th frequer Hz) and so rmonic of 10 MHz O 2 MHZ O 3 MHz O 6 MHz

Q: 3. Let's throw some actual values on here. Let's set R = 1kN, C = 1μF, V = 10V, and the frequency of…

A: Given:  R = 1kΩ, C = 1μF, Vs 10V Input squarewave frequency = 1kHz Need to sketch the output…

Q: Calculate the total current in the circuit below if the circuit is powered by the sine wave voltage…

A:

Q: +1 V Vin 0- -1 V R Vout (d)

A:

Q: . In the figure below, calculate the output voltage Vout, and the phase angle with VIN at 150 KHz.…

A: In this question, we need to determine the output voltage Vout  and the phase angle with voltage…

Q: In the figure below, calculate the output voltage Vout, and the phase angle 8 with V₁ at 300 KHz.…

A: Given dataTo findOutput voltage ,Phase angle ,

Q: Vin o Cy Vcc = 12 V RC 3.3 K RB 270 K www Please determine the AC voltage gain. Cx 1.5 K RE Vout Cy…

A: Step-1-> In this question we need to find the AC voltage gain

Q: toe HA yov find UCH) for to

A: The solution can be achieved as follows.

Q: find and draw excitation table

A: Let the D Flip flop is used Present Variables  = A B Present Input = X Output = Y

Q: The frequency at which this circuit operates is O 50 Hz 120 Hz 100 Hz O 60 Hz

A: In this question, we need to determine the frequency of the given waveform.we know, frequency f=1/T

Q: Vin Cy Vcc = 12 V RC 3.3 K RB 270 K Please determine the AC voltage gain. Cx 1.5 K RE Vout Cy beta =…

A: In this question we need to find the AC voltage gain

Q: lc = 14mA. Determine the lower cutoff frequency f (in kHz) ? Vee 250 R 200 Vee MF 100 PF Asoume

A: The given circuit diagram is shown below, From the above figure,RE=R2=200ΩRS=250ΩC=12πμFand IC=14…

Q: Problem A: Given the time domain current i(t) = %3D 20cos(200t + 20 deg) A. Convert the current into…

A: Given time domain current is, i(t)=20cos200t+200 A

Q: The crystal oscillator of Arduino UNO * is O 11.0592 MHz O 16 MHz O 14.7456 MHz 8 MHz

A: We need to select correct option for crystal oscillator .

Q: Please read the question from the picture  X = 03

A: According to question: The pulse width is calculated as

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.