Electrical/Electronics Principles

Calculations in AC Networks

Calculate the parameters of AC equivalent circuits using transformation theorems  

Use Thèvenin's theorem to determine the power dissipated in the 48Ω resistor of the network shown in Figure 1.

Answer:

R_C = 400 ∠ -90° = 0 - 400j Ω     R₃₀₀ = 300 ∠ 0° = 300 + 0j Ω                R_L = 144 ∠ 90° = 0 + 144j Ω
R₄₈ = 48 ∠ 0° = 48 + 0j Ω         V_s = 50 ∠ 0° = 50 + 0j V                       P₄₈ =?

Z_c = (300 + 0j) – (0 + 400j) = 300 – j400
i_Zc = V_S / Z_C = 50 + 0j / 300 – j400 = 0.06 + j0.08 A
V_T = i_T · R₃₀₀ = (0.06 + j0.08) · (300 + 0j) = 18+ j24 V
Z_T = R_C · R₃₀₀ / Z_C = 0 + 400j · 300 + 0j / 300 – j400 = j120000 / 300 – j400 = -192+ j144 Ω

R_T = (-192+ j144) + (48 + j144) = 240+ 0j Ω
i_T = V_T / R_T = (18+ j24) / 240+ 0j = 0.075 + j0.1 A
P₄₈ = i² · R₄₈ = (0.075 + j0.1)² · 48 + 0j = (-0.004375 + j0.015) · 48 + 0j
                = (-0.21 + j0.72) = 0.75 ∠ 106.26° = 0.75 W

Use Norton’s theorem to determine the current I flowing in the 4Ω resistance shown in Figure 2

FIG 2

Answer:

    R_L=4Ω     I_N = I₁ + I₂
R_N = ((1/2) + (1/1)) ^-1 = 2/3 Ω
I_N = (4/2) + (2/1) = 4A
R_T = (2/3) + ((2/3)+4) = 1/7 Ω I_4Ω = R_T · I_N =((2/3)/((2/3)+4))·4 = 0.571429A

Apply circuit theory techniques to the solution of AC circuit problems  

Using the principal of superposition obtain the current flowing in the (4+ j 3) impedance of Figure 3.

FIG 3

Answer:

       R₁ = 4Ω R₂ = 4Ω                 R_C = 10Ω              R_L = 3Ω

when V₂ removed

	R1	R2	RC	RL	R2+RL	R2+RL+RC	Total
V	14.7783	8.6981	15.2217	6.52357	15.2217	15.2217	30
I	3.69565	2.17452	1.52217	2.17452	2.17452	3.69565	3.69565
R	4	4	10	3	7	4.11765	8.11765

when V₁ removed

	R1	R2	RC	RL	R2+RL	R2+RL+R1	Total
V	6.08618	-3.47782	23.913	-2.60836	-6.08618	-6.08618	30
I	1.5225	-0.86946	2.3913	-0.86946	-0.86946	-2.3913	2.3913
R	4	4	10	3	7	2.54545	12.5455

with V₁ & V₂ in place (adding both)

	R1	R2	RC	RL	R2+RL	Total
V	20.8645	5.22028	39.1357	3.91521	9.13552	30
I	5.21815	1.30506	3.91347	1.30506	1.30506	5.21815

Current flowing in the (4+j3) = 2.17452 + (-0.86946) = 1.30506 A

The network of Figure 4 use nodal analysis to determine:

(a) the voltage at nodes 1 and 2

(b) the current in the j 4Ω inductance

(c) the current in the 5Ω resistance

FIG 4

Answer:

For Node V1 0 = (V1 – 25)2Ω + V14Ω + (V1 - V2)5Ω 0 = (25 – V1)/2 - (V1 - V2)/5 - V14 0 = (25 – V1)0.5 - (V1 - V2)0.2 - V10.25 0 = 12.5 – 0.5V1 - 0.2V1 - 0.2V2 - V10.25 0 = 12.5 – 0.95V1 - 0.2V2

