Question

Question 1. (a) Consider the waveform below that has a period of T of 0.04 seconds. What is the rms voltage of this waveform? V(t) 15 V 0.01 s 0.02 s T=0.04s (b) The figure below shows a network of resistances and reactances having values as shown connected to a voltage source with a voltage of (50+0j) volts rms and an angular frequency of 1000 rad/s. А. 32 3 mH V 52 200 uF 2 mH 1 mH (i) Redraw the figure in your answer booklet and on it put the complex impedances of the capacitors and inductors at the 1000 rad/s angular frequency. Find the complex impedance operator of this circuit as seen between the two terminals A and B. (iii) Calculate the total source current I given out by the voltage source in symbolic notation. (iv) Give the angle (in radians) that the source voltage phasor leads or lags the source current phasor. Remember to state whether current leads or lags the voltage phasor. (v) Calculate the current passing through the 5 2 resistor in symbolic notation. (c) Use the principle of Linear Superposition to calculate the total voltage across the 322 resistor. 392 12 4V 222 8 V (d) A piezoelectric transducer has the Thevenin equivalent circuit of an ideal voltage source in series with a capacitor. At one particular frequency the following observations are made on the transducer: (i) The open-circuit output voltage between the transducer's terminals is 10 V rms. When a 42 resistor is connected between the terminals of the transducer, the voltage across the transducer terminals drops to 8 volts. Find the value of the transducer's Thevenin equivalent voltage source and impedance at this frequency. Draw its equivalent circuit.

Answer 1

defined Question Thermo value of the quantity TV6 vadt 1/2 ao- 15 stat 002 0:04 Now the ware form to given up.ool Now Voltage mi methenetical form!- 1500t oft cool V€ 2= -tsooft45 -1500-t-0.01)+1570 0:01Stro.02 0-0250.04 Sefmally the function > ost.so.cl ve -150067 30 0:015190'02 0.025t50.04 1500t 0.02 1/ o of NOW Vimos (i 500+)?dt + 55-15806730’dky odt O'O2 ( 20:04 dt 1/2 Ool goopth is sy metrical about 7= 0'Olot (1500tedt - (21 4 .) 2 t3 3 2x1500) 2 x 004 0.04 s ( 2x13 xI5Y2 = 6.12 v Ano (4x2

6 Vomo of the circuit = 50v and ongulas frequency w=SIN loro rad /o) - 3 3 DXR=3 3ml 3XL=3j 200 MT Xc = -xs PIMH; XL=0 XL=21 TamH 4 B 2 Complex impedence of mouctor XL=jc2 Capacher XC = -1/wc Complex impedence of total arcuit m branch 3-4 Resistance , Capacitan 2 In dulrance are comneted miseries. 334 Eguberce) abent 3-4 = XR+ XL+ Xc = 3 +4-5) = 3-4 Now about 182 212 = 3; +5+23 = 5759 Now 234 8fe are one connected miparabler. Now Zz 4 in ealer's fomi- -|*53° 53 -5e 9

Similarly 212 = Stei74 5 5/2 $ Now for paralle compriation .-1853 1:45 À Sage Zest=se 1959 Se-1x53 + Suce ③ C 3-4]) (5+61) 3-4j+5+5 - 13 (3<259) 3+1 56721)=5674/2C2-1) 2+1 (0+1)(0-1) =$(55-19) 742 Ano 6513 Zeff= S 55-15) 55-15) - 5 (11-31) 13 Now V=50 volts sms' Now Z in puler's from » 3=5&dizoe -j tent CH) 13 5x130 -jų 15-26 ஓ' (ii) 13 bence current will lead the voltage by anangle of 15-26 (iv) part e Auss

(iii) Now smo ceissent by Source Irmo = Vo 121 5d130 = 50x13 - 1130 A. ñ 11-4A. Now current Canbe written as I = 2130 Cept + 15-26) Veneet ☺ Ao the Brancho are pcerallel.. and mopedenee of branch 1820 5+5] =5a2ejna so current in the be brance V 50 Cep cot 712 502 ejta = 5d2 Galeot-774) so current through sro = siz Amp. @ sur ces (w t- 7714) lag by an ample I = of ITA

12 32 AV 22 or I 1st Cenoldering only 4V battery 12 It 3-2 Il I, 1.02A 10 4r I2=073A I3 = 1091A le 22 tana Considering only 8V Battery Iz' $22 Izl or 2 & fI3' TIL Iz' = 2-10 A Iz!= 0.73 A. Il' =1.45A Now superimpove o 2% n 12 = Current Itt I'= £82+1-45. 3.27 A. Ano - 1-09-0:73 036 A Ano I3-Iz! mga =) I2 + Iz' = 818 +0.73 321 2.9 1 A Doe