Electrical, Electronic and Digital Principles - Assignment Example

Principles of Electric and Electronics Assignment September 9, INTRODUCTION The determination of the correct measurement parameters of a circuit is of paramount importance and needs to be put into consideration. This parameters are expressed in different forms such as the instantaneous, peak or maximum value, average, RMS, among others. The load, current source voltage and current shapes are considered when considering this. Principles of Electric and Electronics Instantaneous voltage The voltage between two points in a circuit at a given time is the instantaneous voltage at that point. Instantaneous voltage varies at different points in a circuit, i.e. instantaneous voltage is not constant (Serway and Vuille). Using the instantaneous velocity, the potential difference of a single point can be determined. Q1. What is the instantaneous voltage across a 2 µF capacitor when the current through it is: . Explain your chosen approach to the solution of the problem. Justify your arguements. Solution Where C is the capacitance Maximum Current, Maximum voltage, Obtaining the values of maximum current and maximum voltage is easier using the above method. Consequently, this values are applied in getting the instantaneous velocity with ease. Instantaneous current The current flowing at a point at a given moment is the instantaneous current (Matthews). The voltage across a 4 MH inductor is: v = 60 cos (500t – 650) V. Find the instantaneous current through it Q2: X is calculated to be 4.41, thus Maximum current, From the above, we can conclude that Q3: Find VL in the opposite figure. Explain your chosen approach to the solution of the problem. Justify your arguments. Solution This approach involves finding the reactance and the total impedance. The values obtained are then used to the voltage across the inductor. Q4: Find ix in the opposite figure. Explain your chosen approach to the solution of the problem. Justify your arguments Solution Since the value of phase angle in voltage equation (instantaneous violtage) is zero, it implies that the value of instantaneous current phase angle provided that current leads voltage by 90 Q5: Calculate the voltage Vab and the current iy in the circuit shown in the opposite figure Solution In the two port network show, iy equals the source current, implying that iy=10A Q6: Find ix (t) and vL(t) in the opposite circuit. Evaluate the effects on circuit performance if you change the values of the impedances. Simulate this circuit in the transient analysis. Justify your arguments. Solution Therefore; Current across branch this implies that Changing the value of impedances causes a sudden increase or decrease in the value of current and voltage (transients) flowing across the load Q7. Determine the value of RL for maximum power transfer to RL, and determine this maximum power. Explain your chosen approach to the solution of the problem. Justify your arguments. Solution Voltage sources are: 4V, 1.3 V and 2V: R1=2ohms, R2=8ohms Shorting 2V and 1.3V R’=8+2=10 ohms So, for maximum power delivered, RL must have the same value, RL’ = 10k Shorting 4V and 1.3V, R’=10ohms Now short 4Vand 2V, R’=1.6ohms For RL to receive maximum power it must have the same value as the total sum of all series resistance in its path to power source 8) A two-port network can be modeled as a load supplied from an a.c. supply. The load consists of (a) a heating load of 15 kW, (b) a motor load of 40 kVA at 0.6 power factor lagging, (c) a load of 20 kW at 0.8 power factor lagging. Calculate the total load from the supply in kW and in kVA and its power factor. What would be the value of a capacitor to bring the power factor to 0.96, and how would the capacitor be connected? Simulate your circuit on PSpice to verify your answer. Comment on the simulation results. Solution Q9. Simulate and analyze the operation of the following amplifier: The Common-Emitter Amplifier is used to achieve high voltage gain and employs a bi-junction transistor (BJT). Parameter definitions: The AC analysis of the common emitter amplifier involves replacing the transistor with its AC equivalent circuit. The equivalent AC circuit of the amplifier is given below: Q10: Design combinational and sequential digital electronic house alarm circuit which can be used to monitors 2 doors and six windows. Your circuit should have the flexibility to either include or exclude any of the doors or windows in setting off the alarm. Solution The inputs to the circuit are A, B for doors and the windows are shown as: C, D, E, F, G and H The output of the logic circuit w=ZY CONCLUSION With the application of the above principles, circuit characteristics can easily be determined hence circuit analysis. Application of the basic principles of electrical and electronic engineering has been greatly achieved when solving the above engineering problems References Matthews, John. Introduction to the Design and Analysis of Building Electrical Systems. Springer, 1993. Print. Serway, Raymond A. and Chris Vuille. College Physics. Cengage Learning, 2011. Print. Read More