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

Network Analysis - Research Paper ExampleA machine consumes a business leader of 10 kW and a reactive power of 4 kvar at a current of (6 + j4)A. line up the applied voltage, expressing your closure in complex form. Solution Here as given, Average precedent,P = 10 kW, Reactive Power, Q = 4 kvar, Current I = (6+j4) A let S be the Apparent Power accordingly we retire that, Apparent PowerS = ReVI* - ImVI* S = P - jQ, substituting the determine of P & Q S = 10 10 - j4 10 S = (10 j4) 10. Eq. Since S = VI*... Eq. 2(2) Where V is the applied voltage and I* is the conjugate of I. As we know that if z = a + jb is a complex number then z* = a jb Therefore, I* = 6 - j4 Now equating Eq. 1(2) & 2(2) and substituting the value of I* we have, V (6 - j4) = (10 j4) 10 V = (10 j4) 10/ (6 - j4)After rationalizing, V = (6 + j4)(10 j4) 10/ (6 + 4)V = (76 + j16) 10/ 52 V = (1.46 + j0.30) 10Network 5 Hence,V = 1.46 10 + j300 . Eq. 3(2)Which is the applied voltage expressed in complex form. Solut ion 3(a) let is the applied voltage & be the resulting current through the given circuit then for complex underground circuit is given as, = Expj . Eq. 1(a3) Where . Let be the phase difference between voltage and current than current = Expj( + . Eq. 2(a3) Since impedance in season domain is defined as, = . Eq. 3(a3) From equations 1(a) & 2(a) we have, = as R=1 (given) = Or in polar form, . Eq. 4(a3) Multiplying by 1 /to yield effective value we have, Z= or Z= 0.707. Eq. 5(a3)Equation 5(a). is the required impedance in polar form. entry is the reciprocal of impedance so, if Y is admittance then Y = 1/Z. Eq.1(b3)... admittance 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)Equation 2(b) gives the expression of admittance for RLC parallel circuit impedance. Network 42. A machine consumes a power of 10 kW and a reactive power of 4 kvar at acurrent of (6 + j4)A. Determine the applied voltage, expressing your answer in complex form.Solution Here as gi ven,Average Power,P = 10 kW,Reactive Power,Q = 4 kvar, CurrentI = (6+j4) A Let S be the Apparent Power then we know that,Apparent PowerS = ReVI* - ImVI* S = P - jQ, substituting the values of P & QS = 1010 - j410S = (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 know that if z = a + jb is a complex number thenz* = a - jbTherefore,I* = 6 - j4 Now equating Eq. 1(2) & 2(2) and substituting the value of I* we have, V (6 - j4) = (10 - j4) 10V = (10 - j4) 10/ (6 - j4)After rationalizing, V = (6 + j4)(10 - j4) 10/ (6 + 4)V = (76 + j16) 10/ 52V = (1.46 + j0.30) 10 Network 5Hence,V = 1.4610 + j300. Eq. 3(2)Which is the applied voltage expressed in complex form.Solution 3(a) Let is the applied voltage & be the resulting current through the given circuit then for complex impedance circuit is given as,= Expj . Eq. 1(a3)Where. Let be the phase difference between voltage and current then