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Basic Concepts of A.C Physicsminds.blogspot.com A waveform is a representation of how alternating current varies with time. The most familiar AC waveform is the sine wave, which derives its name from the fact that the current or voltage varies with the sine of the elapsed time. Other common AC waveforms are the square wave, the ramp, the sawtooth wave, and the triangular wave. Their general shapes are shown below. various types of output waveforms of ac sources The principles you have learned for direct current, such as Ohm’s law, Kirchhoff’s laws and the principles used for resistors in series and parallel are also applicable to alternating current. The Sinusoidal AC Waveform The most common AC waveform is a sine (or sinusoidal) waveform. Basic Concepts of A.C Physicsminds.blogspot.com The vertical axis represents the amplitude of the AC current or voltage, in amperes or volts. The horizontal axis represents the angular displacement of the waveform. The units can be degrees or radians. A complete change of current or p.d. from a particular value and back to the same value in the same direction is known as a cycle. The time taken for a complete cycle is known as the period T of the alternating current or p.d. The number of complete cycles in one second is known as the frequency of the alternating current or p.d. In most countries, the frequency of the alternating current is 50 Hz. Peak amplitude is the height of an AC waveform as measured from the zero mark to the highest positive or lowest negative point on a graph. Also known as the crest amplitude of a wave. Peak-to-peak amplitude is the total height of an AC waveform as measured from maximum positive to maximum negative peaks on a graph. Often abbreviated as “P-P”. Basic Concepts of A.C Physicsminds.blogspot.com Phasor Diagram To account for the different phases of the voltage drops, vector techniques are used Represent the voltage across each element as a rotating vector, called a phasor The diagram is called a phasor diagram Phasor diagram-to visualize the phase of alternating quantities I and V. A phasor is a rotating vector used to represent a quantity that varies sinusoidally with time. Resistors in an ac circuit An AC circuit consists of a combination of circuit elements and an AC generator or source Consider a circuit consisting of an AC source and a resistor. Basic Concepts of A.C Physicsminds.blogspot.com The sinusoidal current in the circuit can be represented by the equation I I 0 sin 2ft The potential difference across the resistor of resistance R is V IR ( I 0 sin 2ft ) R ( I 0 R) sin 2ft V0 sin 2ft V V0 sin 2ft The graph shows the variation of the current I and the potential difference V with time t. Graph (b) shows variation of I and V with time Both achieve max at same time =in phase with each other Phasor diagram (c) -to visualize the phase of an alternating quantities I and V. I and V are in phase, hence their phasors are drawn parallel to each other and they rotate about the point O with the same frequency f. The power dissipated in a resistor of resistance R when a current I flows is given by, P I 2R For a complete cycle, mean Power = mean value of I2R = (mean value of I2) x R mean P I rms R 2 where I rms mean value of I 2 2 Hence I rms mean valueof I 2 Irms=root mean square current of the ac Basic Concepts of A.C Physicsminds.blogspot.com 2 I0 I I rms 0 2 2 V02 V0 Vrms 2 2 AC ammeters and voltmeters are designed to read rms values Example An ac voltmeter, when placed across a 12 resistor, reads 117 V. What are the maximum values for the voltage and current? Solution: Vmax 2Veff 1.414(117 V); Vmax = 165 V 117 V I max 2 I eff 1.414 ; Imax = 13.8 A 12 Example An ac current of rms value 4.0 A and frequency 50 Hz flows in a circuit containing a resistor. What is the (a) peak current and (b) the value of the current 6 x 10-4 s after it changes direction? Answer: (a) Using I 0 2I rms I0 =5.66A (b) Using I I 0 sin 2ft I=1.06 A Capacitors in an ac circuit Pure capacitor – no resistor & no self induction o The current starts out at a large value and charges the plates of the capacitor There is initially no resistance to hinder the flow of the current while the plates are not charged Basic Concepts of A.C Physicsminds.blogspot.com o As the charge on the plates increases, the voltage across the plates increases and the current flowing in the circuit decreases o The current reverses direction o The voltage across the plates decreases as the plates lose the charge they had accumulated o The voltage across the capacitor lags behind the