14 V 1.4 N A ww B F i1 2.7 N F E ww Fi- 14 V 1.4 N C ww iz nd the current i. nswer in units of A

Q: 33KO Rs 2.2KN C. * m_ 10uf I G 10µf Ys = 50mV 1kHz R3.3k2 RĘ3300 1k

A: Introduction While drawing ac equivalent figure of a bjt amplifier you must short circut all the…

Q: 5 A 20 2 w- 25 2 160 -W- a 472 18 Q b

A: Given circuit,

Q: 4.) จงหา Ly ของวงจรตามรูป 0.4 µH ll ll 7 μΗ Le 12 µH 5 µH 5 µH ell ell 0.8 UH 2 µH

A: According to the question we have to find the value of Leq.

Q: 3Ω 2Ω ww 25 V t = 0 4 H

A:

Q: 3.0k2 430k2 Si B=50 2.0k2

A: Given

Q: I, 18k2 6kse 36V 12K2 Cakulate I.

A: By converting Delta network containing 6 kΩ,12 kΩ and 18 kΩ to star network After conversion, Let R1…

Q: .Q1: Find V 30 2 ww 16Ω 15 2 10 Ω 100 V 35 2 12Ω 20 Ω ww

A:

Q: a a 12.5 2 10Ω 52 ww 120 V 30 2 15 2 20 2 b b ww ww

A:

Q: 122 a W 10Ω b e 4Ω d. 32

A: Find the equivalent resistance between points a and e ?

Q: 16 V 430 kf) 4,7 kfN B= 200 Bay =20 µS 120 kN (1,2 kn

A: In DC model analysis all the capacitors must be open circuited .. In the given question they are…

Q: Q1/B:-Determine VBI B= 40 6.2 V VB1 3 k2 186 k2 12 V 12 V

A: DC analysis of transistor is done by finding the finding the DC biasing values also known as Q…

Q: 20 V 470 К ОНм |2.2 КОНМ lo C3 V out I in 1K OHM C1 hfe = 120 Zo hie = 1 K ohm 1nF 24 КОНМ hoe = 25…

A: Considering the hybrid equivalent model of the BJT(Bipolar Junction Transistor) Input impedance =Zi…

Q: OWhat is the value A (Rab)? 20.n 20n Rab. 18n 1.

A: Since you have asked multiple questions in a single request, we will be answering only the 1st…

Q: Home Work 2 9 H ll 10 H ell 3 H 12 H 4Н 6 H a ll ll

A:

Q: - 4s* + 5s³ + 20s² + 4s + 16

A: The solution is given below

Q: 20 Ω 30 Ω 10Ω i2 iz 30 Ω 40 Ω v2 V1 200 V 100 V i4 400 V What is iz in amperes? Round your answer to…

A:

Q: 42 2 82 ww 3. A C 42 2i, (a)

A: From the Nodal Analysis

Q: 20 V 470 K OHM 2.2 КОНМ lo C3 V out I in 1K OHM C1 hfe = 120 Zo hie = 1 K ohm 2.4 KOHM hoe = 25 uS…

A: Given: Vcc=20V......(Vcc is the dc bias voltage)RB=470kΩ.....(RB is the base resistance…

Q: I2 13 I4 10 Α 10 Ω 20 Ω 30 Ω 2 A 60 Ω

A: For the calculation of power in current source, firstly we need to find the voltage across the…

Q: 3. Calculate the following: a) voltage drop across Rs; b) current flow through Rs; and c) power…

A: Circuit diagram is given as,

Q: ab 2Ω 20 v(+ d 423 V ac ww

A: so from the given circuit we will apply mesh analysis to get Vab and Vdc

Q: 14 N 10 N 5 V V i2

A: Here i have solved it using mesh analysis.

Q: 6. Find V1 100 50 24Y V 40

A:

Q: 0.7 Si (A R2 3.3k Si1 Kt 0.7 R1 E1 1k 10 A) 1 E2 10

A:

Q: 36 A 96 SOK vzi=8.6V 38V v22=11.2v S100K

A:

Q: 20 Ω 50 μF 10 mH ww- ell 10 cos 10°r V 202 4i, 30 2 ww

A:

Q: 52 10 2 V1 A 20 v(+ 10 2 4 A VTh

A:

Q: 4s + 32 L-1 (s² – 16). 2

A: we will use laplace transform properties to get the inverse of the given laplace transform.

