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

Network Analysis - Research Paper Example A machine consumes a power of 10 kW and a reactive power of 4 kvar at a current of (6 + j4)A. Determine the applied voltage, expressing your answer in complex form. Solution: Here as given, Average Power, P = 10 kW, Reactive Power, Q = 4 kvar, Current I = (6+j4) A Let S be the Apparent Power then we know that, Apparent Power S = Re{VI*} - Im{VI*} S = P - jQ, substituting the values of P & Q S = 10 10 ³ - j4 10 ³ S = (10 –j4) 10 ³ †¦. 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 ³ V = (10 – j4) 10 ³/ (6 - j4) After rationalizing, V = (6 + j4)(10 – j4) 10 ³/ (6 ² + 4 ²) V = (76 + j16) 10 ³/ 52 V = (1.46 + j0.30) 10 ³ Network 5 Hence, V = 1.46 10 + 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 than current = Expj( + †¦. Eq. 2(a3) Since impedance in time 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. admittance 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 4 2. A machine consumes a power of 10 kW and a reactive power of 4 kvar at a current of (6 + j4)A. Determine the applied voltage, expressing your answer in complex form. Solution: Here as given, Average Power, P = 10 kW, Reactive Power, Q = 4 kvar, Current I = (6+j4) A Let S be the Apparent Power then we know that, Apparent Power S = Re{VI*} - Im{VI*} S = P - jQ, substituting the values of P & 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 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) 10 Network 5 Hence, 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