Q.4/ Find Rf and fr: R, 4.7 kf2 0.022 uF 0.022 uF 0.022 uF 4.7 k. 4.7 kM

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Q: VCC 12V VCC „R1 2.3MO Q1 Xsci 0.47UF 2N2221A Q2 R2 2.3MQ XFG1 2N2221A 4 C2 R3 3.3k0 0.47µF

A: For calculating the required parameter, we will do the DC analysis of the circuit. In the DC…

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Q: Find Vout in terms of waveform generator.

A: The derivation of Vout in terms of waveform generator has been derived. A detail explanation has…

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Q: V R3=5 2, 0A 10V R2=8 N -8Ω VTA

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Q: Vcc Vcc = 12 V Rs = 1 k2. %3D R1= 50 k2. RL =14 k2 B = 100, VBE(ON) = 0.7 V Rc R3 C3 Icq2 C2 Q2 RL…

A: Here dc parameters are given so we will do DC analysis, then we will apply simple ohm's law and bjt…

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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.