6.1 Introduction

Transcription

1 6. Introduction A.C Circuits made up of resistors, inductors and capacitors are said to be resonant circuits when the current drawn from the supply is in phase with the impressed sinusoidal voltage. Then. the resultant reactance or susceptance is zero.. the circuit behaves as a resistive circuit. 3. the power factor is unity. A second order series resonant circuit consists of, L and C in series. At resonance, voltages across C and L are equal and opposite and these voltages are many times greater than the applied voltage. They may present a dangerous shock hazard. A second order parallel resonant circuit consists of ; L and C in parallel. At resonance, currents in L and C are circulating currents and they are considerably larger than the input current. Unless proper consideration is given to the magnitude of these currents, they may become very large enough to destroy the circuit elements. esonance is the phenomenon which finds its applications in communication circuits: The ability of a radio or Television receiver () to select a particular frequency or a narrow band of frequencies transmitted by broad casting stations or () to suppress a band of frequencies from other broad casting stations, is based on resonance. Thus resonance is desired in tuned circuits, design of filters, signal processing and control engineering. But it is to be avoided in other circuits. It is to be noted that if 0in a series circuit, the circuits acts as a short circuit at resonance and if in parallel circuit, the circuit acts as an open circuit at resonance.

2 45 Network Analysis 6. Transfer Functions As ω is varied to achieve resonance, electrical quantities are expressed as functions of ω, normally denoted by F (jω) and are called transfer functions. Accordingly the following notations are used. Z(jω) V (jω) I(jω) Impedance function Y (jω) I(jω) Admittance function V (jω) G(jω) V (jω) Voltage ratio transfer function V (jω) α(jω) I (jω) Current ratio transfer function I (jω) If we put jω s then the above quantities will be Z(s),Y(s),G(s),α(s) respectively. These are treated later in this book. 6.3 Series esonance Fig. 6. represents a series resonant circuit. esonance can be achieved by. varying frequency ω. varying the inductance L 3. varying the capacitance C Figure 6. Series esonant Circuit The current in the circuit is I E + j(x L X C ) E + jx At resonance, X is zero. If ω 0 is the frequency at which resonance occurs, then ω 0 L ω 0 C or ω 0 resonant frequency. The current at resonance is I m V maximum current. The phasor diagram for this condition is shown in Fig. 6.. The variation of current with frequency is shown in Fig Figure 6. Figure 6.3

3 esonance 453 It is observed that there are two frequencies, one above and the other below the resonant frequency, ω 0 at which current is same. Fig. 6.4 represents the variations of X L ωl; X C and Z with ω. ωc From the equation ω 0 we see that any constant product of L and C give a particular resonant frequency even if the ratio L is different. The frequency of a constant frequency source C can also be a resonant frequency for a number of L and C combinations. Fig. 6.5 shows how the sharpness of tuning is affected by different L C ratios, but the product remaining constant. Figure 6.4 Figure 6.5 For larger L C ratio, current varies more abruptly in the region of ω 0. Many applications call for narrow band that pass the signal at one frequency and tend to reject signals at other frequencies. 6.4 Bandwidth, Quality Factor and Half Power Frequencies At resonance I I m and the power dissipated is P m I m watts. When the current is I Im power dissipated is P m I m watts. From ω I characteristic shown in Fig. 6.3, it is observed that there are two frequencies ω and ω at which the current is I Im. As at these frequencies the power is only one half of that at ω 0, these are called half power frequencies or cut off frequencies. The ratio, current at half power frequencies Maximum current When expressed in db it is 0 log 3dB. I m Im

4 454 j Network Analysis Therefore! and! are also called 3 db frequencies. I m E As p p, the magnitude of the impedance at half-power frequencies is p j + j(xl X C )j Therefore, the resultant reactance, X X L X C. The frequency range between half - power frequencies is!!, and it is referred to as passband or band width. BW!! B: The sharpness of tuning depends on the ratio, a small ratio indicating a high degree of L selectivity. The quality factor of a circuit can be expressed in terms of and L of the inductor. Quality factor Q! 0L Writing! 0 f 0 and multiplying numerator and denominator by I m, we get, Q f 0 LI m I m Selectivity is the reciprocal of Q. As and since! 0 p,wehave Q! 0L and! 0L! 0 C ; Q! 0 C Q r L C LI m 6.5 Expressions for ω and ω, and Bandwidth At half power frequencies! and!, I E p jx L X C j i:e:; I m T Maximum energy stored total energy lost in a period E f +(X L X C ) g At!!,! L! C Simplifying,!! C 0!L!C

