this Should the Function Proble m you implem ent F = A. 5. C +A B.C.D vsing only half adder and one umber of 2:1 multi plexers where A, B, c,D mini mu m are the in puts. Yov not all owed to Use are an y logic gates.
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- 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?Qu / MA 20 Using the afind the Ruh Sind the veAih's Voltage Vthi Bfind the value df load resislance? Gfind The max imu m Power at the Condidlon delivered to lead2The 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
- Graph the foll owing functio and show domains and ranges them ot 3X %3Dgoogle.com/forms/d/e/1FAlpQLScF2Wzu92iVI-tEM2nxdhqXrGdR460G4QCTxaUZZeC678PqTg/formResponse مطلوب /06 *:The potential at point P due to Q1 and Q2 is Q, = 1x10-12C 0.5 m 0.5 m Qz = 1x10-12C %3D 50 cm 0.01797 V Ov O -0.01271 V O 0.01271 V O Insert PIC F10 F11 F12 F6 F7 FB Backspace 9 € P Enter H2, 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
- In the circuit given below determine the power assosciated with each independent voltage source : p14v = ? and pev = ? indicating also if it is power generated or power ansorbed. ty %3D -M- 14VProve this formula plzDraw a seven-segment display, show the number highlighted in figure 1 (d3 -d0 are the input, a -g are the outputs). III. Simulation Wavetorm Editor - Cers/Ahmad Fainuz/Desanop/DLD Lab Practice/bedseg - beaeg - bcaseg 2018o03121936sim.vwwt Ee ep Ea w smutation search altera.com Haster Time Bar: ops Pointer: s29 us Start End iterval 529 us o ps 0 ps २ 2.56 us S12 us 75 us 10.24 us 12 s 15.36 us 172 us value at Name Ops do Figure 1
- 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.7 The eir fiçare isa unit foedbach Ris) crs) 5(s+2 with we arinad feenbreh. loop- Ja the absemße of Lerivadine Reedb ack la=0) La har mine t he damping foebon and reoulhing from nm farab fropusney, Alow dederminne ofde stamdy- a uit sgpp eiprt. slede erer 6) D edermne the derivatice Red back conshant of which will imepéase the domping facler the sys hene to o,7. what is the s kemndy-s Lude erron Lo nit romp inpnt withe theé setting of the derivahiue feedback c) conslaut. Telustra kee how the sheady-sLaker errer of the sye ten with derivedine Loadback do unit romp impat can be reoluced to the nahuta as in part (a), while the odomping fachon is at 0,7' Same monin LainedFor the circuit diagram in figure Q22: which of the following is the total impedance? 12 Ri 10R R2||5R Figure Q22 V 220V40 50HZ L1 315mH C1=350uF O a. 3.33/0° O b. 4.71290°2 O c. 7.3304-19.66° O d. 10.37261.19°2