10 kN C, 560 pF C2 560 pF 8V f 20 kHz Figure 3. Diagram of the RC electrical circuit

Q: 5. The voltage across a coil when di/dt = 20 mA/ us and L = 8 µH is O(A) 16 mV O(B) 160 mV O(C) 1.6…

A: Given a coil with, Rate of change of current, didt=20 mAμs Inductance, L = 8 μH

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A:

Q: Circuit Analysis. 4. If the minimum frequency is reduced to 100 Hz, what value of bypasscapacitor C2…

A:

Q: Q-2. For the circuit given below Given: Vat = 12V 22 n 30 LO 10 n 1 2 as vo w R 1.0 kn 1) Identify…

A: The solution can be achieved as follows.

Q: Solve the approximate value of Rc of Figure 44. VCC 12V V=7.6V V = 2.4V V₂ RB 2.2 ΚΩ 2.7 ΚΩ 1.8 ΚΩ…

A: We need to find out collector resistance value for given circuit

Q: 1- What is the input impedance of the circuit shown in the figure below 15 V 560 k2 B-65 0.56 k2 33…

A:

Q: +20 V 10 ΚΩ C2 22 ΜΩ Ci VGSCTH) = 3.5 V V; a k= 0.3 x 10-3 8os = 30 µS

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Q: Q3. For the circuit shown below, draw the expected output voltage and current waveform and determine…

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Q: 3. Draw the output voltage waveform of the circuit with explanation: /- 1000 Hz C = 1 µF 10

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Q: Choice the correct answer: Operational amplifier work in …. 1) DC only 2) AC only 3) both AC and DC

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Q: 3. Determine a. Ic. b. Rc- с. Rв. d. VCE- 12 V Ic RC RB O Vc = 6 V + VCe B= 80 IB = 40 µA

A:

Q: %D Calculate the output impedance for the circuit shown in figure. Assume VA = 0. Circuit parameters…

A: Given data:

Q: 18. Complete the following table with the proper equivalent temperatures °C °F K °R -40 77.0 698…

A: Given data, Thee values of temperature in different units,

Q: Find lo in the circuit shown in the figure below. -j1 N j1 N j2 N j2 N 4/0° A (1 310 -j2 N O.

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Q: i. To amplify the signals we use [ ] a) Attenuators b) Compensator c) Bandwidth d) None

A: The basic problem on amplifier.

Q: Q/ An a.c bridge is in balance with the following constants : Arm AB , R=200 2 in series with C=…

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Q: The value of input frequency is required to cause again (Vo/Vs) equal to 0.3. The value is: 2 k2 60…

A: Given circuit,

Q: Q/ An a.c bridge is in balance with the following constants: Arm AB , R1=200 2 in series with C…

A:

Q: PROBLEM 13.6 In the circuit in Figure 13.82, the switch has been in Position (1) for a long time. At…

A: Given  In the circuit in Figure 13.82, the switch has been in Position (1) for a long time. At t =…

Q: The amplifier circuit below has a single ac input and two ac outputs. Assuming transistor parameters…

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A:

Q: (a) Calculate the time constant t and the pole frequency fp (in Hz) of the circuit below V out + 20…

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Q: R Vn Vout 1. Find H(w) = Vo/Vi. 2. Derive an expression for the cutoff frequency. 3. Calculate the…

A:

Q: The converter in the Figure is connected to a 450V, 60HZ supply and has a purely resistive load of R…

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Q: Find R and the frequency of operation [rad/sec] for the circuit below: 100KO 5ΚΩ Vo 0.01 10 1KO µF…

A:

Q: Design Boost Converter Calculate the value of the capa inductance and resistance according to the…

A: Given boost converter  Vin=100 vVout=200 vfsw=20 kHzΔV=5 % ΔI=10 %

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Q: Q3. For the circuit shown below, draw the expected output voltage and current waveform and determine…

A:

Q: The baud rate of 2ASK is ____________________ than baud rate of 4ASK. Select one: a. less b. high…

A: Baud Rate = Bit Rate / n    RS = RB / n  n = Log2( N )     Given  N1= 2 N2 =4

Q: Vcc اشتقاق القوانین RC RE C2 Figure 8.30 Collector fee configuration with an emitt sistor Rp. RE RE…

A: Dear student as per our guidelines we are supposed to solve only one question in which it should…

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A:

Q: Vin (t) pin 3 1.0 ΚΩ www 1.0 μF Op Amp quivalent circuit (buffer configuration) 1 v. pin2 and 6 2.2…

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Q: 9. Determine the low cut-off frequencies of the figure with following parameters: (Remodel and draw…

A:

Q: Compute the cutoff frequency. C C 4.7 nF 4.7 nF 44 O2 kHz O 1.4 kHz O2.82 kHz 03.12 kHz R₂ 12 kQ www…

A:

Q: If V esa = 7 V, C, = 86 nF and C2 = 126 nF in the circuit in the figure, what is the low cut-off…

A: Given circuit: VGSQ=7 V C1=86 nF C2=126 nF To find: Low cut-off frequency?

Q: for the given figure answer the followings w=100 rad/s 60 Q 2/0° A 60 2 4 m H ell 0.01 F bo 4

A: The solution can be achieved as follows.

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A:

Q: o 15V R 10 k2 8. 5 566 220 pF

A:

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Q: V RB VCC 12V RC Beta - 80 RE 1.2kQ Figure 44 V=7.6V V₂ = 2.4V

A:

Q: The equivalent resistance RAB (in the figure) is: 12. BO 10.

A:

Q: are used to increase the temporary voltages. over Static Var compensators Select one: O True O False

A: SVC tends for Static Var Compensator.

Dielectric Breakdown

A dielectric is a material with low electrical conductivity and so is classified as an insulator. Dielectric materials are used to block the flow of electricity through them, allowing them to behave as an insulators. They can also be used to store electric charge or improve capacitance.Dielectrics have no free electrons for conduction because electrons are tightly bound to the nucleus. Because dielectrics contain a small number of electrons for electrical conductivity, they have a dipole. The two primary roles of dielectrics in electrical devices are insulation and electrical energy storage. An electric field polarizes the dielectric, and the polarization of the molecules allows dielectric materials to store charge.