5 esonance j 455 Solving, we get! C + p C +4 L + s L + Note that only + sign is taken before the square root. This is done to ensure that! is always positive. At!!,! C! L Solving; )! +! C 0! C + p C +4 s L + + L While determining!, only positive value is considered. Subtracting equation(6.) from equation (6.), we get!! Band width. L Since Q! 0L, Band width is expressed as B!! L! 0 Q : and therefore Q! 0! 0!! B Multiplying equations (6.) and (6.), we get!! 4L + 4L! 0 or! 0 p!! The resonance frequency is the geometric mean of half power frequencies. Normally L << p ; in which case Q 5 Then;! ' L + r and! ' L + p! 0! +! L +! 0 and! L +! 0 Arithmetic mean of! and! Since L! 0 Q, Equations for! and! as given by equations (6.) and (6.) can be expressed in terms of Q as!! 0 Q + s!0 +! 0 Q (6.) (6.)

6 456 j Network Analysis s 3! 0 4 Q Q s Similarly!! 0 4 Q + Normally, L << p and then Q>5. Consequently! and! can be approximated as +! ' L + r L +! 0 B +! 0! ' L + r + L +! 0 B +! 0 so that! 0! +! : 3 5 Q 6.6 Frequency esponse of Voltage across L and C As frequency is varied, both the voltages across L and C increase with frequency upto! 0 and they are equal at! 0 : But their maximum values do not occur at! 0 :V c reaches its maximum at!<! 0 and V L reaches its maximum at!>! 0. This can be verified by calculating the frequency at which each occurs. 6.7 Expression for ω at which V L is Maximum Current in the circuit shown in Figure 6. is Voltage across L is Squaring This is maximum when dv L I V L!LI V L d! 0 E q +!L!C E!L q +!L!C E! L +!L!C

7 esonance j 457 That is, E L "( +!C!C +!L!C )!!!L!C!L!C +! L +! C L C! L! C! C +! or! ( C )! C C L Let this frequency be! L. Then;! L! 0 Q That is,! L >! 0.! L! 0 s Q : 6.8 Expression for ω at which V C is Maximum Now V C V C This is maximum when d d! (V C)0: That is, E C "! (!L +!C E q!c +! L!C E! C +!L!C L +! C # 0 (!L L!C! +! +!L )# 0 C!C +!L!L!L +!C!C!C

8 458 j Network Analysis +! L +! C L C! C! L Let this frequency be! C i:e:;! C <! 0! C! 0 r Q! L + L C! L C L L C! 0 L Q Variations of V C and V L as functions of! are shown in Fig Figure 6.6 We know that V C E r! C n + (! )! C o E p f! C +(! ) g (6.3) Consider! C +(! ) and at!! C. Then equation(6.3) becomes! C C +(! C )! 0 Q C +! 0 Q! 0 since Q Q + Q Q +! 0 and! 0 C Q Q! 0 4Q 4 Q Q 4 + 4Q 4 Q 4Q Substituting the above expression in the denominator of equation (6.3), we get 6.9 Selectivity with Variable L EQ V cm q 4Q In a series resonant circuit connected to a constant voltage, with a constant frequency, when L is varied to achieve resonance, the following conditions prevail:

9 esonance j 459. X C is constant and I q E when L 0. + X C. With increase in L; X L increases and I m V at X L X C 3. With further increase in L; I proceeds to fall. All these conditions are depicted in Fig. 6.7 V C max occurs at! 0 but V L max occurs at a point beyond! 0. L at which V L becomes a maximum is obtained in terms of other constants. EX L V L f +(X L X C ) g V L E X L +(X L X C ) Figure 6.7 This is maximum when dv L dx L 0. Therefore; +(X L X C ) X L X L f(x L X C )g + X L + X C X L X C X L X L X C Therefore; Let the corresponding value of L is L m. X L + X C X C Then; L m C( + X C) and L 0 value of L at! 0 such that 6.0 Selectivity with Variable C! 0 L! 0 C : In a series resonant circuit connected to a constant voltage, constant frequency supply, if C is varied to achive resonance, the following conditions prevail:. X L is constant.. X C varies as inversely as C when C 0, I 0. when!c!l, I I m V. 3. with further increase in C; I starts decreasing as shown in Fig. 6.8, where C m is the value of capacitance at maximum voltage across C and C 0 is the value of the capacitance at! 0.

