Electronic Principles Assignment Sample

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Introduction to Electronic Principles Assignment

Part A - Laboratory Assignment

Question 1

Electronic Principles Assignment Sample

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Lilian Barker 8 Years | PhD

• Here the graph between voltage and current for the 510 ohms resistors and 910 ohms resistor have been plotted on the same graph paper.

Figure 1: Voltage and Current plot of 510 ohm and 910-ohm resistor
(Source: Self-created)

For the 510-ohm resistor the gradient or the slope can be made out from the formula of straight line which is y = mx (Petralia, 2021). Here the intercept is zero.
Therefore, slope m = (y/x)
For 510-ohm resistor,
Slope m = 50/80 = 0.625.
Resistance = ΔV/ΔI = 4.62 / 0.025 = 184.8 Ohm
Percentage difference between the given the resistor and the practical resistor can be calculated as,
Percentage difference = (510 - 184.8/ 184.8) = 1.75
For 910 Ohm resistor,
Slope or gradient (m) = 40/ 90 = 0.444
Resistance = Δ V / ΔI = 5.94 / 0.045 = 132 Ohm
Percentage difference = (910 - 132)/132 = 5.89
The resistance which has been given in the question, might have been calculated on the basis of ideal cases (Rosa et al. 2019). The practical values are calculated based on the practical results where the ideal cases are not considered (Yikui, 2021). The values on the capacitors and resistances are marked but in the practical cases the values get changed from the marked values.

Part B- Theoretical Work

Question 1.

• Let just assume the equivalent resistance of the circuit be Req
  Req = 3 Ohm||12 Ohm||48 Ohm + 12 Ohm+ 14 Ohm = (3*12/15) Ohm||48 Ohm + 26 Ohm = (2.4 Ohm||48 Ohm) + 26 = 2.28 Ohm + 26 Ohm = 28.28 Ohm
• Current (I) = 9/28.28 = 0.318 = 318 mA.
• Power dissipation across 14 Ohm resistor = 14 Ohm*.318A = 4.452 Volts.
  
  

Figure 2: Circuit Diagram
(Source: Self-created)

Question 2.

• Let just assume the overall equivalent capacitance of the circuit is Ceq.
  12 u F and 14 u F are in series.
  The total capacitance between them = (12*14/12+14) = 168/26 = 6.46 u F.
  As 5 u F, 10 u F, 14 u F capacitors are in parallel, then the total capacitance between them is = 5+10+14 = 29 u F.
  The equivalent capacitance of the circuit = 29 u F + 6.46 u F = 35.46 u F.
• As per the formula Q = CV, where Q is the stored charge, C is the capacitance and V is the voltage.
  The total charge stored in the circuit,
  Q = Ceq* V = 35.46*9 = 319.14 Coulomb
• Below the circuit has been shown.

Figure 3: Circuit Diagram
(Source: Self-created)

Question 3.

According to the formula that resistance (R) = resistivity (p) length(l) / cross sectional area of the wire(A)
Cross sectional area of the wire = Pi r^2.
Resistivity = 19n Ohmm = 19 10^-9 m
Length = 20000*10^-3 m = 20 m
Resistance = 1.1 Ohm.
According to the formula,
R = p l / A
A = p l/ R
Pi r^2 = p l/ R
r = p l/ R / Pi = (19 10^-9 m 20) / 1.1 * 3.14)) = 1.729 m^2
D = r/2 = 0.8645 m

Question 4.

v = 20sin(628t)
The simplified form of the given equation,
v = Vp sin( W t) = 20sin(628t)

• The peak voltage = 20 Volts
• The effective voltage is the root mean square value of the signal which gives the DC value which is equivalent to deliver the same power.
  The effective voltage = √ (20) = 4.47 Volts.
• Frequency can be calculated as,
  f = W / 2 Pi = 628/(6.28) = 100 Hz
• Periodic time is the time after which the sine wave returns to its fundamental value.
  According to the formula,
  T = 2 Pi / W = 6.28/628 = 0.01 ms.
• Instantaneous voltage at 3.2 ms,
  v = ∫o3.220sin(628t) = 20 ∫03.2 sin(628t) = (20/628) [(-cos 628*3.2) - (-cos 628*0)] = (20/ 628) [0.86 + 1] = 0.031*1.86 = 0.05923 volts.
• Below the waveform has been shown.

Figure 4: Waveform
(Source: Self-created)

Question 5

• Below the circuit diagram of the RLC circuit has been given.

Figure 5: RLC Circuit Diagram
(Source: Self-created)

• To determine the overall impedance of the circuit, the reactance of the circuit has to be determined.
  XL = W L = 2 Pi f*L = 6.28*100*10 = 6280 Ohm
  Xc = 1/(2 Pi f*C) = 1/(628*10*10^-6) = 1/(6280*10^-6) = 159.23 Ohm
  Let the total impedance of the circuit is Z.
  Z = √ (10^2) + (6280 - 159.23) ^2) = 6120.77 Ohm
• The supply current = (supply voltage / impedance of the circuit) = (120/6120.77) = 19.60 mA
• Voltage across the Resistor = (Supply current resistance) = 19.60*10 = 196 mV
  Voltage across Inductor = (Supply current reactance) = 19.60*6280 = 123.088 Volts
  Voltage across Capacitor = (Supply current * reactance) = 19.60*159.23 = 3120.908 Volts.
• The power factor of the circuit = R/Z = 10/ 6120.77 = 0.00163378
  Therefore, cos-1(0.00163378) = 89.90°

Figure 6: Phasor Diagram of RLC Circuit
(Source: Self-created)

Reference list

Journals

• Rosa, P., Sassanelli, C. and Terzi, S., 2019. Circular Business Models versus circular benefits: An assessment in the waste from Electrical and Electronic Equipments sector. Journal of cleaner production, 231, pp.940-952.
• Ab Rahman, A., Mustadza, N. and Yusof, A.M., 2021. Development of Teaching and Learning Module for Basic Electrical and Electronic Course. Research and Innovation in Technical and Vocational Education and Training, 1(1), pp.239-243.
• Osmani, M., Pollard, J., Forde, J., Cole, C., Grubnic, S., Horne, J. and Leroy, P., 2021. Circular economy business model opportunities, challenges, and enablers in the electrical and electronic equipment sector: stakeholders’ perspectives.
• Egea, J.A.L., IA confiable del Instituto de Ingeniería Eléctrica y Electrónica| Reliable AI of the Institute of Electrical and Electronic Engineering.
• Petralia, S., 2021.
• GPTs and growth: evidence on the technological adoption of electrical and electronic technologies in the 1920s. European Review of Economic History, 25(3), pp.571-608.
• Yikui, H., 2021, July. Design and Implementation of Virtual Experiment System Platform for Electrical and Electronic Engineering in Vocational College. In Journal of Physics: Conference Series (Vol. 1982, No. 1, p. 012176). IOP Publishing.

Electronic Principles Assignment