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- Ererrises: IA platinum oil has a resifance of 3-14k r at go°C and 3.747JV at 100°C.- Fin'd the resistance at o'C ond the kmp -creff. of resifonce af 40°C 2. A pofential difference of r5oV is apphed to a field wnding at 15°C and the currend is SA. What will be the mean fempouafure of the winiding when currend has falen to 3.91A, apphedvrltage being constant Aasume s =p7.5. 3. A cril has a resi fance of 182 when its mean femperature is 20°C and 20v when ik mean femperature s 50°C. Frid is mean femperature rise when ik resotonce is 21JL and the sorrandinig femperature is i5°C. A cril of a relay takes a curre of 0.12 A when it is at the vem insbi o al 4. A jemperature of is'C ond canneckerdacms a oV supply If the minimum operating current of the relay is 0.1 A, colculate the femperature above ohich the relay will fail to operate when connected ti the same supply. Resi tncenfemp.coefficrenf of the cril makria] is o.0043 per °C at 0°C.igtht Login .. ILUVEPDF ulg3l P.. Q/ Given the information provided in figure below, determine: RB Vcc 32.1 μΑ 32.5 KN Rg VCE = 7.3 V 2.1 V 20.42 K2 RB = 952.5 KO RB = 925.5 KO a RB = 592.5 KO1. Express the following in Phasor or Instantaneous form : Find Phasor Form: a.A = 5sin(2t +5x) ŷ b.B = 3cos (wt- 2z)x- 3cos (wt- 2z)x c. C = 50 sin (wt-2z)p Find Instantaneous Form: d. D, = - j50e -6z e. E, = e* (i-2)sin (zy)
- :D A docs.google.com/forms/d/1 o المرحلة والشعبة: * A non-sinusoidal voltage (e=20+15sin1000t+ 10sin3000t+5sin4000t) Volt is applied to the circuit shown. Find the ?total power iT 42 İL -J92 J42 44.03 W 86.08 W 88.06 W 44.04 W n-sinusoidal wave can be طلب الإذن بالتعديل of sine waves with .harmonic frequencies IIQuistion.4. a) A sinusoidal voltage is applied to the resistive circuit in figure.Q.4. a. Determine the following: i. Ims ii. Iavg iii. Ip iv. Ipp 10V 1.0 ΚΩ Figure.Q.4.aConsider the following circuit used to provide power for an induative RL load. The input voltage is V-100V and the load has a 50 impedance value. The thyristor is working at a frequency fs = 2 kHz. The discharge current is to be limited to 40A and the required dv/dt is 40V/us. If the value of Ce is equal to 0.14uF, then the snubber losses are equal to: R. V. Select one: O a 5.4W O b. 7.4W 1.4W Od. 3.4W
- s are annected in series across Four curtuit elements a sinuscidal alternatue veltoge guien ky e = 10 sin (ut + 30°). The ins tantaneaus velbae acoss three of the elements art gwen by Y, = 30sinwt, Vz =60sinlwt ¥60°) and Va Determine the expression for the fourth form V = Asin (ut tB What is r.m.s Yalue of V4. - 30sio (et -30). veltage in the %3D %3D ). %3DAdd : V, = 30 cos (w t + 50°) Vz= 50 cos (wt -b0°) a. Per for m Vi +Vz =V and finally expre so V as signal b. Find v and write v in both forms of phasor a CoS"If lab is equal to 595.8252306 and lb is 344296, the phase sequence is:" O POSITIVE O NEGATIVE
- Exercise-1 Vs = 240V 60Hz Is=10A/....... Power system Is ww R 1k0 Xc=-j/(2pi60x0.00016)=-1994 XL=ZL=j2pifl=j2pi60x0.142=j53.54 142mH Is=Ir+IL+Ic=.........=Vs/Ztotal=....... 160μFVR. VR2 R2 R. Ve YL AIs Ic 6ト0°Y f: 50 H2 2mfC 15mH3 L Find= Is, Ic, I-, Vai, Vaz, Ye, VL (ANSWER IN RMS)Ministry of Manpower Directorate General of Technological Education Salalah College of Technology Zlectrical Inginearina Problem - 8 Refer to the circuit below and -25V C2 N. compute the following: i) Total capacitance ii) Voltage across C, iii)Charge across C, iv)Voltage across C5 5 uF C1 5 uF 5 uF C4 5 uF C6 5 uF -16-65V. C5 5 uF (Take V, as 25V) 8-35V VT 25V