ANSWER. (a) Let Q 1 be the charge on C 1 and Q 2 charge on C 2,

Transcription

1 Q. (a) Pove fom fist pinciples that two capacitos an connecte in seies can be eplace with an equivalent capacitance of / ( ). [ maks] 4 F F 4 F 6 F Figue Q-(b) Figue Q-(c) (b) Fin the voltage acoss each capacito of the cicuit in Figue Q-(b). [4 maks] (c) Fin an epession fo the chaging time constant of the capacito shown in Figue Q-(c). Also fin the steay state voltage acoss the capacito. [4 maks] ANSWE (a) et Q be the chage on an Q chage on, then Q an Q. Fo the equivalent capacito, we have Q Since the two capacitos ae in seies, Q Q Q, Hence we have, Q Q Q, ompaing with Q, we get o (b) The equivalent cicuits ae given below, Using the voltage ivision ule, we get the following

2 (c) Get the Thevenin s equivalent acoss the capacito. th ( ) ( ), theefoe, th.. /( ) Also, E th E E Hence, The time constant th.... /( ) Final oltage E th. Q. (a) Wite a iffeential equation elating output cuent i(t) to the input voltage v(t) fo the cicuit in Figue Q. [ maks] (b) By means of the iffeential equation eive above, answe the following. i. What is the oe of the iffeential equation? How is the oe of the iffeential equation elate to the cicuit in question? [ mak] ii. f the input voltage v(t) is a D souce of magnitue E, what is the cuent at steay state? [ mak] iii. f the input voltage v(t) is an A souce of angula fequency, eive an epession fo the input impeance of the cicuit. [ maks] iv. aliate you answe in pat (iii) above using A theoy. [ mak] ANSWE (a) We can wite the following equations fo, an. () i i i () i () i. (4) i i () i Diffeentiating equation (4) twice yiels, an combining with equation (), i i i i we get,, o (6) i i Also, iffeentiating equation () twice yiels (7) Aing (6) an (7), i i i (8)

3 Substituting fom equation () an its eivatives, (8) becomes i i i i i e-aanging the tems, i i i i (b) Using the above iffeential equation, i. The oe of the iffeential equation is. This is because the cicuit has inepenent enegy stoing elements. ii. To fin steay state cuent, substitute i/ an / (all highe eivatives ae also zeo). Thus, E/. iii. Substitute /, (an swapping left an ight han sies) [ ] iv. Using A theoy, the impeance is

4 Q. (a) Deive an epession fo the total enegy stoe in the couple cicuit in Figue Q-(a). [ maks] (b) The cicuit in Figue Q-(b) is supplie by a sinusoial voltage souce s of vaiable fequency. The output out is taken acoss the inucto. i. Wite an epession fo the output voltage out as a function of the input voltage s. [ mak] ii. Fin an epession fo the esonance fequency of the cicuit. [] iii. Show that half-powe fequencies ae given by, [ maks] ANSWE f, f ± Figue Q-(b) 4π π iv. f the cicuit is to be esigne with a esonance fequency of. khz an a banwih of khz, calculate the appopiate values fo an, given that. kω. [ maks] v. With the cicuit paametes as foun in pat (iv) above, etemine the output voltage at the fequencies of f an f, when s. [ maks] s out (a) Total enegy stoe W v i i i i i v i i i i i i i i i ( i i i i ) i i (b) (i). The output voltage is given by, out s,whee is the impeance of an in paallel. ( - ) s ( - ) s s ( ) (ii). The esonance fequencies ae given by unity powe facto, ( ) (iii). At half-powe fequencies, 4

5 s ( ) i.e., ( ) ± Taking the positive sign, Taking the nagative sign, f, f ± 4π π (iv). esonnance fequency, f Hz π π f Banwih is given by, f f Hz 4π 4.9 π (BW) Thefoe,.8 µf an mh. s, Since at esonance the output voltage iss, Note that the othe solution yiels a negative fequency an hence it is ignoe., Note that the othe solution yiels a negative fequency an hence it is ignoe. (v). The output voltage at half-powe fequencies must be / of peak output at esonance. Since the peak output is, the output at half-powe fequencies is 7.7.

