Calculations revised

calculationrevised

The problen was just the second diagram i paste from paint. I paste the wrong one. All others are correct. This is for the calculations document

Verification of simulated values for IL , IC, and IR. :

Our reference circuit:

Using superposition theorem:

Find IL , IC, and IR. :

First study the effects of E1 (remain E1 then short E2):

E1 = 2 ∠0

E2 =6∠0

R1 = 4 ∠0 Ω

XL = 2 ∠90 Ω

XC = 1∠-90 Ω

ZT = 2 ∠90 + 4 ∠0 *1∠-90 / (4 ∠0 +1∠-90)

ZT = 1.08∠77.47 Ω

IT = i1’ = E1/ ZT

i1’ = 2 ∠0/ 1.08∠77.47

i1’ = 1.85 ∠-77.47 A

V2 = 2 ∠0 – 1.85 ∠-77.47 (2 ∠90)

V2 = 1.8∠-153.53

i2’ = 1.8 ∠-153.53/ 4 ∠0

i2’ = 0.45∠-153.53 A

i3’ = i1’ – i2’

i3’ = 1.85 ∠-77.47 – 0.45∠-153.53

i3’ = 1.8∠-63.39 A

Second, we study the effect of E2 (remain E2 and short E1)

E1 = 2 ∠0
E2 =6∠0
R1 = 4 ∠0 Ω
XL = 2 ∠90 Ω

XC = 1∠-90 Ω

ZT = 1∠-90 + 2 ∠90 * 4 ∠0/ ( 2 ∠90 +4 ∠0 )

ZT = 1∠36.87 Ω

IT = i1’’ = E1/ ZT

i1’’ = 6 ∠0/ 1∠36.87

i1’’ =6∠-36.87 A

V2 = 6∠0 – 6∠-36.87 * (1 ∠-90)

V2 = 10.73∠-26.57

i2’’ = 10.73∠-26.57 / 4 ∠0

i2’’ = 2.68∠26.57 A

i3’’ = i1’’ – i2’’

i3’’ = 6∠-36.87 – 2.68∠26.57 A

i3’’ == 5.37∠-63.4 A

Solving for the currents:

Current should be the algebraic sum of the currents provided when E1 was acting along and when E2 was acting alone

Inductor current:

iL = 5.37∠-63.4 – 1.85 ∠-77.47

iL = 3.604∠-56.23 A

The magnitude is actually the same value that we obtained in Multisim. Thus for this current, simulation is verified.

Resistor current:

iR = 0.45∠-153.53 + 2.68∠26.57

iR = 2.23∠26.59 A

The magnitude is actually the same value that we obtained in Multisim. Thus for this current, simulation is verified.

Capacitor current:

iC = 6∠-36.87 – 1.8∠-63.39

iC = 4.46∠-26.49 A

The magnitude is actually the same value that we obtained in Multisim. Thus for this current, simulation is verified.

Conclusion:

Simulation in Multisim for currents were all verified using superposition theorem.

Using Mesh Analysis:

KVL around loop 1:

In this case it’s more convenient to use rectangular form:

2-(4+j2) i1 + 4i2 = 0

i2 = [-2+ (4+j2) i1]/ 4 — > eq. 1

KVL around loop 2:

-6- (4-j) i2 + 4i1 = 0 —- > eq. 2

Substitute eq.1 to eq. 2:

-6- (4-j) [[-2+ (4+j2) i1]/ 4] + 4i1 = 0

-24 + 8 -2j – (18+j4) i1 + 16i1 = 0

(-2-4j) i1 = 16+2j

i1= 16.12∠7.13/ 4.47∠-116.57

i1 = 3.606∠123.7 A

i2 = [-2+ (4+j2) *3.606∠123.7]/ 4

i2 = 4.47∠153.44 A

Solving for the currents:
Inductor current:

iL= i1 = 3.606∠123.7 A

The magnitude is actually the same value that we obtained in Multisim. Thus for this current, simulation is verified.
Resistor current:

iR= 4.47∠153.44 – 3.606∠123.7

iR= 2.23∠-153.38 A

The magnitude is actually the same value that we obtained in Multisim. Thus for this current, simulation is verified.
Capacitor current:

iC= i2 =4.47∠153.44 A

The magnitude is actually the same value that we obtained in Multisim. Thus for this current, simulation is verified.

Solution:

Simulation in Multisim for currents were all verified using Mesh Analysis theorem.