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## 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:

Figure: A Circuit containing a combination of ReactancesWith f = 1 kHz ω = 2πf = 2 x 3.14 x 10

^{3}= 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 [(10^{6})/(6.28 x 0.47 x 10^{3})] = -j 339 ΩThen the impedance as observe by the source voltage is as follows:

Current drawn from the source is provided by:

i(t) = V(t)/Z = (10∠0

^{o})/(304 ∠21.6^{o}) = 32.9 ∠-21.6^{o}mAVoltage across the resistor in parallel with the inductor is as:

V

_{R}= V_{L}= i(t) Z_{R}//L = i(t) (283 + j451)By using the above computations. With,

|Z

_{R//L}| = √283^{2}+ 451^{2}= 532? ∠Φ_{R//L }= Tan^{-1}(451/283) = 57.9^{o}V

_{R}= V_{L}= i(t) Z_{R//L}= 32.9∠-21.6^{o}x 10^{-3}x 532∠57.9^{o}V

_{R}= V_{L }= 17.5 ∠36.3^{o}VThe currents flowing via each of the individual branches are as:

i

_{R}= V_{R}/R = (17.5 ∠36.3^{o})/10^{3}= 17.5∠36.3^{o}mAi

_{L}= V_{L}/jωL = (17.5 ∠36.3^{o})/628∠90^{o}= 27.9 ∠-57.3^{o}mAThe only outstanding quantity that remains unknown is the voltage across capacitor that is given as:

V

_{C}= V – V_{L }= 10 ∠0^{o}– 17.5 ∠36.3^{o}The evaluation of V

_{C}requires V_{L}expressed in the complex terms from below figure gives:Figure: Argand Diagram for VLa = Real(V

_{L}) = |V_{L}|cosΦ_{L}jb = Imag (V

_{L}) = j|V_{L}|sinΦ_{L}V

_{L}= |V_{L}| cosΦ_{L }+ j|V_{L}|sinΦ_{L}V

_{L}= 17.5 cos 36.3 + j 17.5 sin 36.3Then,

V

_{L}= 17.5 x 0.8059 + j17.5 x 0.5920V

_{L}= 14.1 + j 10.36Then,

V

_{C}= V – V_{L}= 10 - 14.1 – j10.36 = - 4.1- j10.36And hence finally:

|V| = √ (4.1)

^{2}+ (10.36)^{2}= 11.14 V∠Φ

_{C}= Tan^{-1}(-10.36/-4.1) = 68.4^{o}– 180^{o}= -111.6^{o}The fact that both real and imaginary coefficients of VC are negative should be accounted for which is done by comprising the -180

^{o}rotation to place this vector in third quadrant. The whole phasor diagram exhibiting all currents and voltages is given in figure shown below.Figure: A Complete Phasor Diagram for the Circuit of first figureKirchhoff’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.

Figure: An AC Circuit Requiring Complex Phasor AnalysisjωL = 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 [(10

^{9})/(2 x 3.14 x 50 x 80)] = -j2K ΩDetermine both the potential at Node A and the current via the inductor, in terms of phase and magnitude.

(i) Allocate the currents i

_{1}, i_{2}and i_{3}at Node A as shown(ii) By applying the Kirchhoff’s Current Law at Node A gives:

i

_{1}+ i_{2}= i_{3}(iii) Substituting in terms of potential drops across the components provides:

{(V

_{1}- V_{A})/R} + {(V_{2}- V_{A})/-j(1/ωC)} = V_{A}/jωL{(V

_{1}- V_{A})/10^{3}} + {(V_{2}- V_{A})/(-j2 x 10^{3})} = V_{A}/(j3 x 10^{3})Replacing and multiplying across by 10

^{3}[(10 - V

_{A})/1] + [(j5 – V_{A})/-j2] = (V_{A}/j3)10 – V

_{A}+ j2 (5/2) – j (V_{A}/2) = -j (V_{A}/3)10 - 2.5 = V

_{A}+ j 0.5 V_{A}– j 0.33 V_{A}7.5 = V

_{A}(1 + j 0.17)V

_{A}= 7.5/(1+ j 0.17)Rationalising:

V

_{A }= [7.5 (1- j 0.17)]/[ (1+ j 0.17) (1- j 0.17)] = [7.5 (1- j 0.17)]/[1 + (0.17)^{2}]V

_{A}= (7.5 – j 1.275)/(1.03)V

_{A}= 7.28 – j1.24Then, |V

_{A}| = √(7.28)^{2}+ (1.24)^{2}= 7.38∠ΦA = Tan

^{-1}– (1.24/7.28) = -9.7^{o}The current via the inductor is i

_{3}that can be obtained as:i

_{3}= V_{A}/jωL = (7.28 – j1.24)/(j3 x 10^{3}) = - j2.43 – 0.41mA|i

_{3}| = √(0.41)^{2}+ (2.43)^{2}= 2.46 mA∠Φ

_{3}= Tan^{-1}(-2.43/-0.41) = 80.3^{o }– 180^{o}= -99.7^{o}Phasor diagram:Latest technology based Electrical Engineering Online Tutoring AssistanceTutors, at the

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