Topic 2b ac_circuits_analysis
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1) An AC generator with an RMS voltage of 110 V is connected in series with a 35-Ω resistor and 1-μF capacitor. To maintain a current of 1.2 A, the generator must operate at a frequency of 1.9 kHz. 2) An AC generator with an emf of 22.8 V at 353 rad/s is connected to a 17.3 H inductor. When the current is maximum, the emf is 22.8 V. When the emf is -11.4 V and increasing, the current is 4.11 A. 3) A series RLC circuit with a 148-Ω resistor, 1.50-μF capacitor
Topic 2b ac_circuits_analysis
- 1. 1 Topic 2b AC Circuits Analysis
- 2. 2 Example An ac generator with rms voltage of 110 V is connected in series with a 35-Ω resistor and 1-µF capacitor. At what frequency must the generator operate if it is to maintain a current of 1.2 Amps.
- 3. 3 Example Z = V/I = 110/1.2 = 91.7 Ω Z2 = R2 + (1/ ωC)2 ; f = ω/2π = 1.9 kHz
- 4. 4 Example An ac generator emf is ε = ε0∙sinωt, with ε0∙ = 22.8 V and ω = 353 rad/s. It is connected to a 17.3 H inductor. When the current is a maximum, what is the emf of the generator? When the emf of the generator is -11.4 V and increasing in magnitude, what is the current?
- 5. 5 Example mAXVI L 73.3)3.17353/(8.22/maxmax =⋅== )30180cos( 3sin 3018030360 2/18.22/4.11sin max +⋅= ⇒ +−=⇒ −=−= II quadrantrdmagnitudegincrea or t ϕ ω ϕ
- 6. 6 Example An ac generator emf is ε = ε0∙sin ωt, with ε0∙ = 28.6 V and ω = 333 rad/s. It is connected to a 4.15 µF capacitor. When the current is a maximum, what is the emf of the generator? When the emf of the generator is -14.3 V and increasing in magnitude, what is the current?
- 7. 7 Example ϕ mAXVI C 5.39)1015.4333/1/(6.28/ 6 maxmax =×⋅== − )30180cos( 3sin 3018030360 2/16.28/3.14sin max +⋅= ⇒ +−=⇒ −=−= II quadrantrdmagnitudegincrea or t ϕ ω
- 8. 8 Example A 0.4-H inductor and a 220-Ω resistor are connected in series to an ac generator with an rms voltage of 30 V and a frequency of 60 Hz. Find the rms values of the current, voltage across the resistor, and voltage across the inductor.
- 9. 9 Example Irms = Vrms/Z = 0.112 A (VL)rms = IrmsXL= 17 V (VR)rms = IrmsR = 24.6 V
- 10. 10 Example An ac generator has a frequency of 1200 Hz and a constant rms voltage. When a 470-Ω resistor is connected between the terminals of the generator, an average power of 0.25 W is dissipated in the resistor. Then, a 0.08-H inductor is connected in series with the resistor, and the combination is connected between the generator terminals. What is the average power dissipated in the series combination?
- 11. 11 Example
- 12. 12 Example A person is working near the secondary of a transformer, as shown. The primary voltage is 120 V at 60.0 Hz. The capacitance Cs , which is the stray capacitance between the hand and the secondary winding, is 20.0 pF. Assuming the person has a body resistance to ground Rb = 50.0 kΩ, determine the rms voltage across the body.
- 13. 13 Example
- 14. 14 Example A typical "light dimmer" used to dim the stage lights in a theater consists of a variable inductor L (whose inductance is adjustable between zero and Lmax) connected in series with a lightbulb. The electrical supply is 110 V (rms) at 62.5 Hz; the lightbulb is rated as "110 V, 900 W.“ What Lmax is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 900 W? Assume that the resistance of the lightbulb is independent of its temperature. Could a variable resistor be used instead?
