Circuit Theorems and Analysis Assignment Help - HND Electrical and Electronic Engineering

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Programme: HND Electrical & Electronic Engineering

Unit Number and Name: Unit 66 Electrical, Electronic and Digital Principles T/601/1395

Level: QCF Level 5

Assignment Title - Circuit Theorems and Analysis

Task Introduction -

You are a consultant electrical engineer. You have been asked by your line manager, Mithun, to analyse a series of AC networks and document your findings in a technical report.

Task 1 -

Part 1 - For the circuit below using Theveins theorem, determine:

(a) The value of R to provide maximum power to the load.

(b) The maximum power transferred in the load.

Part 2 - For the network shown below determine the current with phase angle and power dissipated in the load connected between A and B using Nortons transformation theorem.

Opportunity for Merit - Repeat Task 1 (Part1 and Part2) using an alternative circuit theorem. M1

Task 2 -

For the network below, show correct method of analysis in an attempt to determine the magnitude and phase angle of the currents I1, I2 and I3 using one of the following circuit analysis theorems.

Nodal
Mesh
Superposition

Opportunity for Merit - Produce the same results for Task 2 using all three methods with a high degree of accuracy.

Opportunity for Distinction - Provide a table of results like the one below including simulated together with an explanation of any discrepancies D1

Analysis Method

I1

I2

I3

MESH

 

 

 

NODAL

 

 

 

SUPERPOSITION

 

 

 

SIMULATED

 

 

 

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Task 1: (LO2.1)

Part 1 -

Solution -
The value of R for maximum power to the load

As per the Thevenin theorem -

For finding the Thevenin equivalent resistance -

R_TH=7||(4+j15)
R_TH=7*(4+j15)/(11+j15)=(28+j105)/(11+j15)=5.4422+j2.1243 ohm
For maximum power transfer -
R_TH=R_L
R_TH=5.4422+j2.1243 ohm
R_L=R_0-j0.8
5.4422+j2.1243=R_0-j0.8
R_0=5.4422+j2.1243+j0.8
R_0=5.4422+j2.9243 ohm

Now the Thevenin Voltage -
V_TH=V_S*(4+j15)/(7+4+j15)
V_S=120 ∠40=91.925+j77.1345 V
So that,
V_TH=(91.925+j77.1345)*(4+j15)/(7+4+j15)
V_TH=(91.925+j77.1345)*(4+j15)/(11+j15)
V_TH=48.06+j87.865 V

The maximum power transferred in the load -
P_max=((V_TH^2)/(4R_L )) watts
P_max=((48.06+j87.865)^2/4(5.4422+j2.1243) )
P_max=((48.06+j87.865)^2/4(5.4422+j2.1243) )=((-5410.49+j8445.58))/(21.7688+j8.4972)
P_max=-84.265+j420.859 watts
P_max=429.2124 ∠101.32° watts

Part 2 -
Solution -

Start from the Thevenin conversion -

Equivalent Resistance or Thevenin resistance -
R_TH=(2+j3)||(1+j2)
R_TH=[(2+j3)*(1+j2) ]/(3+j5)
R_TH=(-4+j7)/(3+j5)=0.67647+j1.2
Equivalent Voltage or Thevenin voltage -

I=((10∠0-5∠0))/((3+j5) )=5/(3+j5)
I=0.4411-j0.735 A
Now,
V_TH=10-(2+j3)I
V_TH=10-(2+j3)(0.4411-j0.735)
V_TH=10-3.0872-0j.1467
V_TH=6.9128-j0.1467
V_TH=6.91435 ∠-1.2157° V
Thevenin equivalent circuit -
The current
I_TH=(6.9128-j0.1467)/((0.67647+j1.2)+(2-j3) )=1.8038+j1.1583 A
I_TH=2.1436∠37.7° A

Now Thevenin equivalent circuit can be given as thevening resistance in series with load resistance and driven by thevenin voltage.

Now Thevenin to Norton Conversion -

Norton resistance -
R_N=R_TH= 0.67647+j1.2 ohm
Norton current -
I_N=V_TH/R_TH =(6.9128-j0.1467)/(0.67647+j1.2)
I_N=2.37154-j4.4237 A
I_N=5.02∠-61.80° A
Norton equivalent circuit -

Current in the load across AB -
R_AB=2-j3 ohm
I_AB=((0.67647+j1.2))/((0.67647+j1.2+2-j3) )*I_N
I_AB=(-0.03359+j0.42576)*(2.37154-j4.4237) A
I_AB=1.80378+j1.1583 A
I_AB=2.1436 ∠32.7065° A
Hence the current in the load AB is determined by Norton and also its verified by the Thevenin theorem.
Now,
Power dissipated in the load -
P_AB=I_AB^2*R_AB
P_AB=(2.14)^2*(2) watts
P_AB=9.1592 Watts
 