10 460 j Network Analysis C at which V C becomes maximum can be determined in terms of other circuit constants as follows. V C EX C p +(X L X C ) V C E X C +(X L X C ) Figure 6.8 For maximim V C ; dv C 0 dx C Then; +(X L X C ) X C X C f(x L X C )( )g 0 + X L + X C X L X C X L X C + X C Let the correrponding value of C be C m. X C + X L X L Then; C m L + X : L 6. Transfer Functions 6.. Voltage ratio transfer function of a series resonant circuit and frequency response For the circuit shown in Fig. 6.9, we can write H(j!) V 0(j!) V s (j!)!l +j!c +j +jq + j!l!c n! 0 L! 0!! 0!! 0 C h h +Q i!! 0! 0! o i!! 0! 0!!,tan Q! 0! 0! Figure 6.9

11 esonance j 46 Let be a measure of the deviation in! from! 0. It is defined as Then!! 0! 0!!! 0! 0!! 0 For small deviations from! 0 ; <<: Then, Then; H(j!) ( +) + ( +) + + +!! 0! 0! ' +jq p tan Q +4Q The amplitfude and phase response curves are as shown in Fig Figure 6.0 (a) and (b): Amplitude and Phase response of a series resonance circuit 6.. Impedance function The Impedance as a function of j! is given by Z(j!) + j!l!c!l +j!c! +jq! 0! 0! s +Q!! 0! 0!, tan Q!! 0! 0! For small deviations from! 0, we can write Z(j!) ' [ + jq] p +4Q tan Q

12 46 j Network Analysis 6. Parallel esonance The dual of a series resonant circuit is often considered as a parallel resonant circuit and it is as shown in Fig. 6.. The phasor diagram for resonance is shown in Fig. 6.. The admittance as seen by the current source is Y (j!)y + Y L + Y C + j!c G + jb!l Figure 6. Parallel esonance Circuit Figure 6. Phasor Diagram If the resonance occurs at! 0 ; then the susceptance B is zero. That is, or! 0 C! 0 L! 0 p rad sec : and At resonance, I C0 I L0 j! 0 CI I I C0 + I L0 0 The quality factor, as in the case of series resonant circuit is defined as Maximum energy stored Q Energy dissipated in a period CV m V m T f 0 C! 0 C: Since! 0 C! 0 L ; Q! 0 L :

13 esonance j 463 On either side of! 0 there are two frequencies at which the voltage is same. At resonance, the voltage is maximum and is given by V m I and is evident from the response curve as shown in Fig At this frequency, p p m V m watts. The frequencies at which the voltage is p times the maximum voltage are called half power frequencies or cut off frequencies, since at these frequencies, Figure 6.3 V p m p At any!, At! and! ; Squaring, Therefore; At!!, V m half of the maximum power. Y + j!c!l s jy j p +!C!L +!C!L!C!L! C! L!! L!! L 0 Hence;! L + p L +4 Note that only positive sign is used before the square root to ensure that! is positive. s Thus;! C + + C Similarly;! C + So that, bandwidth s C B!! C +

14 464 j Network Analysis and!! Thus;! 0 p!! + C C! 0 As! 0 p and Q! 0 C! 0 L Q L p r C L Since C B! B + s B +! 0 and! B + s B +! 0 Using B! 0 Q, s 3!! 0 4 Q Q s and!! 0 4 Q Q 6.3 Transfer Function and Frequency esponse The transfer function for a parallel circuit shown in Fig is H(j!), the current ratio transfer function. H(j!) I 0(j!) I (j!) Y (j!) + j!c!l!! +j 0 C! 0! 0! 0! L +j!c!l +jq As in the case of series resonance, here also let!! 0! 0!!! 0! 0!! 0 Figure 6.4 Parallel Circuit

15 esonance j 465 then,!! 0! 0! + + For <<, for small deviations from! 0!! 0! 0! ' Therefore, H(j!) +jq 6.4 esonance in a Two Branch L C Parallel Circuit Consider the two branch parallel circuit shown in Fig Let E be the voltage across each of the parallel circuit shown in the figure. The vector diagram at resonance is shown in Figure 6.. Figure 6.5 Two branch Parallel Circuit Figure 6.6 The admittance of the circuit is Y (j!)g L jb L + G C + jb C For resonance, If this occurs at!! 0, then B L B C! 0 L L +! 0! 0 C L C +! 0 C! 0 C C! 0 C + L( +! 0 C C)C( L +! 0 L )! 0( C L C) L C L