For Node V2 0 = (V2 - 25)2.5Ω + V24Ω + (V2 – V1) 5Ω 0 = (V1 - V2)/5 + (V2–25)/2.5 - V2/4 0 = (V1 - V2)0.2 + (V2–25)0.4 - V20.25 0 = 0.2V1 - 0.2V2 + 0.4V2–10 - V20.25 0 = -0.05V2 + 0.2V1 –10

i₁ = i₂+i₄ and i₂ = i₅ - i₃,

Voltage at node 1 = 11.24 V
Voltage at node 2 = 9.12 V
Current at 4Ω = V₁/4 = 11.24 /4 = 2.81 A
Current at 4Ω = V₂/4 = 9.12/4 = 2.28 A

Analyse the operation of magnetically coupled circuits

Write a brief description of Magnetically Coupled circuits, describe Mutual inductance using a transformer as an example. Describe Dot Notation and its use in magnetically coupled circuits, providing an example where this is used to good effect.

Answer:

Magnetically coupled circuits are where two conductors, and when current moves through one conductor magnetic field generated on the other, this induced voltage in the second conductor is called mutual conductance. Most uses of mutual inductance are in transformers where they are used to step-up or step-down voltage, those are the two types of transformers.

In a transformer, you have insulated two wraps of coils opposite each other, a mutual conductance formed between two coils, the number of wraps decides the levels of step down or step up and the type. We can donate number of wraps in the first & second coil as N_p and N_s.

Here is an example of basic magnetically coupled circuit,

The Dot Notation
When positive current enters the dot on one side of the conductor the current direction will be opposite direction on the second conductor. If positive current exits the dotted line in the first conductor, then the current direction in the second coil will be entering the dot. A dot example shown above; the dot entering the first conductor therefore exiting in the second conductor.

A suitable transformer designed for operation at 50 Hz has 220 primary turns and 110 secondary turns. The primary and secondary resistances of 2.0 and 1.0 W respectively, whilst the primary and secondary leakage inductances are 25 mH and 15 mH respectively. Assuming core losses are negligible, determine the equivalent impedance referred to the primary circuit.  

Answer:

       f = 50Hz  N_p= 220  N_s = 110  R_p = 2Ω  R_s = 1Ω  L_p = 25E-3H  L_s= 15E-3H  Z_pe. = ?

k = N_p/N_s = 220/110 = 2

R_pe. = R_p + R_s (k)² = 2 +1 ·2² = 6Ω
L_pe = L_p + L_s (k)² = 25E-3 +15E-3 ·2² = 0.085H = 85mH

Z_pe = R_pe. + jω L_pe = 6 + j 2 π 50 · 0.085 = 6 + j26.7035 Ω

Use circuit theory to solve problems relating to series and parallel R-L-C tuned circuits  

Use circuit theory to solve problems relating to series (Figure 5)  and parallel Figure 6) R-L-C circuits shown below. 

 a) Calculate the impedance for the series circuit and the admittance for the parallel circuit

 b)Current drawn from the supply.

 c) Resonant frequency

 d) Q-factor at resonance                

 e) Dynamic Resistance at resonance 

 f) Bandwidth

FIG 5

FIG 6

Answer:

For Fig 5 (in series),

R = 10∠0°Ω  L =20E-3∠90°H  C=4E-6∠-90°F  V = 15∠0°V  f = 1E3Hz

a)       X_L = 2πfL=125.664Ω         X_C =(2πfC)^-1=39.7887Ω
Z = R+j(X_L-X_C) = √R²+(X_L-X_C)² = √10²+(125.664-39.7887)²= 86.4553Ω
ϴ = tan^-1(X_L-X_C)/R = 83.3579°
Z = 86.4553∠83.3579°Ω

b)      I = V/Z = 15∠0°/86.4553∠83.3579° = 173.5∠-83.3579°mA

c)       f_r=(√LC)^-1·2π = (√20E-3·4E-6)^-1 ·2π = 22214.4Hz = 22.2144MHz

d)      Q = ω_R(L/R) =(√LC)^-1(L/R)= 3535.53(20E-3/10) = 7.07107

e)      N/A

f)        BW = f_r/Q = 22214.4/7.07107 = 3141.59Hz = 3.142MHz

For Fig 6 in (parallel),

R = 10∠0°Ω  L =20E-3∠90°H  C=4E-6∠-90°F  V = 15∠0°V  f = 1E3Hz

a)       X_L = (2πfL)^-1=0.007958S X_C =(2πfC) =0.025133S
Z = (√(R^-1)²+(X_L^-1-X_C)²)^-1= (√(10^-1)²+(0.007958-0.025133)²)^-1 = 9.85569Ω
ϴ = tan^-1(X_L^-1-X_C)R = -9.74547°
Z = 9.85569∠-9.74547°Ω
Y = 1/Z = 0.101464S