- 16. 16 Modern dimmer
- 17. 17 Phasors for a series RLC circuit Max or rms: V0 2 = VR 2 + (VL - VC )2 Vrms = Irms Z VR = Irms R, VC = Irms XC , and VL = Irms XL
- 18. 18 Power factor for a series RLC circuit
- 19. 19 Example A series RLC circuit contains a 148-Ω resistor, a 1.50-μF capacitor, and a 35.7-mH inductor. The generator has a frequency of 512 Hz and an rms voltage of 35 V. What is (a) the rms voltage across each circuit element and (b) the average electric power consumed by the circuit.
- 20. 20 Example
- 21. 21 rms current in RL and RC circuits depends on frequency
- 22. 22 When the phase difference between current and voltage is zero (Xc=XL), the circuit is in resonance
- 23. 23 How does the variable air capacitor work?
- 24. 24 Phasors • Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions. • A phasors is a complex number that represents the amplitude and phase of a sinusoid.
- 25. 25 Complex number • A complex number z can be written in rectangular form as z = x + jy where : x is the real part and y is the imaginary part of z. • The complex number z can also be written in polar form or exponential form as where r is the magnitude of z and φ is the the phasor of z. 1−=j φ φ j rerz =∠= φ φ j rerjyxz =∠=+=
- 26. 26 Complex number • Rectangular form • Polar form • Exponential form • Given x, y can get r and φ • Know r and φ, can obtain x and y jyxz += φ∠= rz φj rez = .tan, 122 x y yxr − =+= φ .sin,cos φφ ryrx ==
- 27. 27 Important operation • Addition: • Subtraction: • Multiplication: • Division: • Reciprocal: • Square root: )()( 212121 yyjxxzz −+−=− )()( 212121 yyjxxzz +++=+ )( 212121 φφ +∠= rrzz )( 21 2 1 2 1 φφ −∠= r r z z ( )φ−∠= rz 11 ( )2/φ∠= rz
- 28. 28 Complex conjugate • Complex conjugate of z is Real part of e jφ . Imaginary part of ejφ . ( ) φ φ j rerjyxz −∗ =−∠=−= j j −= 1 ( ) ( )φ φ φ φ φ φφ j j j e e je Imsin Recos sincos = = ±=±
- 29. 29 Real and imaginary part • can be express as • V is the phasor representation of the sinusoid v(t). )cos()( φω += tVtv m ( ) ( ) ( ) φ φω φω ωφ φω ∠=== = =+= + m j m tj tjj m tj mm VeVetv eeVtv eVtVtv VV whereRe)( Re)( Re)cos()( )(
- 30. 30 Sinusoidal-phasor transformation Time –domian representation Phasordomian representation )cos( φω +tVm )cos( φω +tIm )sin( φω +tVm )sin( φω +tIm φ∠mV )90( o mV −∠ φ φ∠mI )90( o mI −∠ φ ( ) tion)representadomain-(Phasortion)representadomain-Time( cos)( φφω ∠=⇔+= mm VtVtv V
- 31. 31 Derivative and Integral • The derivative of v(t) is transformed to the phasor as jωV. • The integral of v(t) is transformed to the phasor as V/ jω. domain)(Phasordomian)(Time j Vω⇔ dt dv domain)(Phasordomian)(Time jω V ⇔∫vdt
- 32. 32 Differences between v(t) and V 1. v(t) is the instantaneous or time domain representation, while V is the frequency or phasor-domain representation. 2. v(t) is time dependent, while V is not 3. v(t) is always real with no complex term, while V is generally complex.
