R a 0 Х R ли 2 л чее L= 10m H If V₂ = ±7√ when x goes to terminal a, what would be VL when t = 25ms?
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- r cut 1oum 20gm 1ocm 个 //// wax poper/ A capacitur is mad by sandwiching a sheet of wax psper between two sheebs of aluminum fail, as shoumm What is the figue. capacitance! I the capacitoris aut into twopieces, alme the hme, what oil be the expacitenel of each Diece ? wiggly What lo be the capactncebelow triminal that e the nnin calculate obtain to the circult by Seen OF R the eQvivalent as a-b, Hence Svitable Value So %3D 20V aIf the voltage V, = 800 mV, then the current flowing in the load I, is: %3D -j20n Select one: O a. None of these O b. Is 100 <-90° mA O c. Is 5.0<.-90° A O d. Is 5 <90° mA ell
- Take R1= 2ohmsمهند زعيتر تحويل مقا AEENG250 AEENG 250 Series an in EENG250 ttempt=196811&cmid%3D11615 My Courses This course 2.J, 50 O d. 11 = 2.5 A, 12 = 5 A, V=25 V Question 5 Find Rab- Not yet answered 20 2 Marked out of 5.00 * Flag question 16 Q 18 2 20 Q 9Ω. ww 20 Select one: O a. 16 Ob. 20 C. 34 TOSHIBASolve for ro- I₁ Z₁ 2 Mohms 50 kohms O 40 kohms 100 kohms www 330 ΚΩ Vcc To 2.7kQ2 HA hfe = 120 hje = 1.175 kΩ hoe = 20 μA/V Zo
- 4. Solve for I, I2, Iz and I4 in the ckt below using Cramer's Rule. ISMAT169NUMI 252 METHO 169NUMERICALIM SANDANALYSISM THOANDANALY RICLMET 169NUICA MAT169NUMERICA 幸 DDSANDA MAT169NUMERI THODSANDANALYSSMAT ANALYSISMAT 9NUM METHODSA 12V (1) I2 1 YSISMAT169N CALMETHODSANDA 351 ETHO SISMATI6 OSANDANAL RICALMETHODSAND NDAN SAND ANALYSISMAT 丰 RICAM MERICALME NDANALYSISM NUMERICAL METHOSAN SAND ALYSI METH JALYS 69NUMERI ISN I) I3 ERICALM ooDANAL TRICALMET NDANASA SMAT169NUAALYSISMAT169NUMERICALMETHODSA AT 169NU SANDA MERICLMETHODSANDAIf the voltage Vs = 800 mV, then the current flowing in the primary winding Ip, is: 1:8 -j202 Select one: a. Ip =40<0° mA O b. Ip =320<-90° mA O c. Ip = 0.320<90° A %3D O d. None of these llDevre Teorisi I| II. öğretim 7. Hafta Ders Videosu.mp4 Nachste Loschen Board Internet Dokumente Desktop anzeigen OpenBoard Örnek A m I aA a! a J2n mm. Seite 1 J 28 n Seite 2 | 240L0° Io b' 3r J182 Seite 3 172 240L120° Seite 4 :-J42n 240L-120° Seite S Icc J 2n 17n a) 1o=? b) VAB=? c)Vca=? Seite 8 1/ 13 15:37 YS TD AG KD L 0:58 / 2:18:17 YAĞIZ KAAN SERT FATİH GÜNDOĞDU TUGÇE DÖNER ABDULLAH GÜNES AGAH CEYLAN
- Solve for computed VA IR3 using NODAL ANALYSIS R1 – 4.7 kΩ • R2 – 6.8 kΩ • R3 – 10 kΩ • R4 – 22 kΩ • R5 – 33 kΩIn the figure: R1 = 1439.170, R2 = R1 and R3 =2R2, then determine the value of the voltage source E. a 8 mA 10 mA 2 mA RT R1 R2 R3 O 25.33 11.51 O 17.27 15.43 13.82 +P1. 24. Determine the value of a so that A = 2i + aj + k and B = 4i -2j - 2k are perpendicular. %3D %3D Select one: О а. 3 O b. 4 O c. 5 O d. 2