16 466 Network Analysis ω 0 L C C L C L C L ( C C L) L L C (C L ) C Therefore, ω 0 L L C C L C This is the expression for resonant frequency. It is to be noted that. resonance is not possible for certain combination of circuit elements unlike in a series circuit where resonance is always possible.. resonance is also possible by varying of L or C. Consider the case where or C < L C < L L < L C < C In both these cases, the quantity under radical is negative and therefore resonance is not possible. The admittance at resonance of the above parallel circuit is Y 0 ( L L + X L 0 + C C + X C 0 where X L0 and X C0 are the inductive and capacitive reactances respectively at resonance. L If L C C then ω 0 as in, L, C series circuit. ) S If L C which means L C L C L C X LX C. X L X C Then, B L B C L + X L C + X C 0 X L + X C X L + X C

17 esonance j 467 In this case, the circuit acts as a pure resistive circuit irrespective of frequency. That is, the circuit is resonant for all frequencies. In this case the circuit admittance is L C Y L + + X L C + X C + X L + + X C 4 + (X L + X C )+X L X C + X L + X C 4 + (X L + X C ) + X L + X C + X L + X C r C L r L or Z C 6.4. esonance by varying inductance If resonance is achieved by varying only L in the circuit shown in Figure 6.5 but with constant current constant frequency source, then the condition for resonance is X L Therefore; L C B L B C X C ) L + X L C + X C X C Z where C Then; X L X C X L Z C + X C L 0 q Solving, forx L we get X L Z C Z 4 C 4X C L X C Z C The following conditions arise: q Z 4 C 4X C L. If Z 4 C > 4X C L ;Lhas two values for the circuit to resonate.. For Z 4 C 4X C L, L CZ C for reasonance. 3. For Z 4 C < 4X C L, No value of L makes the circuit to resonate. Z C C + X C since X L X C L C

18 468 j Network Analysis 6.4. esonance by varying capacitance As in the previous case, we have at resonance ) Simplifying we get, X C C + X C B L B C X L Z ; where Z L L + X L L X C X L X C Z L + C X L 0 q X C Z L Z 4 L 4X L C X L L Therefore; C Z L qz 4 L 4X L C The following conditions arise:. For Z 4 L > 4X L C, there are two values for C to resonante.. For Z 4 4 4X L C L, resonance occurs at C Z. L 3. For Z 4 L < 4X L C, no value of C makes the circuit to resonate esonance by varying L or C It is often possible to adjust a two branch parallel combination to resonate by varying either L or C. This is because, when the supply is of constant current and, constant frequency, these resistors control inphase and quadratare components of the currents in the two parallel paths. From the condition B L B C,weget X L X C L + X L C + X C L X L C + X L X C X L X r C XL L C X + X LX C X L (6.4) C This equation gives the value of L for resonance when all other quantities are constant and the term under radical is positive. Similarly if only C is variable, keeping all other quantities constant, the value of C for resonanace is given by r XC C L X + X LX C X C L provided the term under radical is positive.

19 esonance Practical Parallel and Series esonant Circuits A practical resonant parallel circuit contains an inductive coil of resistance and inductance L in parallel with a capacitor C as shown in Fig It is called a tank circuit because it stores energy in the magnetic field of the coil and in the electric field of the capacitor. Note that resistance C of the capacitor is negligibly small. Condition for parallel resonance is shown by the phasor diagram of Fig I C I L sin φ That is, B C B L ωl + ω L ωc. Let the value of ω which satisfy this condition be ω 0. Then, + ω0l L ( C ) L ω0 C L ω 0 C L ( ) C L (6.5) (6.6) Figure 6.7 Figure 6.8 Admittance of the circuit shown in figure 6.7 is Y (jω) + jωl + jωc At ω ω 0, Y (jω) is purely real. Hence, Y(jω 0 ) + ω L jωl + ω L + jωc + ω 0 L (6.7)