current by 90° o The impeding effect of a capacitor on the current in an AC circuit is called the capacitive reactance and is given by: Suppose that the alternating p.d. applied to the capacitor is given by V V0 sin 2ft On capacitor, Q CV CV0 sin 2ft d CV0 sin 2ft CV0 sin 2ft CV0 (2ft ) cos(2ft ) dQ d From I dt dt dt The current in the circuit I I 0 sin(2ft ) 2 where I 0 CV0 (2f ) is the peak current. Compare (6.7) and (6.8): V and I not in phase Phase difference between V and I radians 2 Current I lead V by rad or V lag behind the current by radians 2 2 From graph (a) Imax at t = 0 and Vmax at t = T/4 Basic Concepts of A.C Physicsminds.blogspot.com phase diff = rad 2 From phasor diagram (b) when both the phasors are rotated in the direction shown at the same frequency f I always ahead by rad 2 From I0 =CV0 ( 2f ) V0 1 where V0 = 2 Vrms and I0 = 2 Irms I 0 2fC V0 2Vrms Vrms 1 X c reactance of the capacitor I0 2 I rms I rms 2fC 1 Xc 2fC Xc frequency Variation of Xc with frequency From Ohm’s law, V =IR Ohm’s Law for a capacitor in an AC circuit Vc I rms X c …(6.10) Example A 6-F capacitor is connected to a 40-V, 60 Hz ac line. (a) What is the reactance? (b) What is the effective ac current in the circuit containing pure capacitance? Solution: 1 1 (a) X C ; XC = 442 2 fC 2 (60 Hz)(6 x 10-6 F) Basic Concepts of A.C Physicsminds.blogspot.com Veff 40 V (b) I eff ; Ieff = 90.5 mA XC 442 Inductors in an ac circuit o Consider an AC circuit with a source and an inductor o The current in the circuit is impeded by the back emf of the inductor Current flow in an inductor: 0I I sin 2ft When a charging current flows in the inductor, a back emf is produced. Basic Concepts of A.C Physicsminds.blogspot.com dI back L dt Applying Kirchoff’s 2nd Law to the circuit, dI back L 0 dt dI d V L L ( I 0 sin 2ft) 2fLI0 cos(2ft) dt dt V V0 sin 2ft 2 Where the peak voltage, V0 2fLI 0 Compare (6.11) and (6.12); I I 0 sin 2ft V V0 sin 2ft 2 Voltage V leads the current I by rad 2 V0 Vrms From (6.13); V0 2fLI 0 X L 2fL I 0 I rms V f rad I 2 The effective resistance of a coil in an AC circuit is called its inductive reactance and is given by XL Ohm’s law for inductor: Vrms I rms X L Basic Concepts of A.C Physicsminds.blogspot.com Example A 50-mH inductor of negligible resistance is connected to a 120-V, 60-Hz ac line. (a) What is the inductive reactance? (b) What is the effective ac current in the circuit? Solution: (a) X L 2 fL 2 (60 Hz)(0.050 H) 18.85 ; ; XL = 18.9 Veff 120 V (b) I eff ; Ieff = 6.37 A XL 18.85 Resistors and Capacitors in Series R & C in series the same I flows in R & C Across R VR=IR VR and I in-phase Across C VC=IXC VC lags behind I by rad 2 VC lags behind VR by rad 2 Resultant V across R & C, V 2 V R2 VC2 I 2 ( R 2 X C ) 2 1 V I R2 2fC2 V 1 R2 Z = impedance of the circuit I 2fC2 1 Z R2 Unit: (ohm) 2fC2 Basic Concepts of A.C Physicsminds.blogspot.com Figure 16 shows the variation of the impedance Z with frequency. For a very high frequency (f ), the impedance Z equals the resistance R of the resistor. Phase difference between I & V : VC IX C 1 tan VR IR 2fCR Example An ac current of rms value 1.5 mA and angular frequency =100 rad/s flows through a resistor of 10 k and a 0.50 F capacitor joined in series. Calculate (a) the rms voltage(potential difference) across the resistor (b) the rms voltage across the capacitor (c) the total rms voltage across the resistor and the capacitor (d) the rms voltage across the resistor when the current is maximum (e) the impedance in the circuit Solution (a) VR =IR=15 V (b) VC = IXC =30 V (c) Using V 2 V R2 VC2 Vtotal = 33.5 V (d) Across resistor, V and I in phase, where as across the capacitor, V lags behind the current by rad. 2 Therefore, when the current is maximum, the voltage across the capacitor is zero and the voltage across the resistor is maximum. Hence the voltage across the resistor when the current is maximum V=I0R= 2I rms =21.2 V V (e) Impedance, Z rms 2.23 10 4 I rms Basic Concepts of A.C Physicsminds.blogspot.com 6.6 Resistors and Inductors in series Across R VR=IR VR and I in-phase Across L VL=IXL=I(2 fL) VL leads I by rad 2 VL leads VR by rad 2 Resultant V across R & L, V 2 VR2 VL2 I 2 ( R 2 X L ) 2 V I R 2 (2fL) 2 V Impedance of the circuit Z R 2 (2fL) 2 I V IX 2fL Phase angle between I & V : tan L L VR IR R Basic Concepts of A.C Physicsminds.blogspot.com Example A solenoid has inductance 2.5 H and resistance 6.0 . (a) Sketch graphs to show how the resistance R and the reactance X of the solenoid varies with the frequency. (b) At what frequency is the resistance of the solenoid equal to 1% of its reactance? Solution: (a) XL R=constant frequency (b) When R=1% of reactance 1 6= 2 f 2.5 f=38.2 Hz 100 Basic Concepts of A.C Physicsminds.blogspot.com Resistor, Capacitors and Inductors (RCL) in series Figure 18 When a resistor of resistance R, a capacitor of capacitance C and an inductor of inductance L are connected in series with an ac supply, the potential difference across R, C and L are, respectively, Across R VR=IR VR and I in-phase Across C VC=IXC VC lags behind I by rad 2 Across L VL=IXL VL leads I by rad 2 The magnitude of VL can be greater or less than VC. Phasor Diagram for RLC Series Circuit: o The voltage across the resistor is on the +x axis since it is in phase with the current o The voltage across the inductor is on the +y since it leads the current by 90° Basic Concepts of A.C Physicsminds.blogspot.com o The voltage across the capacitor is on the –y axis since it lags behind the current by 90° Magnitude VL can be greater or less than VC 1 2 Resultant V across R, C & L, V 2 VR2 (VL VC ) 2 I 2 [ R 2 (2fL ) 2fC V 1 2 Impedance of the circuit Z R 2 (2fL ) I 2fC (VL VC ) 2fL 2fC 1 Phase angle between I & V : tan … VR R Ohm’s Law can be applied to the impedance ΔVmax = Imax Z 6.8 Resonance In the RLC circuit, if XL = XC then the impedance becomes Z=R. 1 Since X L 2fL and X C , there would be a certain frequency known as the 2fC resonance frequency f0 where the value of XL = XC. When this happens, the impedance Z will have the smallest value and the current will be maximum. The resonance frequency can be determined as follows: 1 When XL = XC 2f 0 L 4 2 f 02 LC 1 2f 0 C 1 f0 2 LC resonance frequency of the RLC series. Basic Concepts of A.C Physicsminds.blogspot.com Figure 19 Figure 19 shows the variation of the impedance with the frequency f and (b) shows the 1 variation of the current I with the frequency. When f 0 , the impedance is 2 LC least and the current is maximum. The frequency f0 is known as the resonance frequency of the series RLC circuit. Basic Concepts of A.C Physicsminds.blogspot.com Current amplitude vs angular frequency for series RLC circuit with various R values. Example A 300- resistor, a 3-F capacitor, and a 4-H inductor are connected in series with a 90- V, 50 Hz ac source. What is the net reactance of the circuit? What is the impedance? What is the resonant frequency? Solution: 1 XC 1061 ; X L 2 (50 Hz)(4 H) 1257 ; 2 (50 Hz)(3 x 10-6 F) Net Reactance: XL – XC = 196 Z R 2 ( X L X C )2 (300 )2 (1257 1061 ) 2 ; Z = 358 1 1 fr ; fr = 45.9 Hz 2 LC 2 (4 H)(3 x 10-6 F) Basic Concepts of A.C Physicsminds.blogspot.com Power in an ac circuit o No power losses are associated with capacitors and pure inductors in an AC circuit o In a capacitor, during one-half of a cycle energy is stored and during the other half the energy is returned to the circuit o In an inductor, the source does work against the back emf of the inductor and energy is stored in the inductor, but when the current begins to decrease in the circuit, the energy is returned to the circuit o The power in any component in an electric circuit is given by:P=IV From I I 0 sin 2ft V V0 sin 2ft P IV I 0V0 sin 2 (2ft ) o Mean power, P = mean value of I 0V0 sin 2 ( 2ft ) 2 1 Vrms P I 0V0 I rms R 2 2 R In a resistor, the power dissipated in the form of heat. Basic Concepts of A.C Physicsminds.blogspot.com Example An inductor dissipates heat at a rate of 10 W when an alternating current of 0.5 A flows in it. If the reactance of the inductor at the frequency of the supply is 30 , what is its impedance? Solution: Since heat is dissipated from the inductor, the inductor is not a pure inductor and has a resistance R. Power is dissipated due to this resistance. Using P I R R= 40 2 Impedance Z R 2 X L 50 2 Example A series ac circuit consists of a 100- resistor, a 0.2-H inductor, and a 3-F capacitor connected to a 110-V, 60 Hz ac source. What are the inductive reactance, the capacitive reactance and the impedance for the circuit? Solution: 1 XC 884 ; X L 2 (60 Hz)(0.2 H) 75.4 2 (60 Hz)(3 x 10-6 F) Z R 2 ( X L X C )2 (100 )2 (75.4 884 ) 2 ; Z = 815 Example When a 6- resistor and a pure inductor are connected to a 100-V, 60 Hz ac line, the effective current is 10 A. What is the inductance? What is the power loss through the resistor and what is the power loss through the inductor? Solution: 110 V Z 11.0 ; X L Z 2 R 2 (11.0) 2 (6 ) 2 ; 10 A XL (9.22 ) XL = 9.22 L ; L = 24.5 mH 2 f 2 (60 Hz) P = I2R = (10 A)2 (6.00 ); P = 600 W; No Power lost in inductor: 0 Basic Concepts of A.C Physicsminds.blogspot.com Basic Concepts of A.C Physicsminds.blogspot.com

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