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Electric Circuits Assignment Sample

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Electric Circuits Assignment Sample

INTRODUCTION

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1.0 Question 1

Answer:

Figure 1: Circuit 1

(Source: Given)

a. Here from this figure one can see that the elements that are connected in series are D and E   apart from A and C

b. Resistance of A = RA

    Resistance of B = RB

    Resistance of C = RC

    Resistance of D = RD

    Resistance of E = RE

    Resistance of F = RF

    Let the equivalent resistance of the following circuit is R. 

    From the figure one can see that the equivalent resistance of the circuit is,

“R = ((RA+RC) || RB || RF || (RD+RE))”

   Value of RB = 20 Ohm

   Value of RF = 20 Ohm

   Equivalent resistance of RB and RF is = (20*20) / (20+20) = 400 /40 =10 Ohm

   Total resistance of RA and RC = (30+10) Ohm = 40 Ohm

   Total resistance of RD and RE = (40+50) Ohm = 90 Ohm

   Equivalent resistance of the circuit,

“1/R = 1/10 + 1/40 + 1/90”

Or, “1/R” = 0.1 + 0.025 + 0.011

Or, “1/R” = 0.136

Therefore, R = 7.352 Ohm

The equivalent resistance of the circuit is 7.352 Ohm. 

 

2.0 Question 2

Answer:

Figure 2: Circuit 2

(Source: Given)



Here six resistors have been given with the values of RA, RB, RC, RD, RE, RF, RG, RF, RG, RH, RI, RJ, RK.

All the values of the resistors have been given below.

RA = 10 Ohm

RB = 20 Ohm

RC = 30 Ohm

RD = 40 Ohm

RE = 50 Ohm

RF = 20 Ohm

RG = 30 Ohm

RH = 4 Ohm

RI = 5 Ohm

RK = 10 Ohm

Total resistance of RA and RC = (20+10) Ohm = 30 Ohm

Total resistance of RI and RK = (5+10) Ohm = 15 Ohm

Total resistance of RG and RH = (30+4) Ohm = 34 Ohm

Let The equivalent resistance of RAC, RB and RF is R.

“1/R = (1/RAC) + (1/RB) + (1/RF)”

Or, 1/R = 1/30 + 1/20 + 1/20

Or, 1/R = 0.333 + 0.05 + 0.05

Or, 1/R = 0.333 + 0.1

Or, 1/R = 0.433

Therefore, R = 2.309 Ohm

Equivalent resistance of RIK and RJ = (15*6)/(15+6)

= 90/21

= 4.28 Ohm

Then R is in series connected with RD. As well as the equivalent resistance of RIK and RJ is connected in series with RGH (Ezawa, 2018).

Total resistance of R and RD = (2.309 + 40) = 42.309 Ohm

Let the equivalent resistance of RIK and RJ is R1 

Total resistance of R1 and RGH = 4.28 + 34 = 38.28 Ohm

Let the total equivalent resistance of the remaining circuit is REQ

“1/REQ = 1/38.28 + 1/42.309 + 1/50”

1/REQ = 0.026 + 0.023 + 0.02

REQ = 14.49 Ohm 








3.0 Question 3

Answer: 

Figure 3: Circuit 3

(Source: Given)

The following diagram shows a circuit which consists of some resistors of values of 10 ohm and 5 ohm. Also here is a current source and a voltage source of value 10 volts (Barbero. and Lelidis, 2017).

An unknown current Ix flows through the voltage source of 10 volts.

Let the value of the unknown current source be I. 

The voltage across 5 ohm is 10 volts (Zhang, et a ,2019).

According to Ohm's law, the voltage across an active and passive element is equal to the multiplication of current through it to the magnitude of that element.

So, the current through 5 Ohm resistor= 10/5 Ampere= 2 Ampere.

Let the voltage across the 10 ohm resistor is Vx and the other 10 Ohm resistor has Vy voltage across it. 

 According to KCL, 

“(Vx-0)/10+ ix-(Vy-0)/10=0”

Or, “(Vx-Vy)/10 + (10/10) ix = 0”

Or, (10/10) + 1*ix=0

Or, - (1/1) = ix

Therefore, ix= -1 Ampere.

According to KVL, 

“-(10*1)-10-(10*(I+1)) = 0”

Or, -10-10-10I-10=0

Or, -30=10I

Therefore, I=-3 Ampere.