6 Q4. (a) Fo the cicuit in Figue Q4, fin the input impeance acoss XY. [ maks] (b) Fin the shot cicuit cuent though XY. [ maks] (c) Fin the Noton s equivalent cicuit acoss XY. [ maks] () f the voltage souce is move to the teminals XY, What woul be the Noton s equivalent cicuit acoss AB? [ maks] (e) Ae the two Noton s equivalent cicuits ientical? omment on you esults. [ maks] ANSWE (a) The impeance acoss XY, is XY XY YXY.9.4 XY (b) Fist, fin the total impeance when XY ae shot cicuite Using cuent ivision ule successively, 4 6 sc , this is the s/c cuent at the souce sc (c) Hence, the Noton equivalent cicuit has a cuent souce of ( ) A an a paallel amittance of Y XY.9.4 () When the voltage souce is move to XY, using the ecipocity theoem, the shot cicuit cuent acoss AB must be the same as the shot cicuit cuent sc foun ealie. By inspection, the input impeance acoss AB, must be the same as sc calculate in pat (b) above. (e) The two equivalent cicuits ae not ientical. Howeve, they have the same cuent souce but iffeent amittances in paallel. 6

7 7 Q. (a) An altenating voltage souce s has an intenal impeance of. Fin an epession fo the puely esistive loa p which when connecte to the souce, maimises the eal output powe. [4 maks] (b) Deive an epession fo the souce efficiency fo the above cicuit when it elives maimum powe. [ maks] (c) You nee to esign a, A souce such that fo puely esistive loas, it elives a maimum of.w at a souce efficiency of 7 pecent. Detemine the intenal impeance of the souce. [ maks] ANSWE (a) When a puely esistive loa is connecte, the powe output is given by P Fo maimum powe, P/ [ ] ) ( P p (b) The souce efficiency at maimum powe, Powe at souce Total Output Powe p m p m η (c) The maimum powe elivee to the loa, ) ( P p m Fom the equation fo efficiency, η.7, we get (/) 8 o /.88. By substituting fo / in equation fo P, we get,.ω. Hence,.88Ω.

8 8 Q6. (a) Fin the ABD paametes fo the tansfome cicuit in Figue Q6-(a). [ maks] (b) Fin appopiate values fo, an of the T-netwok in Figue Q6-(b) such that it is equivalent to the tansfome cicuit in Figue Q6-(a). [ maks] ANSWE (a) Witing the Kichhoff s law fo the two loops, Using the above equations togethe with the efinitions fo A, B, an D, [ ] D B B A But fom secon equation above, (b) Fo the T-netwok, Y Y Y Y Y D B A By compaing the two sets of paametes, Figue Q6-(b)

9 Fo paamete, Y Fo paamete A, ( Y ) Substituting fo Y, ( Y ) which yiels Substituting fo Y, Similaly fo paamete, ( ) ( ) Q7. (a) Using the fist pinciples an geneal popeties of aplace tansfom, fin the aplace at tansfom of causal time function f ( t) e sint. [ maks] (b) n the cicuit of Figue Q7, the capacito has an initial chage of µ. f the switch S is close at t, using aplace tansfom metho, fin an epession in time t fo the subsequent vaiation of cuent i(t). [ maks] (c) f mh, 4Ω, 6Ω an µf, etemine the following values elate to i(t), [ maks] iv. initial cuent, v. time at which cuent becomes zeo momentaily, vi. peak value of cuent an the time it occus, vii. steay state value of cuent. () Sketch, appoimately to scale, the cuent wave fom i(t) fom t to ms. [ mak] ANSWE (a) See class notes fo the aplace tansfom of sin t. ultiplying by e -at in time omain will have the effect of shifting s (sa) in fequency omain. ombining these two esults, we get F ( s). ( s a) (b) When convete to fequency omain the cicuit becomes (the initial chage is equal to step input in seies with the capacito), /s (s) s (s) /s /s 9

10 An applying Kichhoff s law s, ( s) ( s) s s () s () Equation () yiels, s ( s) s s ( s) s s ( s) Substituting fo fom (), s ( s) ( s) s ( s) s s s s ( s) s s Solving fo (s), 8 s () s s s The above epession has a secon oe enominato an a fist oe numeato. n geneal this epession can be factoe as, a b () s k ( s a)( s b) Hence the time omain epession fo cuent i(t) is, at bt i( t) k e k e (c) Substituting values, 8 s () s.s 6s 4 4..s 7 s.4 s.6 Using patial faction epansion, we get.4.4 ( s) ( s 66.7) ( s.4 ) n time omain, i( t).4 e 66.7t.4 e.4 t

11 The initial cuent: substitute t; i().4.4 ma. The at which cuent cosses the zeo ais:.4e.4e 66.7t 66.7t.4e.4e Taking log on both sies,.4 t.4 t (.4) 66.7t ln(.4).4 t ln Thus t.8-6 sec. Peak value of cuent: i e 66.7t.99 t t e.4 (.4 ) 686 e.4 t t ln.4 t t 4.7 oesponing cuent woul be i( t).4e ma e e.4 t The steay state value of cuent is zeo.