* $ ZT- 4 O R The magnitude of the total impedance is........... ohm LI F % Z₁ www R = 60 5 Q Search T ^ 40 6 Y GH & + 7 moo XL = 10 Ω U * 4+ J Z₂ 8 00 - 144 ( K ➤11 ) O HE Xc = 1202 O Z3 L A4 { + [ (1 prt sc } delete ← backspace
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- Questions 16 to 20 (Write True or False) A16 A larger inductor produces a smaller rate of change of current. A17 KVL states that E IR + E EMF = 0 for any open circuits. A18 For a sine waveform, the value of form factor is 1.414. A19 In a purely inductive circuit, the current lags the voltage by II/4 rads. In an ideal transformer, the rate of change of flux is the same for both primary A20 and secondary.The average value of sinusoidal varying voltage is more than its r.m.s. value. Select one: O True O False Which of the following formula is CORRECT? a. E= - L d0/dt O b. E = - N dl/dt O . L= NI / O O d. W = 0.5 12L In a step down transformer, number of turns in the primary winding is lesser than the number of turns in the secondary winding. Select one: O True O FalseIn the given circuit, a step down transformer with an RMS voltage as 110V /11-0-11 V is used with RL = 1 KO. Assume diode as Germanium and the input frequency is F= 50HZ. D, CT Vin RL D2 1- Name the given circuit. 2- Find The PIV. 3- Find Current trough load resistance. 4- Draw the Input and Output wave forms. all
- 2_53528980697... -> For the circuit shown below, detemine the voltage of the -j10 2 capacitive reactance. -j5 0 ILOA(T) ==j102 jlon is Q 102 O 0.5L-90 52 A.Ro. 8. For unit- slep responses A,4 find the homafor the system. chowa ahe figure , 20 see tlJae) fimetion of4) Therenin's Equinlena Circuit and Marimum Poer truns fer Thiorem a) For the Circuit derive the given in the figure, therenin's eq uivalence Circuit acrass the branch witn the loud impedanu. b) Culculate to deliver the volue of the load Impedanca makimum power to the load, c) caleula te the Power dissipa ted through the loud, 72
- From step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?Exeras: 1. Be rebistance of a sm length of oire is cön sr. Detia) the resitance of an8m longth of the same wire, b) the ength of the seme wine when the res tfonce is 4W s. 2. A Piece f re of crss-cfisndd area 2 muy go0 sr. Finid the rebutance of a wie of the some length and maferial if the crss secfional area s SMm. b). the crosss so area of the wire of the some length ond maeriol of resisfance soD2 3. A oril cons o turns of oopperwirehaving a cm-seton area of 0.8 mmn, The maan Lengtb perfurn is s0 čm anithe nosiãfivity of Cu s 0.02 um. Finh the resistonca of the coil ond pwer absorbed 4. An aluminum wire 7.5m. Lang is consacfed parallel writha copper 201ire de m larg. uhen a curent of rdix passd throngh the Cmibination, iř in fannl tht the curent the aluminung wire ik 3Aup The diamefer of the alumsinum wire is 1,0 mm. Defermane the diamefer of the copper wire. has rebi tance of by the cril when camecfed acm8 110V&c supply?The dioces are ideal. write the Fransfer Characteidie equati ons ( &, as a funchiar g &) Plot &o agaiwet &, imdicaling all intrcepts, slopes and Volt ope lerels. Sketch bo f bq =40 sin wt. Indicate al voltage lezels - %3D 1OK lok lov
- M 46 ll Asiacell O Mid electro... -> e: sketh tha oudiput veltage for t followling circut Vsi z0.7 kzo-8 aka ス0 -20 ç: Draw the oulput voltuge for tha following cirent Cand dertermine de Voltuse N: - 152-A resultant current is made of two components: a 10 A D.C. component and a sinusoidal component of maximum value 14.14. The average value of the resultant current is amperes (a) 0 (b) 24.14 (c) 10 (d) 4.14 and r.m.s. value is - amperes (e) 10 (f) 14.14 (g) 24.14 (h) 1002, Zs Z7 Solve for: Solve for: h.) S5 i.) P6 j.) Q7 Given: E = 1002 – 90 volts Z4 = 42450 Z1 = 1200 Z5 = 5453.13N Z2 = 2436.87N Z6 = 62 – 90N Z3 = 32 - 30N Z, = 72750