20 470 Network Analysis Substituting for ω 0 in equation (6.7), Y (jω 0 ) + ω 0 L L C C L and the circuit is a pure resistive with 0 L, which is called the dynamic resistance of C the circuit. This is greater than if there is resonance. However, note that if C >, there is L no resonance. Fig. 6.9 shows a practical series resonant circuit. The input impedance as a function of ω is G Z(jω)jωL + G + ω C j ωc G + ω C Condition for resonance is ωc ωl ( G + ω C ) C ω L G C C ( L ) C ω ( L ) C Impedance at resonance is Z 0 G G + ωc G C L L C The circuit at resonance is a purely resistive, and Z 0 0 L also resonance is not possible for C >. Figure 6.9 (6.8) L. However, note that here C In both the circuits, shown in Figs 6.8 and 6.9, resonance is achieved by varying either C or L until the input impedance or admittance is real and this process is called tuning. For this reason these circuits are called tuned circuits. Series circuits EXAMPLE 6. Two coils, one of 0.5 Ω, L 3mH, the other of.3 Ωand L 5 mhand two capacitors of 5 μf and 6 μf are all in series with a resistance of 0.4 Ω. Determine the following for this circuit (i) esonance frequency (ii) Q of each coil

21 esonance 47 (iii) Q of the circuit (iv) Cut off frequencies (v) Power dissipated at resonance if E 0 V. SOLUTION From the given values, we find that (i) esonant frequency: s Ω L s 3+547mH 5 6 C s μf 7.86 μf 87 ω 0 Ls C s rad/ sec (ii) Q of coils: For Coil, Q ω 0L For Coil, Q ω 0L (iii) Q of the circuit: Q ω 0L s s (iv) Cut off frequencies: Band width is, B ω 0 Q Considering Q>5, the cut off frequencies, ω, ω 0 ± B 09.8 ±.856 Therefore, ω 5 rad/ sec and ω 07 rad/ sec.

22 47 j Network Analysis (v) Power dissipated at resonance: Given E 0V We know that at resonance, only the resistance portion will come in to effect. Therefore P E 0 :05 48:78 W EXAMPLE 6. For the circuit shown in Fig. 6.0, find the out put voltages at (i)!! 0 (ii)!! (iii)!! when v s (t) 800 cos!t mv. SOLUTION Figure 6.0 For the circuit, using the values given, we can find that resonant frequency! 0 p p :5 0 :6 06 rad sec Quality factor: Band width: As Q>5, Hence; and Q! 0L : : B! 0 Q : : 0 6 rad sec! ;! 0 B (:6 0:)0 6 rad sec! :7 0 6 rad sec! :5 0 6 rad sec

23 esonance 473 (i) Output voltage at ω 0 : Using the relationship of transfer function, we get H(jω) ωω0 50I m 6.5I m 0.8 / 0 Since the current is maximum at resonance and is same in both resistors, v o (t) cos( t)mv 640 cos( t)mv At ω and ω, Z in s / ±45. Therefore, and H(jω) ωω out Z in / 45 H(jω) ωω / / 45 (ii) Out put voltage at ω ω v o (t) cos( t +45 )mv cos( t +45 )mv (iii) Out put voltage at ω ω v o (t) cos( t 45 )mv EXAMPLE 6.3 In a series circuit 6Ω, ω rad/sec, band width 0 5 rad/sec. Compute L, C, half power frequencies and Q. SOLUTION We know that Quality factor, Also, Therefore, Q ω 0 B Q ω 0L L Q ω μH and Q ω 0 C Hence, C ω 0 Q pf 4 6

24 474 Network Analysis As Q>5, and That is, ω, ω 0 ± B ± 05 ω rad/ sec ω rad/ sec EXAMPLE 6.4 In a series resonant circuit, the current is maximum when C 500 pf and frequency is MHz. If C is changed to 600 pf, the current decreases by 50%. Find the resistance, inductance and quality factor. SOLUTION Case Given, C 500 pf I I m f 0 6 Hz ω 0 π 0 6 rad/ sec We know that Therefore, Inductance, ω 0 L ω 0 C 0 (π 0 6 ) mh Case When C 600 pf, I I m E Z + X X 3 X X L X C π Ω 3 0 π

25 esonance j 475 Therefore resistance, Quality factor, 53:3 p 3 30:77Ω Q! 0L 38:56 30:77 0:35 EXAMPLE 6.5 In a series circuit with 50 Ω, L 0.05 H and C 0 F, frequency is varied till the voltage across C is maximum. If the applied voltage is 00 V, find the maximum voltage across the capacitor and the frequency at which it occurs. epeat the problem for 0Ω. SOLUTION Case Given 50Ω, L 0.05 H, C 0F We know that! 0 p 0 3 p 0: rad/sec Q! 0L 03 0:05 0 Using the given value of E 00 V in the relationship QE V Cm q 4Q we get V Cm 00 q 4 and the corresponding frequency at this voltage is! C! 0 r Q 5:5 V 0 3 r 707 rad sec Case When 0Ω, Q 03 0:05 0 V Cm 5 00 q :5 V r! C rad sec 50