- 33. 33 Example Transform these sinusoids to phasors (a) (b) Solution (a) (b) ( )o ti 4050cos6 −= ( )o tv 5030sin4 +−= ( )o I 406 −∠= ( ) ( ) ( ) o o oo o t t tv 1404 14030cos4 905030cos4 5030sin4 ∠= += ++= +−= V sin (ωt±180o ) = −sin ωt Cos (ωt±180o ) = −cos ωt sin (ωt±90o ) = ±cos ωt cos (ωt±90o ) = sin ωt
- 34. 34 Example Find the sinusoids represented by these phasors: (a) I = -3 +j4 (b) Solution (a) I = -3 +j4 = 5∠(126.87o ) (b) o j ej 20 8 − =V ( )o tti 87.126cos5)( += ω ( )( ) ( )o o ooj ttv ej o 70cos8)( 708 )20(89018 20 += ∠= −∠∠== − ω V o j 901∠=
- 35. 35 Phasor relationships for circuit elements • For resistor R , V = RI I = V/R • For inductor L, V = jωLI • For capacitor C, I = jωCV Cjω I V = Ljω V I =
- 36. 36 Example The voltage is applied to a 0.1-H inductor. Find the steady-state current through the inductor. Solution ω = 60, V = 12∠45o ( )o tv 4560cos12 += ( )o o oo jLj 452 906 4512 )1.0)(60( 4512 −∠= ∠ ∠ = ∠ == ω V I ( )A4560cos2)( o tti −=
- 37. 37 Impedance and Admittance of passive elements • Impedance Z of a circuit is the ratio of the phasor voltage V to the phase current I, measured in ohms. • The admittace Y is the reciprocal of impedance, measured in siemens. Element Impedance Admittance R Z = R L Z = jωL C Cjω 1 =Z R 1 =Y Cjω=Y Ljω 1 =Y
- 38. 38 Impedance Z • The complex quantity Z may be represented by rectangular form as Z = R +jX. where R = ReZ is the resistance and X = Im Z is the reactance. θθ θ θ sinX,cos tan, 1-22 ZZ Z ZZ == =+= ∠=+= R R X XR jXR
- 39. 39 Example Find v(t) and i(t) in the circuit shown. +_ i 5Ω 0.1F v + _ vs=10cos4t
- 40. 40 Solution ∀ ω = 4. Vs = 10∠0o . • Voltage across the capacitor ( ) o o oo s j j j 57.26789.1 57.2659.5 010 5.25 010 5.25 )1.0)(4( 1 5 ∠= −∠ ∠ = − ∠ == −=+= Z V I Z )43.634cos(47.4)( )57.264cos(769.1)( )43.63(47.4 904.0 57.26789.1 )1.0)(4( 57.26789.1 o o o o oo c ttv tti jCj −= += −∠= ∠ ∠ = ∠ === ω I IZV
- 41. 41 Impedance combination • Consider N series-connected impedance, V = V1 + V2 +… +VN = I(Z1+Z2+…+ZN) • Consider N parallel-connected impedance, N21eq Z...ZZ I V Z +++== N21eq N21eq N21 N21 Y...YYY Z 1 ... Z 1 Z 1 V I Z 1 Z 1 ... Z 1 Z 1 VI...III +++= +++== +++=+++=
- 42. 42 Example Find the input impedance in the circuit shown. Assume that the circuit operates at ω = 50 rad/s. Zin 2mF 0.2H 8Ω 3Ω 10mF
- 43. 43 Solution • Impedance of 2mF capacitor • Impedance of the 3-Ω resistor in series with 10 mF capacitor • Impedance of the 8-Ω resistor in series with 0.2 H inductor Ω−=== 10 )002.0)(50( 11 1 j jCjω Z Ω−=+=+= )23( )01.0)(50( 1 3 1 32 j jCjω Z Ω+=+=+= )108()2.0)(50(882 jjLjωZ
- 44. 44 Solution • The input impedance is 07.1122.3 07.122.310 811 )811)(1444( 10 811 )108)(23( 10 )108()23( )108)(23( 10 22 j jj jj j j jj j jj jj j −= −+−= + −+ +−= + +− +−= ++− +− +−=+= 321in ZZZZ
- 45. 45 Example • Determine vo(t) in the circuit. +_ + _vo 5H10mF 60Ω 20sin(4t-15o )
- 46. 46 Solution • Vs = 20∠(-105o ) • Impedance of 10 mF capacitor • Impedance of 5-H inductor ZL= j(4)(5)=j20Ω • Impedance of the parallel combination of 10-mF capacitor and 5-H inductor Ω−== 25 )01.0)(4( 1 j j cZ V)04.744cos(15.17)( 04.7415.17)105(20)(96.308575.0( )105(20( 10060 100 60 100 5 500 )20()25( )20)(25( 2025 o o ooo o so ttv j j j jjj jj jj −= −∠=−∠∠= −∠ + = + = Ω= − = +− − =−= V Z Z V Z
- 47. 47 Effective or RMS Value • The idea of effective value arises from the need to measure the effectiveness of a voltage or current source in delivering power to a resistive load. • The effective value of a period current is the dc current that deliver the same average power to a resistor as the periodic current.
