Given the following circuit with Is-40<30° mA 12:1 V 10002 Select one: O a. V, =120<60° V O b. V,-480<30° V O c. None of these O d. V,-480<-30° V ll
Q: The Norton current at terminals a-b in the circuit -j2 Q a 6/0° V j4 Q ll
A: The solution is given below
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- The manimum value of the altenating wotaye and curent are 400 U and 20A respectvely s a cieritoanected to a svita supply . The ristantanerus values of the veltnge and current are 283V and 10 A respectvely of fime t=0, beth icreasing pesitively. 4) Write dan the gapressom for vwltaga and curend at pmet. b). Defermaned tha prver omsumed in the cranit. Take the velye and cunent to be ginusrida). fie P= VI end\ P = YI (PF)For the circuit shown in figure. For the circuit shown in Figure IF IB= 40 MA , VBB= 6v, RE= | Ksr, P= 80o FRc Re VeB Re The value of IE is . . mA..* О 3.24 O 4.24 O 5.24 O 6.24 O 1.24 O 2.024 The value of RB is . Kohm.. * O 52.5 O 55.5 O 47.5 O 57.5 O 25.5 O 48.5 The value of Vce is .. volt. NOTE: USE Rc= 3 Kohm, and Vcc= 15 volt. * O 5.04 O 2.04 1.04 O 4.06 O 4.04 O 3.04 O O O O O ООО О оExplain what each component is in the design below: DD 7Ro PQ5 8Eo DD 9 So 10 - type of circuit? 1. 6. 2. 7. 3. 8. 4. 9. 5. 10. CO
- 4) Label magnitude and direction for the currents through R1, R2, R3, R4 & R5. Show all work.In the following circuit, VGG = 3 V, Vp = -8 V, loss = 10 mA. What is the current lp? V₁0 C₂₁ Select one: O A. 18.90 mA OB. 18.60 mA O C. 10 mA OD. 3.80 mA OE. 3.90 mA VDD os 45Find the open circuit voltage between terminals a & b. 160 m 120° 100/0° V Select one: O Vab 402 -156.25238.75⁰ j8Q2 b. Vab125238.75⁰ O Vab156.252-38.75° Od Vab1252-38.75° a
- The ratio of the forward voltage to the forward current is called the characteristic impedance? What is forward voltage? What is the forward current? What is characteristic impedance?for The Circuit shown in figure if CB = 50) find The values of CIe). and of Then express The c hamge in value by apercentage + 15レ こ RB2 = so KQ RE 3K522. The circuit given below is using a transformer of 230V/30V. The load resistance is 10KS. Find Vin (rms and peak values), Vdiode and VLOAD (Peak and dc values). Draw the input and output voltage waveforms. (Si diode) Vin: Vin rms= Vin peak = VDIODE Vdiode = 4 Vin VLOAD: VLOAD Peak =. Load VLOAD VLOAD DC= 2031 Supply wwwwww
- Find the open circuit voltage between terminals a & b. 402 -/302 j6 Q ww HH m 100/0° V j8Q2 Select one: O a Vab O b. Vab125238.75⁰ OC. Vab156.252-38.750 O d. Vab 1252-38.750 156.25238.75° /20 aMATLAB R2022b-academ HOME New Open Save P |999999941, FILE ++ ADD¯ ► Cu... O Editor - C:\Us +7 MP5_5 N = n_sp J = kB = T1 = T2 = n_mc n_pr D... A W... Ⓒ Name col delt HE E_a Hi HJ kB kB_ kB latt latt M M_ |N n_n 1 2 3 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 8 PLOTS 25 26 27 28 29 CL fx >> latt latt % In M = E = M_ar E an % MO tic; for Command Wind >> SW8 Tr Magnetiza Magnetiza Total tim Total tim .....3、、、、...2... L File 1 Page 2 of 2 Home R1 Insert Design R2 R3 6. Which of the following resistors has R = 2 k? a) A b) B c) C d) D 9 8 Effective resistance (kn) 7 6 3 2 1 0 Layout References 0 8.00 4.00 (9) I (mA) V (Volt) (7) (8) A -B -C D 0.5 1 1.5 0 words English (Philippines) 1 2 3 1 (10) Mailings . For numbers 7 to 10, complete the table of missing information. 4.00 2 V (volt) 1 I V1 12V Review 2.5 R1 Document1 - Word (Product Activation Failed) Tell me what you want to do... View CAT NOOISTOT T 1kQ2 3 3.5 4 2 R2 ww 1k92 R3 1k92 * ****.. 4 5 6・・・・・7・・・・・・8・・pa + 1 For…. For the circuit shown in figure. For the circuit shown 16ke Rc 18V in Figure. If VeE =-|4V -14v RB Vec=18 V, VeE = 8 V, I8= 8MA, B= 100 %3D The value of RB is Kohm. . * ... ... ......