Voltage across 10 Ohm resistor = 10*1 = 10 Volts

Voltage Across the other 10 Ohm resistor = 10*(-2) = -20 Volts

Voltage across 5 Ohm resistor = 5*2 = 10 Volts

Current flowing from 10 Ohm resistor = 1 Ampere

Current flowing from right 10 Ohm resistor = -2 Ampere

Current flowing from right 5 Ohm resistor = 2 Ampere

Power of the 5 Ohm resistor = 10*2 = 20 Watts

Power of the leftmost 10 Ohm resistor = 10*1 = 10 Watts

Power of the rightmost 10 Ohm resistor = 10*(-2) = -20 Watts 

4.0 Question 4

Answer:

Figure 3: Circuit 4

(Source: Given)

The following diagram shows that some elements have been connected with each other and the value of the voltages across those elements has to be found out (Nilsson. and Riedel, 2020). 

Value of Va = 5 V

Value of Vb = 7 V

Value of Vf = - 10 V

Value of Vh = 6 V

Calculation of Vd,

As per KVL, ()

“- Va+Vd-Vb = 0”

Or, -5+Vd-7 = 0

Or, “Vd-12” = 0

Therefore, Vd = 12 Volts

 

Calculation of Vg:

Again as per KVL, 

“+ (Vf) + Vd-Vg = 0”

Or, + (-10) + 12 = Vg

Or, Vg = 12-10

Therefore, “Vg” = 2 Volts

Calculation of Ve:

Again as per KVL,

             “Vg- Ve + Vh = 0”

            Or, 2- Ve+6 = 0

Or, Ve = 8 V

Calculation of Vc:

Again as per KVL,

“Vb - Ve- Vc = 0”

Or, 7-8-Vc = 0

Or, Vc = -1 V

 











5.0 Question 5

Answer:

Figure 5: Circuit 5

(Source: Given)

  The following diagram shows a circuit that consists of some resistors, voltage source and current sources. 

Value of iID Ampere = 1 A

Using the source changing technique the current source has been converted into a voltage source of 2 Volts with series connection with the 2 Ohm resistor.

Voltage across the 20 Ohm resistor is 20 Volts (Zhang and Franz, 2020.). According to Ohm's law the current through the 20 Ohm resistor is 1 Ampere.

The current through the 10 Ohm resistor is i1 A.

The voltage across the 10 Ohm resistor is 10i1 Volts.

The current passes through the 7 Ohm resistor (Since the 2 Ohm and the 5 Ohm resistor are in series) = 1-i1 A

According to Kirchhoff's Voltage Law, 

“2-7(1-i1)-10i1=0”

Or, 2-7+7i1-10i1 = 0

Or, -5 -3i1 = 0

Therefore, i1 = -1.66 A

Current across the 10 Ohm resistor = - 1.66 A

Current across the 5 Ohm resistor = (1-(-1.66)) A = 2.66 A

Current across the 2 Ohm resistor = 2.66 A

Current across the 20 Ohm resistor = 1 A

Voltage across the 10 Ohm resistor = 10*(-1.66) = -16.6 A

Voltage across the 20 Ohm resistor = 20 Volts

Voltage across the 5 Ohm resistor = 5*2.66 = 13.3 Volts

Voltage across the 2 Ohm resistor = 2*2.66 = 5.32 Volts

Power Across the 10 Ohm resistor= (-16.6)*(-1.66) = 27.556 Watts

Power Across the 20 Ohm resistor = 20*1 = 20 Watts

Power Across the 5 Ohm resistor = 13.3*2.66 = 35.378 Watts 

Power Across the 2 Ohm resistor = 5.32*2.66 = 14.1512 Watts 

Reference List

Journal

Zhang, X.X. and Franz, M., 2020. Non-Hermitian exceptional Landau quantization in electric circuits. Physical review letters124(4), p.046401.

Nilsson, J.W. and Riedel, S.A., 2020. Electric circuits. Pearson Education Limited.

Zhang, Z.Q., Wu, B.L., Song, J. and Jiang, H., 2019. Topological Anderson insulator in electric circuits. Physical Review B100(18), p.184202.

Ezawa, M., 2018. Higher-order topological electric circuits and topological corner resonance on the breathing kagome and pyrochlore lattices. Physical Review B98(20), p.201402.

Barbero, G. and Lelidis, I., 2017. Analysis of Warburg's impedance and its equivalent electric circuits. Physical Chemistry Chemical Physics19(36), pp.24934-24944..

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