTER – 7 ALTERNATING CURRENT ALTERNATING CURR ENT
i.e. I = I0 sint and E = E0 sint Since I0= E0 R , hence I0 √2 = E0 R√2  Irms= Erms R AC circuit containing pure inductance A circuit containing a pure inductance L (having zero ohmic resistance) connected with a source of alternating emf. Let the alternating e.m.f. E = E0 sin ω t When a.c. flows through the circuit, emf induced across inductance = −L dI dt Hence, I = I0 sin (ωt − π 2 ) In a pure inductive circuit current always lags behind the emf by π 2 or alternating emf leads the a.c. by a phase angle of π 2 . Expression I0 = E0 ωL resembles the expression E I = R. This non–resistive opposition to the flow of A.C. in a circuit is called the inductive reactance (XL) of the circuit. XL= L = 2  f L where f = frequency of A.C. Unit of XL : ohm (L) = Unit of L × Unit of (= 2f) = henry × sec–1 = volt ampere/sec × sec−1 = volt ampere = ohm Inductive Reactance XL∝ f Higher the frequency of A.C. higher is the inductive reactance offered by an inductor in an A.C. circuit. For d.c. circuit, f = 0  XL = L = 2  f L = 0 Hence, inductor offers no opposition to the flow of d.c. whereas a resistive path to a.c. AC circuit containing pure Capacitance A circuit containing an ideal capacitor of capacitance C connected with a source of alternating emf as shown in fig. The alternating e.m.f. in the circuit E = E0 sin t. When alternating e.m.f. is applied across the capacitor a similarly varying alternating current flows in the circuit. The two plates of the capacitor become alternately positively and negatively charged and the magnitude of the charge on the plates of the capacitor varies sinusoidally with time. Also the electric field between the plates of the capacitor varies Negative sign indicates that induced emf acts in opposite direction to that of applied emf. Because there is no other circuit element present in the circuit other than inductance so with the help of Kirchhoff’s circuital law E + (−L dI dt) = 0  E = L dI dt so we get I = E0 ωL sin (ωt − π 2 ) Maximum current I0 = E0 ωL × 1 = E0 ωL Note
sinusoidally with time. Let at any instant t charge on the capacitor = q Instantaneous potential difference across the capacitor E = q/C  q = CE  q = CE0 sin  The instantaneous value of current I = dq dt = d dt (CE0 sin ω t) = CE0 cos  I = E0 (1/ωC sin (ωt π 2 )= I0 sin (ωt + π 2 ) where I0 = CE0 In a pure capacitive circuit, the current always leads the e.m.f. by a phase angle of 2. The alternative emf lags behinds the alternating current by a phase angle. Ac Voltage Applied to A Series LCR Circuit A circuit containing a series combination of an resistance R, a toil of inductance Land a capacitor of capacitance C, connected with a source of alternating e.m.f. of peak value of E0, as shown in figure. wt. As L, C and R are joined in series, therefore, current at any instant through the three elements has the same amplitude and phase. However, voltage across each element bears a different phase relationship with the current. Let at any instant of time t the current in the circuit is I. Let at this time t the potential difference across L, C, and R VL= I XL, VC= I XC and VR= IR Now, VR is in phase with current I but VL leads I by 90° While VC lags behind I by 90°. The vector OP: represents VR (which is in phase with I) the vector OQ represent VL (which leads I by 90°) and the vector OS represents VC (which legs behind I by 90°) VL and VC are opposite to each other. If VL > VC (as shown in figure) the their resultant will be (VL –VC) which is represented by OT. Finally, the vector OK represents the resultant of VR and (VL – VC), that is, the resultant of all the three ≃applied e.m.f. Thus E = √VR 2 + (VL − VC) 2 = I√R2 + (XL − XC) 2  I = E √R2+(XL−XC) 2 Impedance Z = √R2+ (XL − XC) 2 = √R2 + (ωL − 1 ωC) 2 The phasor diagram also shown that in LCR circuit the applied e.m.f. leads the current I by a phase angle  tan φ = XL−XC R Resonance A circuit is said to be resonant when the natural frequency of circuit Is equal to frequency of the applied voltage. For resonance both L and C must be present in circuit. Series Resonance (a) At Resonance (i) XL = XC (ii) VL = VC (iii)  = 0 (V and I in same phase) (iv) Zmin= R (impedance minimum) (v) Imax = V R (current maximum) (b) Resonance frequency XL = XC  rL = 1 ωrC  ωr 2 1 LC   = 1 √LC  fr = 1 2π√LC (c) Variation with f (i) If f < fr then XL < XC circuit nature capacitive,  (negative) (ii) At f = fr then XL = XC circuit nature, Resistive,  = zero (iii) If f > fr, then XL > XC circuit nature is inductive,  (positive) (d) Variation of I with f as f increase, Z first decrease then increase as f increase, I first increase then decreases At resonance, impedance of the series resonant circuit is minimum so it is called 'acceptor circuit' as it most readily accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning we receive the desired station by making the frequency of the circuit equal to that of the desired station.
