Analysis of AC Networks, Combined Reactances and Kirchhoff’s Laws

Combined Reactances:

Consider a circuit containing both capacitive and inductive reactance as shown in figure below:

2184_13.1.jpg

Figure: A Circuit containing a combination of Reactances

With f = 1 kHz    ω = 2πf = 2 x 3.14 x 103 = 6280 rad/sec then:

jωL = j 6280 x 100 x 10-3 = j 6.28 x 100 = j 628 Ω

-j (1/ωC) = - j [1/(6280 x 0.47 x 10-6)] = - j [(106)/(6.28 x 0.47 x 103)] = -j 339 Ω

Then the impedance as observe by the source voltage is as follows:

1097_13.2.jpg


Current drawn from the source is provided by:

i(t) = V(t)/Z = (10∠0o)/(304 ∠21.6o) = 32.9 ∠-21.6o mA

Voltage across the resistor in parallel with the inductor is as:

VR = VL = i(t) ZR//L = i(t) (283 + j451)

By using the above computations. With,

|ZR//L| = √2832 + 4512 = 532?  ∠ΦR//L = Tan-1 (451/283) = 57.9o

VR = VL = i(t) ZR//L = 32.9∠-21.6o x 10-3 x 532∠57.9o

VR = VL = 17.5 ∠36.3o V

The currents flowing via each of the individual branches are as:

iR = VR/R = (17.5 ∠36.3o )/103 = 17.5∠36.3o mA

iL = VL/jωL = (17.5 ∠36.3o)/628∠90o = 27.9 ∠-57.3o mA

The only outstanding quantity that remains unknown is the voltage across capacitor that is given as:

VC = V – VL = 10 ∠0o – 17.5 ∠36.3o

The evaluation of VC requires VL expressed in the complex terms from below figure gives:

340_13.3.jpg

Figure: Argand Diagram for VL

a = Real(VL) = |VL|cosΦL

jb = Imag (VL) = j|VL|sinΦL

VL = |VL| cosΦL + j|VL|sinΦL

VL = 17.5 cos 36.3 + j 17.5 sin 36.3
   
Then,

VL = 17.5 x 0.8059 + j17.5 x 0.5920

VL = 14.1 + j 10.36

Then,

VC = V – VL = 10 - 14.1 – j10.36 = - 4.1- j10.36

And hence finally:

|V| = √ (4.1)2 + (10.36)2 = 11.14 V

∠ΦC = Tan-1 (-10.36/-4.1) = 68.4o – 180o = -111.6o

The fact that both real and imaginary coefficients of VC are negative should be accounted for which is done by comprising the -180o rotation to place this vector in third quadrant. The whole phasor diagram exhibiting all currents and voltages is given in figure shown below.

381_13.4.jpg

Figure: A Complete Phasor Diagram for the Circuit of first figure

Kirchhoff’s Laws:

Kirchoff’s laws can be applied to the ac circuits in a similar way to dc circuits. The major difference is that the sources and impedances encompass to be treated as the complex quantities. The values for currents and voltages and can be computed as complex quantities first and transformed into the form of phase and magnitude at the end. Consider the circuit shown in figure below where the frequency and component values have been selected to give round figures for the values of reactance. Such can then be treated as purely imaginary impedances. Such circuit is analysed by using Nodal Analysis by applying the Kirchhoff’s Current Law to node(s) of interest.

1175_13.5.jpg

Figure: An AC Circuit Requiring Complex Phasor Analysis

jωL = j 2 Π f L = j 6.28 x 50 x 470 x 10-3 = j3K Ω

- j (1/ωC) = - j (1/2 Π f C ) = -j [(109)/(2 x 3.14 x 50 x 80)] = -j2K Ω

Determine both the potential at Node A and the current via the inductor, in terms of phase and magnitude.

(i) Allocate the currents i1, i2 and i3 at Node A as shown

(ii) By applying the Kirchhoff’s Current Law at Node A gives: 

i1 + i2 = i3

(iii) Substituting in terms of potential drops across the components provides:

{(V1 - VA)/R} + {(V2 - VA)/-j(1/ωC)} = VA/jωL

{(V1 - VA)/103} + {(V2 - VA)/(-j2 x 103)} = VA/(j3 x 103)

Replacing and multiplying across by 103

[(10 - VA)/1] + [(j5 – VA)/-j2] = (VA/j3)

10 – VA + j2 (5/2) – j (VA/2) = -j (VA/3)

10 - 2.5 = VA + j 0.5 VA – j 0.33 VA
7.5 = VA (1 + j 0.17)

VA = 7.5/(1+ j 0.17)

Rationalising:

VA = [7.5 (1- j 0.17)]/[ (1+ j 0.17) (1- j 0.17)] = [7.5 (1- j 0.17)]/[1 + (0.17)2]

VA = (7.5 – j 1.275)/(1.03)

VA = 7.28 – j1.24

Then, |VA| = √(7.28)2 + (1.24)2 = 7.38

∠ΦA = Tan-1 – (1.24/7.28) = -9.7o

The current via the inductor is i3 that can be obtained as:

i3 = VA/jωL = (7.28 – j1.24)/(j3 x 103) = - j2.43 – 0.41mA

|i3| = √(0.41)2 + (2.43)2 = 2.46 mA

∠Φ3 = Tan-1 (-2.43/-0.41) = 80.3o – 180o = -99.7o

Phasor diagram:

2071_13.7.jpg

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