26 476 j Network Analysis EXAMPLE 6.6 (i) A series resonant circuit is tuned to MHz. The quality factor of the coil is 00. What is the ratio of current at a frequency 0 khz below resonance to the maximum current? (ii) Find the frequency above resonance when the current is reduced to 90% of the maximum current. SOLUTION (i) Let! a be the frequency 0 khz below the resonance, I a be the current and Z a be the impedance at this frequency. Then! a khz! a! 0 980! 0! a : Now the ratio of current, I a I m Z a +j()q j00(40: ) j4:0408 0:40 /76 (ii) Let! b be the frequency at which I b 0:9I m I b I m Then +j()q 0:9 or p +x 0:9 where x ()00 Then; +x 0:8 :346 or x 0:346 and x 0:4843 We know that! b 0:4843! 0 00 Hence! b + 0:4843! 0 00 :004 MHz

27 esonance 477 EXAMPLE 6.7 For the circuit shown in Fig. 6., obtain the values of ω 0 and v C at ω 0. Figure 6. SOLUTION For the series circuit, ω rad/sec At this ω 0,I I m. Therefore, and the circuit equation is V 5I m.5 V +(I m 0 05V )0 + jv L jv C Since V L V C, the above equation can be modified as.5 5I m +0I m.05 5I m Hence, I m A and V c V EXAMPLE 6.8 For the circuit shown in Fig. 6.(a), obtain Z in and then find ω 0 and Q. Figure 6.(a)

28 478 j Network Analysis SOLUTION Taking I as the input current, we get and the controlled current source, V 0I 0:3V 0:3 0I 3I The input impedance can be obtained using the standard formula Z in (j!) Applied voltage Input current For futher analysis, the circuit is redrawn as shown in Fig. 6.(b). It may be noted that the controlled current source is transformed to its equivalent voltage source. V I (6.6) Figure 6.(b) eferring Fig. 6.(b), the circuit equation may be obtained as V 0 + j0 3! j09 30! j3 30! 0 9 I (6.7) Substituting equation (6.7) in equation (6.6), we get Z in 0+j 0 3! ! For resonance, Z in should be purely real. This gives Ω earranging, 0 3! !! :33 0

29 esonance j 479 Solving we get!! 0 p 0: rad sec Quality factor Q! 0L :5 Parallel circuits EXAMPLE 6.9 For the circuit shown in Fig. 6.3(a), find! 0, Q, BW and half power frequencies and the out put voltage V at! 0. Figure 6.3(a) SOLUTION Transforming the voltage source into current source, the circuit in Fig. 6.8(a) can be redrawn as in Fig. 6.3(b). Then;! 0 p 09 p rad sec Q! 0 C B! 0 Q rad sec Figure 6.3(b)

30 480 j Network Analysis As Q>0,! ; B +! 0 Hence; Output voltage, ! 5:05 M rad sec and! 4:95 M rad sec V I 80 kω j :04 90 V EXAMPLE 6.0 In a parallel circuit, C 50F. Determine BW, Q, and L for the following cases. (i)! 0 00;! 0 (ii)! 0 00;! 80 SOLUTION (i)! 0 00;! 0 We know that earraging we get! 0 p!! Band width!! 0! :33 radsec B!! 0 83:3336:67 rad sec Quality factor, We know that Q! 0 B 00 36:67 :73 Q! 0 L! 0C (6.8)

31 esonance j 48 earraging equation (6.8), Q! 0 C : Ω Similarly L! 0 C H (ii)! 0 00;! 80: Solving the same way as in case (i), we get! BW B rad sec Q : EXAMPLE 6. In the circuit shown in Fig. 6.4(a), v s (t) 00 cos!t volts. Find resonance frequency, quality factor and obtain i ;i ;i 3. What is the average power loss in 0 kω. What is the maximum stored energy in the inductors? 0k 40k SOLUTION Figure 6.4(a) The circuit in Fig. 6.4(a) is redrawn by replacing its voltage source by equivalent current source as shown in Fig. 6.4(b). esonance frequency,! 0 p p : rad sec Quality factor, Figure 6.4(b) Q! 0 C eq 4000 :