- 48. 48 Root Mean Square Value • Effective value for current • Effective value for voltage • The effective value of a periodic signal is them root mean square (rms) value ∫= T eff dti T i 0 21 ∫= T eff dtv T v 0 21 ∫= T rms dtx T X 0 21
- 49. 49 RMS value for Sinusoidal function • The rms value of a constant is the constant itself. • For sinusoidal i(t) = Im cos ωt, the effective value is • For v(t) = Vm cos ωt, the effective value is . 2 )2cos1( 2 1 cos 1 2 22 m T o m T o mrms I dtt T I tdtI T I =+== ∫∫ ωω . 2 m rms V V =
- 50. 50 Average power • The average power absorbed by a resistor R can be written as • For instance, the 230 V available at every household is the rms value. • It is convenient in power analysis to express voltage and current in their rms value. R V RIP rms rms 2 2 ==
- 51. 51 Example Determine the rms value of the current waveform as shown. If the current is passed through a 2-Ω resistor, find the average power absorbed by the resistor. 0 10 -10 i(t) t2 4 6 8 10
- 52. 52 Solution • The period of the waveform T = 4. • Can write the current waveform as • The power absorbed by a 2-Ω resistor is At t dtdttdti T I t tt ti T rms 165.8200 3 200 4 1 2 4 100 0 2 3 25 4 1 )10()5( 4 11 42,10 20,5 )( 3 4 2 2 2 0 2 0 2 = += +×= −+== <<− << = ∫∫∫ .3.133)2()165.8( 22 WRIP rms ===
- 53. 53 Example The waveform shown is a half-wave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10-Ω resistor. 10 0 v(t) tπ 2π 3π
- 54. 54 Solution • The period of the voltage is T = 2π, and ( ) W R v P v t t dtttdtv t tt tv rms rms rms 5.2 10 5 isabsorbedpowerAverage 5 250 2 2sin25 02 2sin 4 100 2cos1 2 100 2 1 sin100 2 1 2,0 0,sin10 )( 22 00 22 === = = −−= −= −== << << = ∫∫ π π π π π ππ ππ π ππ
- 55. 55 Maximum Power Transfer • Why use Thevenin and Norton equivalents? – Very easy to calculate load related quantities – E.g. Maximum power transfer to the load • It is often important to design circuits that transfer power from a source to a load. There are two basic types of power transfer – Efficient power transfer: least power loss (e.g. power utility) – Maximum power transfer (e.g. communications circuits) • Want to transfer an electrical signal (data, information etc.) from the source to a destination with the most power reaching the destination. Power is small so efficiency is not a concern.
- 56. 56 Maximum Power Transfer Theorem Given a Thévenin equivalent circuit of a network, the maximum power that can be transferred to an external load resistor is obtained if the load resistor equals the Thévenin resistance of the network. Any load resistor smaller or larger than the Thévenin resistance will receive less power.
- 57. 57 Maximum Power Demonstrated (1/2) PL is max if and Th Th L V I R R = + ( ) 2 2 L L Th Th L R P V R R = +0L L dP dR =
- 58. 58 Maximum Power Demonstrated (2/2) • Maximum Power Transfer Theorem applied when matching loads to output resistances of amplifiers • Efficiency is 50% at maximum power transfer ⇒ and 0L L dP dR = L ThR R=