Saturday, August 22, 2020

Network Analysis Research Paper Example | Topics and Well Written Essays - 1750 words

System Analysis - Research Paper Example A machine devours an intensity of 10 kW and a receptive intensity of 4 kvar at a current of (6 + j4)A. Decide the applied voltage, communicating your answer in complex structure. Arrangement: Here as given, Average Power, P = 10 kW, Reactive Power, Q = 4 kvar, Current I = (6+j4) A Leave S alone the Apparent Power then we realize that, Apparent Power S = Re{VI*} - Im{VI*} S = P - jQ, subbing the estimations of P and Q S = 10 10â ³ - j4 10â ³ S = (10 â€j4) 10â ³ †¦. Eq. Since S = VI*... Eq. 2(2) Where V is the applied voltage and I* is the conjugate of I. As we realize that in the event that z = a + jb is an intricate number, at that point z* = a †jb Therefore, I* = 6 - j4 Presently comparing Eq. 1(2) and 2(2) and subbing the estimation of I* we have, V (6 - j4) = (10 †j4) 10â ³ V = (10 †j4) 10â ³/(6 - j4) In the wake of excusing, V = (6 + j4)(10 †j4) 10â ³/(6â ² + 4 ²) V = (76 + j16) 10â ³/52 V = (1.46 + j0.30) 10â ³ System 5 Hence, V = 1.46 10 + j300 †¦. Eq. 3(2)Which is the applied voltage communicated in complex structure. Arrangement: 3(a) Let is the applied voltage and be the subsequent current through the given circuit then for complex impedance circuit is given as, = Expj †¦. Eq. 1(a3) Where . Let be the stage distinction among voltage and current than current = Expj( + †¦. Eq. 2(a3) Since impedance in time space is characterized as, = †¦. Eq. 3(a3) From conditions 1(a) and 2(a) we have, = as R=1 (given) = Or then again in polar structure, †¦. Eq. 4(a3) Multiplying by 1/to yield compelling worth we have, Z= or Z= 0.707†¦. Eq. 5(a3) Condition 5(a). is the necessary impedance in polar structure. permission is the complementary of impedance along these lines, in the event that Y is induction, at that point Y = 1/Z†¦. Eq.1(b3)... permission Y=1/Z, or, Yeq=1/Zeq 1/Zeq=Yeq From Eq. 1(b1) Hence, Yeq=1/R+j(C - 1/L) . Eq. 2(b1) Condition 2(b) gives the statement of induction for RLC equal circuit impedance. System 4 2. A machine expends an intensity of 10 kW and a receptive intensity of 4 kvar at a current of (6 + j4)A. Decide the applied voltage, communicating your answer in complex structure. Arrangement: Here as given, Normal Power, P = 10 kW, Receptive Power, Q = 4 kvar, Current I = (6+j4) A Leave S alone the Apparent Power then we realize that, Clear Power S = Re{VI*} - Im{VI*} S = P - jQ, subbing the estimations of P and Q S = 1010 - j410 S = (10 - j4) 10 . Eq. 1(2) Since S = VI* . Eq. 2(2) Where V is the applied voltage and I* is the conjugate of I. As we realize that in the event that z = a + jb is a mind boggling number, at that point z* = a - jb Therefore, I* = 6 - j4 Presently comparing Eq. 1(2) and 2(2) and subbing the estimation of I* we have, V (6 - j4) = (10 - j4) 10 V = (10 - j4) 10/(6 - j4) In the wake of legitimizing, V = (6 + j4)(10 - j4) 10/(6 + 4) V = (76 + j16) 10/52 V = (1.46 + j0.30) 10 System 5 Hence, V = 1.4610 + j300 . Eq. 3(2) Which is the applied voltage communicated in complex structure. Arrangement: 3(a) Let is the applied voltage and be the subsequent current through the given circuit then for complex impedance circuit is given as, = Expj . Eq. 1(a3) Where. Let be the stage distinction among voltage and current at that point

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