b)      I = V/Z = 15∠0°/9.85569∠-9.74547° = 1.52196∠9.74547A

c)       f_r=((√LC)·2π)^-1 = 562.698Hz

d)      Q= (√C/L)R = (√4E-6/20E-3)10 = 0.141421

e)      R_D =L/CR = 20E-3/4E-6·10 = 500Ω

BW = f_r/Q = 562.698 /0.141421 =3978.89Hz = 3.98MHz

Select/design and apply appropriate methods/techniques: Complex information/data has been synthesised and processed              

Replace the delta-connected network shown in Figure 7a by an equivalent star connection.

Answer:

Z1= Z1Z2/Z1+Z2+Z3   = 20·(10+j10)/20+(10+j10)-j20   = 200+j200/30-j10   = 4+j8Ω

Z2= Z2Z3/Z1+Z2+Z3  =(10+j10)·-j20/20+(10+j10)-j20  =200-j200/30-j10  = 8-j4Ω

Z3= Z1Z3/Z1+Z2+Z3   =20·-j20/20+(10+j10)-j20   =-j400/30-j10   =4-j12Ω

Determine the delta-connected equivalent network for the star-connected impedances shown in Figure 7b

Answer:

Z1= Z1·Z2+Z2·Z3+Z3·Z1/Z2   = j5·20+20·10+10·j5/20   = j100+200+j50/20   = j150+200/20   = 10+ j7.5Ω

Z2= Z1·Z2+Z2·Z3+Z3·Z1/Z3   = j5·20+20·10+10·j5/10   = j100+200+j50/10   = j150+200/10   =20+ j15Ω

Z3= Z1·Z2+Z2·Z3+Z3·Z1/Z1   = j5·20+20·10+10·j5/j5   = j100+200+j50/j5   = j150+200/j5   =30-j40Ω

Coherent, logical development of principles/concepts for the intended audience:

FIG 8

a) For the circuit shown in Figure 8 determine the current flowing in the inductive branch using Norton’s theorem.

b) Describe the basic premise of Norton’s theorem and superposition theorem and give reasons why circuit theorems are used.

Answer:

R_L = 1.5+j2.95Ω

R_N = (2^-1+3^-1)^-1 = 1.2Ω

I_N = 20/2+ 10/3 = 13.333A

Norton Equivalent circuit becomes,

R_T = 1.2/1.5+j2.95+1.2 = 0.203-j0.221Ω

I_j2.95 = I_N·R_T = 13.333·0.203-j0.221 = 2.70127-j2.95138A

    Norton Theorem allow us to calculate current flowing in components in circuits very easily. It’s is similar to Thevenin’s theorem except current source used instead of voltage source. In a superposition theorem you get to analyse each current individually and simply add them all toghether with sources removed except one. All circuit theorems are to simplify circuit calculations.

The validity of results has been evaluated using defined criteria                    

FIG 9a

a) For the a.c. network shown in Figure 9a determine, using mesh-current analysis (a) the mesh currents I1 and I2, (b) the current flowing in the capacitor, and (c) the active power delivered by the 100∠0◦V voltage source.

b) For the network shown in Figure 9b, determine the voltage VAB, by using nodal analysis.

c) Evaluate Mesh and Nodal analysis against each other making and fully justifying any conclusions reached.

The unforeseen has been accommodated                                             

Given the circuits in task 1 deal with transformation theorems, unforeseen circumstances have developed that require the component values to be changed for a different application.

Determine the power dissipated in the 56Ω and 250Ω resistors by using both Thèvenin's theorem and Norton’s theorem. Compare your results and the different methods to solving this problem in terms of efficiency of calculation and state your preferred method by justifying your reasons