Q: 5 A 42 22 62 10 A ww

A:

Q: 3s+25 5) y=L-1 s2+6s +13

A:

Q: Find R3 (in 22) for a balanced bridge circuit if R1=18.60, R2=230, R4=13.6 0. (Give the answer in 1…

A: Given, R1=18.6Ω R2=23Ω R4=13.6Ω We need to Find here Resistance value R3

Q: (-1)" (-1)" Wi Hp (e") w,[n] W2[n] Wa[n] x[n] - y[n] Hip (el") wĄ[n]

A: Solution- The given system is shown below,

Q: Using Kirchhoff's laws, calculate the value of the Ig current of the following circuit 10 mn 1A(4 10…

A: The solution is given below

Q: 4 A 30kHz 0.⁰ R2 100 400 L1 40pH R1 260 C1 1µF

A:

Q: R1 = 1 k! R2=1 kll Ry = 3 kll Rs 6 kf R 28 V 6 kN R7 = 2 kfl

A: Given V= 28 V R1 = 1K R2 =1 K R3 =3 K R4 =5 k R5 =6 k R6 =6 k R7 =2 k

Q: 13 I4 10 A 10 Ω 20 Ω 30 Ω 2 A 60 Ω

A:

Q: + i1 12 13 10 Ω 6Ω 1 mF w is = 8 cos 200t A 40 mH wwm

A: (b) The given circuit diagram is:

Q: 60 Ω ww 12Ω 50 ww ww , 30Ω 80 2 40 V 15 Q 20 2 Reg

A:

Q: 1- Calculate and Iz 1 kn V, 16 V V RS skn Vz <-10 V P. <-30 mW ZM

A: Need to find zener diode current and load current In Breakdown region zener diode operates as…

Q: i1(t) 1502 liL(t) lic(t) + vi(t)12u(t) A 300 V 300 2 40 H 1 mF

A: Given :   In the above given question they are asking to calculate the value of inductor current at…

Q: 9 V A) I=?? 10 2

A: According to the question we have to find the value of I.

Q: Rs IL + 220 Iz 20V RL VL 180 Vz=10V Pzm=400mW

A:

Q: shown below: find the value of each element in the circuit if w=377 rad/sec? * jB AB=BC A 60° 1.1 G…

A: Admittance Locus Diagram:

Q: Assume A = 106 HOMEWORK R1 R2 Vin = 12V R = 20kN R2 = 200kN Vin A Vout What is Vout? Vout =? What is…

A: Input is applied to the negative terminal of the given op-amp, so it is inverting amplifier. The…

Q: IOU V₁ = 4√ Soa M Z Micke 2502 5005 199 Ix + 2KG V₂

A:

Q: 200 k D,o W R, 0123 4 5 6789 1011 12 1314 15 +5 V 100 kn D, o D. +5 V 10 kf 50 kn D +5 V D, +5 V D,…

A:

Q: 1.0 kfN 1.5 kN 3 4.7 kN 30 V 4.7 kN (c)

A: The solution can be achieved as follows.

Q: ЗА 6 8 V 3 10 2 A ( 16 12 20

A:

Q: R1 = 500 ohms R2 = 10 kilo chms R4 = 8.9 kilo ohms 30 V R3 - 3.4 kilo dhms R5 = 10 kilo ohms

A:

Fault Analysis

In an electrical engineering system, a fault is an abnormal electric current flowing through the system, and fault analysis is the study of these faults. For instance, a short-circuit is a fault where the live wire touches the ground or neutral wire. If a circuit is interrupted by a failure of a current-carrying wire then an open circuit fault occurs. The open-circuit faults are also called series faults. One or more phases or ground may be involved in a three-phase fault or may occur only between two phases. Current flows into the earth in case of a ground fault or earth fault. For most situations, the prospective short circuit current of a fault that is predictable can be calculated. To limit the loss of service because of a failure, the protective devices can detect fault conditions and operate circuit breakers and also other devices. All phases equally can be affected in a poly-phase system which is known as symmetric electrical fault. The asymmetric fault becomes more complicated to analyze if only some phases are affected. By using methods such as symmetrical components, the power system analysis of these types of faults is often simplified. The main objective of power system protection is to design, detect, and interrupt power system faults.