32 48 j Network Analysis At resonance, the current source will branch into resistors only. Hence, v(t) i (t) lags v(t) by 90. Therefore, Average power in 0 kω: (0kΩjj40kΩ) v s(t) cos 4000t volts 80 i (t) sin 40000t sin 4000t ma 80 i (t) cos 4000t cos 4000t ma i 3 (t) i (t) 400 sin 4000t ma 80 p P av :3 W Maximum stored energy in the inductance: E LI m ( ) 4mJ EXAMPLE 6. For the network shown in Fig. 6.5(a), obtain Y in and then use it to determine the resonance frequency and quality factor. Figure 6.5(a) Figure 6.5(b) SOLUTION Considering V as the input voltage and I as the input current, it can be found that 0kΩ I V ) 0 4 I V

33 esonance j 483 The circuit in Fig. 6.5(a) is redrawn by replacing the controlled voltage source in to its equivalent current source by taking s j! and is shown in Fig. 6.5(b). eferring Fig. 6.5(b), I 0V sl V sc + + sl ) I V sc + + sl Input admittance, with s is being replaced by j! is Y in I V j! 0 8 j 03! 4: j! 0 8 j500! At resonance, Y in should be purely real. This enforces that 0 8! 500! Therefore;! 0 p Quality factor: 500 K rad sec Q! 0 C EXAMPLE 6.3 In a parallel circuit, cut off frequencies are 03 and 8 rad/sec. jzj at! 05 rad/sec is 0 Ω. Find, L and C. SOLUTION Given! 03 rad sec! 8 rad sec Therefore esonant frequency, B rad sec! 0 p!! p :45 rad sec

34 484 Network Analysis Quality factor Q ω 0 B Admittance, Y ( + j ωc ) ωl [ ( +j ωc )] ωl [ ( ω0 ωc +j ω )] 0 ω 0 ωω 0 L Since Q ω 0 C we get Note that, Y ω 0 L, [ ( ω +jq ω )] 0 ω 0 ω ω ω 0 ω 0 ω Therefore, Y ( + j7.35( )) get Y ( j0.768) (6.) +(0.768).3 It is given that Z 0and therefore Y. Putting this value of Y in equation (6.9), we 0 From the relationship Q ω 0 C,weget 0.3.3Ω ω 0 C 7.35 Therefore, C μf

35 esonance j 485 Inductance, L! 0 C 0:45 5: :8 mh EXAMPLE 6.4 For the circuit shown in Fig. 6.6(a), find! 0, V at! 0, and V at a frequency 5 k rad/sec above! 0. Figure 6.6(a) SOLUTION Changing voltage source of Fig. 6.6(a) into its equivalent current source, the circuit is redrawn as shown in Fig. 6.6(b). eferring Fig. 6.6(b),! 0 p k p radsec Figure 6.6(b) Voltage across the inductor at! 0 is, V j Quality factor, Given Now; j5 V Q! 0 C ! a! 0 +5k rad/sec : radsec! a! 0 :05! 0! a :05 0:03

36 486 Network Analysis Using this relation in the equation, Y [ ( ωa +jq ω )] 0 ω 0 ω a we get Y ( + j ) / 56.3 The corresponding value of V is V IY jω a Y j / / V EXAMPLE 6.5 A parallel circuit has a quality factor of 00 at unity power factor and operates at khz and dissipates Watt when driven by A at khz. Find Bandwidth and the numerical values of, L and C. SOLUTION Given f khz, P W, I A, Q 00, cos φ B ω 0 Q 03 π 00 P I Therefore Ω L ω 0 Q 0π rad/sec 0π μh C ω0 L 0 (0π) μf

37 esonance 487 EXAMPLE 6.6 For the circuit shown in Fig. 6.7, determine resonance frequency and the input impedance. Figure 6.7 SOLUTION Equation for resonance frequency is ( ) ω L L L C C L C ( ) rad/sec We know that X L ω 0 L Ω and X C ω 0 C Ω

38 488 Network Analysis Admittance Y at resonance is purely real and is given by Y G + G + G 3 +(0.ω 0 ) ( ) 0 3 ω 0 Y S and the input impedance, Z Y 4.35 Ω EXAMPLE 6.7 The impedance of a parallel circuit as a function of ω is depicted in the diagram shown in Fig Determine, L and C of the circuit. What are the new values of ω 0 and bandwidth if C is increased by 4 times? 0.4 SOLUTION It can be seen from the figure that Figure 6.8 ω 0 0 rad/sec B 0.4 rad/sec 0Ω Then Quality factor We know that Q ω 0 BW L ω 0 Q H