Opportunity of merits -

Part 1 -
Solution -
The value of R for maximum power to the load
The alternate Norton theorem -

As per the Thevenin theorem -

R_TH=7||(4+j15)
R_TH=7*(4+j15)/(11+j15)=(28+j105)/(11+j15)=5.4422+j2.1243 ohm
And
V_S=120 ∠40=91.925+j77.1345 V
Thevenin to Norton conversion -
V_TH=48.06+j87.865 V
R_TH=5.4422+j2.1243 ohm
Norton current -
I_N=V_TH/R_TH =(48.06+j87.865 )/(5.4422+j2.1243) A
I_N=13.1321+j11.019 A
I_N=17.1428 ∠40° A
And
R_TH=R_N
So the Norton circuit -

For maximum power transfer -
R_TH=R_L
R_TH=5.4422+j2.1243 ohm
R_L=R_0-j0.8
5.4422+j2.1243=R_0-j0.8
R_0=5.4422+j2.9243 ohm
The maximum power transferred in the load -
P_max=((V_TH^2)/(4R_L )) watts
P_max=((48.06+j87.865)^2/4(5.4422+j2.1243) )
P_max=((48.06+j87.865)^2/4(5.4422+j2.1243) )=((-5410.49+j8445.58))/(21.7688+j8.4972)
P_max=-84.265+j420.859 watts
P_max=429.2124 ∠101.32° watts

Part 2 -
Solution -

By the alternate Thevenin theorem -
Equivalent Resistance or Thevenin resistance -
R_TH=(2+j3)||(1+j2)
R_TH=[(2+j3)*(1+j2) ]/(3+j5)
R_TH=(-4+j7)/(3+j5)=0.67647+j1.2

I=((10∠0-5∠0))/((3+j5) )=5/(3+j5)=0.4411-j0.735 A
Now,
V_TH=10-(2+j3)I=10-(2+j3)(0.4411-j0.735)
V_TH=6.9128-j0.1467
V_TH=6.91435 ∠-1.2157° V
Thevenin current
I_TH=(6.9128-j0.1467)/((0.67647+j1.2)+(2-j3) )=1.8038+j1.1583 A
I_TH=2.1436∠37.7° A

The picture above shows the Thevenin equivalent of the circuit.
Here the voltage 6.91435 V is the Thevenin voltage and 1.8+j1.15 or 2.1436 A is current flowing through the loop. The 0.67647+j1.2 is the equivalent Thevenin resistance and 2-j3 is the load resistance between A and B.
By considering the Thevenin equivalent, the circuit can be easily analysed.
Current in the load across AB -
R_AB=2-j3 ohm
I_AB=2.1436 ∠32.7065° A
Power dissipated in the load -
P_AB=I_AB^2*R_AB
P_AB=(2.1436)^2*(2) watts
P_AB=9.1592 Watts

Task 2: (LO2.2)

To determine the currents I1, I2 and I3 -
By Superposition theorem -
frequency=1 kHz
For
L_2: jωL_2=j*2πfL_2=j*2*3.14*1000*2.7*?10?^(-3)
jωL_2=j16.956 ohm
L_1: jωL_1=j*2πfL_1=j*2*3.14*1000*1.2*?10?^(-3)
jωL_1=j7.536 ohm
C_1:1/( jωC_1 )=1/(j*2πfC_1 )=1/(j*2*3.14*1000*5*?10?^(-6) )
1/( jωC_1 )=-j31.847133 ohm
Z_1=R_1+X_L1=9+j16.956
Z_2=R_2+X_L2=15+j7.536
Z_3=R_3+X_C1=18-j31.847
Now consider only voltage V1=10 V -

The equivalent resistance will be -
R_eq=(9+j16.956)+[(15+j7.536)||(18-j31.847) ]
R_eq=(9+j16.956)+[(15+j7.536)(18-j31.847)/(33-j24.311)]
R_eq=(9+j16.956)+[(15+j7.536)(18-j31.847)/(33-j24.311)]
R_eq=(9+j16.956)+14.9675+j0.66
R_(eq,1)=23.97+j17.617 ohm
Then
I_1^'=V_1/R_(eq,1)
I_1^'=10/(23.97+j17.617) ohm
I_1^'=0.27088-j0.2 A
And then after current division -
I_2^'=((18-j31.847))/((18-j31.847+15+j7.536) )*(0.27088-j0.2)
I_2^'=0.1476-j0.2618 A
And
I_3^'=I_1^'-I_2^'=0.12328+j0.0618 A