Half power frequencies The frequencies at which, power become half of its maximum value is called half power frequencies Band width= f = f2 – f1 Quality factor Q : Q-factor of AC circuit basically gives an idea about stored energy & lost energy. Q = 2π maximum energy stored per cycle maximum energy loss per cycle (i) It represents the sharpness of resonance. (ii) It is unit less and dimension less quantity (iii)Q = (XL)r R = (XC)r R = 2πfrL R = 1 R √ L C = fr Δf = fr band width Magnification At resonance VL or VC= QE (where E = supplied voltage) So at resonance Magnification factor = Q–factor Sharpness of resonance Sharpness  Quality factor  Magnification factor R decrease  Q increase  Sharpness increases Power In an AC Circuit In case of steady current, the rate of doing work is given by, P = VI In an alternating circuit, current and voltage both vary with time, so the work done by the source in time interval dt is given by dw = Vidt Suppose in an ac, the current is leading the voltage by an angle φ. Then we can write V = Vmsin ωt and I = imsin(ωt + φ) dw = VmImsin ωtsin(ωt + φ)dt dw = Vmim(sin2 ωtcos φ + sin ωtcos ωtsin φ)dt The total work done in a complete cycle is W = Vmimcos φ ∫ T 0 sin2 ωtdt + Vmimsinφ ∫ T 0 sin ωtcos ωtdt W = 1 2 Vmimcosφ ∫ T 0 (1 − cos 2ωt)dt + 1 2 Vmimsinφ ∫ T 0 sin 2ωtdt W = 1 2 VmimTcos φ The average power delivered by the source is, therefore P = W/T P = 1 2 Vmimcos φ P = Vm √2 im √2 cos φ P = Vrmsirmscos φ This can also be written as, P = I 2Zcosφ Here, Z is impedance, the term cos φ is known as power factor It is said to be leading if current leads voltage, lagging if current lags voltage. Thus, a power factor of 0.5 lagging means current lags voltage by 60∘ (as −1 cos−1 0.5 = 60∘ ). the product of Vrms and irms gives the apparent power. While the true power is obtained by multiplying the apparent power by the power factor cos φ. (i) Resistive circuit: For φ = 0 ∘ , the current and voltage are in phase. The power is thus, maximum. (ii) purely inductive or capacitive circuit: For φ = 90∘ , the power is zero. The current is then stated as wattless. Such a case will arise when resistance in the circuit is zero. The circuit is purely inductive or capacitive (iii) LCR series circuit: In an LCR series circuit, power dissipated is given by P = I 2Zcosφ were φ = tan−1 ( XC−XL R ) So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. (iv) Power dissipated at resonance in LCR circuit: At resonance XC −XL = 0, and φ = 0. Therefore, cos φ = 1 and P = I 2Z = I 2R. That is, maximum power is dissipated in a circuit (through R ) at resonance LC Oscillations We know that a capacitor and an inductor can store electrical and magnetic energy, respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems. Let a capacitor be charged qm(at t = 0 ) and connected to an inductor as shown in Fig. Q. In an L-C-R A.C. series circuit L = 5H, ω = 100 rad s −1 , R = 100Ω and power factor is 0.5. Calculate the value of capacitance of the capacitor Sol. Power factor cos δ = R √R2+(ωL− 1 ωL) 2 Squaring on both side Cos δ = 0.5 cos2 δ = R 2 R2+(ωL− 1 ωL) 2 1 4 = R 2 R2+(ωL− 1 ωL) 2 R 2 + (ωL − 1 ωL) 2 = 4R 2 (ωL − 1 ωL) 2 = 3R 2 ωL − 1 ωC = √3R ωL − √3R = 1 ωC C = 1 ω ( 1 ωL−√3R ) C = 1 100 ( 1 100×5−√3×100) C = 10−2 500−173.2 = 10−2 326.8 = 30.6 × 10−6 F C = 30.6μF