39 esonance j 489 As Q! 0 C, C :5 F If C is increased by 4 times, the new value of C is Farad. Therefore,! 0 p p 0:04 5 and the corresponding bandwith B C 0: EXAMPLE 6.8 In a two branch L C parallel resonant circuit, L 0:4 H and C 40F. Obtain resonant frequency for the following values of L and C. (i) L 0; C 80 (ii) L C 80 (iii) L 80; C 0 (iv) L C 00 (v) L C 0 SOLUTION As L and C are given separately, we can use the following formula to calculate the resonant frequency. Let us compute the following values v uut!! 0 p L L C C L C 0: p 50 (6.0) (i) L 0; C 80 Using equation (6.0), L C 04! 0 50 r

40 490 Network Analysis As the result is an imaginary number resonance is not possible in this case. (ii) L C 80 (iii) L 80; C 0 80 ω rad/sec 80 ω rad/sec (iv) L C ω As the result is indeterminate, the circuit resonates at all frequencies. (v) L C 0 0 ω rad/sec EXAMPLE 6.9 The following information is given in connection with a two branch parallel circuit: L 0Ω, C 0Ω, X C 40Ω, E 0 V and frequency 60 Hz. What are the values of L for resonance and what currents are drawn from the supply under this condition? SOLUTION As the frequency is constant, the condition for resonance is Solving we get X L X C L + X L C + X C X L XL X L 50X L +000 X L Ω or.087 Ω Then the corresponding values of inductances are L X L ω 0.7 H or mh

41 esonance 49 The supply current is I EG E (G L + G C ) ( ) 0 Thus, I A for X L Ω ( ) 0 or I A for X L.087 Ω Exercise Problems E.P 6. efer the circuit shown in Fig. E.P. 6., where i is the source resistance (a) Determine the transfer function of the circuit. (b) Sketch the magnitude plot with i 0and i 0. Figure E.P. 6. Ans: H(s) V o (s) V i (s) L s s + ( + i L ) s + E.P 6. For the circuit shown in Fig. E.P. 6., calculate the following: (a) f 0, (b) Q, (c) f c, (d) f c and (e) B Figure E.P. 6. Ans: (a) khz (b) 8 (c) 39.3 khz (d) 7.06 khz (e) 3.83 khz

42 49 j Network Analysis E.P 6.3 efer the circuit shown in Fig. E.P. 6.3, find the output voltage, when (a)!! 0 (b)!!, and (c)!! c. Figure E.P. 6.3 Ans: (a) 640 cos( t)mv (b) cos( t +45 )mv (c) cos( t 45 )mv E.P 6.4 efer the circuit shown in Fig. E.P Calculate Z i (s) and then find (a)! 0 and (b) Q. Ans: (a) krad/sec, (b) 36 Figure E.P. 6.4 E.P 6.5 efer the circuit shown in Fig. E.P Show that at resonance, jv o j max q QjVsj. 4Q Figure E.P. 6.5

43 esonance j 493 E.P 6.6 efer the circuit given in Fig. E.P. 6.6, calculate! 0 ;Qand jv o j max Figure E.P. 6.6 Ans: rad/sec, 6.35, 36 volts. E.P 6.7 A parallel network, which is driven by a variable frequency of 4 A current source has the following values: kω, L 0mH, C 00 F. Find the band width of the network, the half power frequenies and the voltage across the network at half-power frequencies. Ans: 0 rad/sec, 995 rad/sec, 0005 rad/sec E.P 6.8 For the circuit shown in Fig. E.P. 6.8, determine the expression for the magnitude response, jz in j versus! and Z in at! 0 p. Ans: (a) jz in j q (! ) +(!L) Figure E.P (!C), (b) jz in j q C L (+ C L ) E.P 6.9 A coil under test may be represented by the model of L in series with. The coil is connected in series with a variable capacitor. A voltage source v(t) 0 cos 000 t volts is connected to the coil. The capacitor is varied and it is found that the current is maximum when C 0F. Also, when C :5F, the current is of the maximum value. Find Q of the coil at! 000 rad/sec. Ans: 5

44 494 j Network Analysis E.P 6.0 A fresher in the devices lab for sake of curiosity sets up a series network as shown in Fig. E.P.6.0. The capacitor can withstand very high voltages. Is it safe to touch the capacitor at resonance? Find the voltage across the capacitor. Figure E.P. 6.0 Ans: Not safe, jv c j max 600 V