Now consider only voltage V2=10 V -

The equivalent resistance will be -
R_eq=(15+j7.536)+[(9+j16.956)||(18-j31.847) ]
R_eq=(15+j7.536)+[(9+j16.956)(18-j31.847)/(27-j14.89)]
R_eq=(15+j7.536)+(19.6455+j11.5225)
R_(eq,2)=34.6448+j19.0588 ohm
Then
I_2^''=V_2/R_(eq,2)
I_2^''=10/(34.6448+j19.0588)
I_2^''=0.22158-j0.12189 A
After the current division -
I_1^''=((18-j31.847))/((18-j31.847+9+j16.956) )*(0.1755-j0.1745)
I_1^''=0.1479-j0.261
And
I_3^''=I_2^''-I_1^''=0.0736+j0.13915 A
Now finally after the effect of superposition -
I_1=I_1^'+I_1^''=0.27088-j0.2-0.1479+j0.261=0.12298-j0.061 A
I_1=0.1372 ∠26.38° A
I_2=I_2^'+I_2^''=0.1476-j0.2618-0.22158+j0.12189=-0.0739-j0.1398 A
I_2=0.1582 ∠-117.86° A
I_3=I_3^'+I_3^''=0.12328+j0.0618+0.0736+j0.13915=0.25868-j0.13915 A
I_3=0.2937 ∠28.276° A

Opportunity of merits -
Nodal analysis -

At node VX -
(V_X-10)/(9+j16.956)+(V_x-10)/(15+j7.536)+V_x/(18-j31.847)=0
V_X [1/(9+j16.956)+1/(15+j7.536)+1/(18-j31.847)]=[10/(9+j16.956)+10/(15+j7.536)]
V_X=(10*[1/(9+j16.956)+1/(15+j7.536)])/[1/(9+j16.956)+1/(15+j7.536)+1/(18-j31.847)]
V_X=(10*[1/(9+j16.956)+1/(15+j7.536)])/[1/(9+j16.956)+1/(15+j7.536)+1/(18-j31.847)]
V_X=(10*[0.077654-0.072756])/[0.077654-0.072756+0.01345+j0.0238]
V_X=(10*[0.077654-0.072756])/[0.0911-0.049] =9.9436-j2.6425 V
V_X=10.288∠-14.88
The currents in the branches -
I_1=(10-9.9436+j2.6425)/(9+j16.956)=0.1230 + j0.0619 A
I_1=0.1377 ∠26.71° A
I_2=(10-9.9436+j2.6425)/(15+j7.536)= 0.0737 +j 0.1392
I_2=0.1575 ∠62.10° A
I_3=(9.9436-j2.6425)/(18-j31.847)=0.1966+j0.2011 A
I_3=0.2812 ∠45.6483° A

Mesh analysis -
Loop 1 -
10=I_1 (9+j16.956)+(I_1-I_2 )(18-j31.847)
10=I_1 (9+j16.956+18-j31.847)+(-I_2 )(18-j31.847)
10=I_1 (27-j14.891)+(-I_2 )(18-j31.847)

Loop 2 -
-10=I_2 (15+j7.536)+(I_2-I_1 )(18-j31.847)
-10=(-I_1 )(18-j31.847)+I_2 (33-j24.311)
Now solve for the equations -
From 1st equation -
[10+I_2 (18-j31.847) ]/((27-j14.891) )=I_1
Then put I1 in the 2ndequation -
-10=-I_1 (18-j31.847)+I_2 (33-j24.311)
-10=-[[10+I_2 (18-j31.847) ]/((27-j14.891) )](18-j31.847)+I_2 (33-j24.311)
-10+[[10+I_2 (18-j31.847) ]/((27-j14.891) )](18-j31.847)=I_2 (33-j24.311)
-10+ 10(18-j31.847)/(27-j14.891)+[(I_2 (18-j31.847))/((27-j14.891) )](18-j31.847)=I_2 (33-j24.311)
-10+10.099-j6.225=I_2 [(33-j24.311)+1.645+j43.37]
0.099-j6.225=I_2 [34.645+j19.059]
((0.099-j6.225))/[34.645+j19.059] =I_2
I_2=-0.07368-j0.13914
I_2=0.15745∠-117.90° A

And
[10+I_2 (18-j31.847) ]/((27-j14.891) )=I_1
[10+(-0.07368-j0.13914)(18-j31.847) ]/((27-j14.891) )=I_1
I_1=0.1229-j0.06196
I_1=0.13768 ∠26.74° A

And then
I_3=I_1+I_2
I_3=0.29513∠-91.16° A

Opportunity for distinction -
Simulated results -

The voltages V1, V2 and VX -

The current I1 -

The current I2 -

The current I3 -

Comparison table -

Analysis Method

I1

I2

I3

Mesh

0.13768 ∠26.74°

0.1574∠-117.90°

0.29513∠-91.16°

Nodal

0.13775 ∠-26.75°

0.15745 ∠62.10°

0.281∠45.64°

Superposition

0.1372 ∠26.38

0.1582 ∠-117.86°

0.2937 ∠28.276°

Simulated

140 mA

=0.140 A

160 mA

= 0.160 A

280 mA

= 0.280 A