Network Analysis - Research Paper Example

Network NETWORK ANALYSIS ENGINEERING PRINCIPLES: (ELECTRICAL NETWORKS) NETWORK ANALYSIS Network 2 1. The three components R, L and C are placed first in series and then in parallel across a supply of angular frequency . Write expressions, in complex form, for: a) the impedance of the series circuit b) the admittance of the parallel circuit. Solution: a) Consider the following (Fig. a(1)) R, L & C series circuit with impedances ZR, ZL & ZC. Let a supply V of angular frequency is applied across this circuit with a current I through it. If VR, VL & VC are voltage drops across R, L & C then according to Kerchhoff's Law then, V=VR+VL+VC IZeq=IZR+IZL+IZC (where Zeq is the equivalent or total impedance of this circuit.) IZeq=I(ZR+ZL+ZC) R jL -j/C Fig. a(1) IZeq=I(R+ jL -j/C) Zeq= R+ jL -j/C Hence Zeq or simply Z the (total) impedance of a RLC series circuit is given by, Z= R+ j(L -1/C) .Eq. 1(a1) Network 3 b)Now consider the circuit with same parameters in parallel configuration as shown in fig. b(1). Now if IR, IL & IC be the currents flowing through R, L and C respectively then according to Kirchcoff's current Law, Fig. b(1) Ieq=IR+IL+IC 1/Zeq=V/R+V/ jL+V/-j/C 1/Zeq=V/{1/R+1/ jL-1/(j/C)} 1/Zeq=1/(1/R+1/ jL-C /j) 1/Zeq=1/R+j(C -1/L) .Eq. 1(b1) Since the admittance of a circuit is defined as the reciprocal of impedance or 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 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. Network 6 3(b) As admittance is the reciprocal of impedance so, if Y is admittance then Y = 1/Z . Eq.1(b3) From equation 5(a), Y =1/0.707 or Y = 1.41 or Y = 1.41() or Y = . Eq. 2(b3) Where G = 1.41& B = be the required expression for admittance. 3(c) Before calculating the values of Rs, Xs & Rp & Xp let find the nature of circuit. If we see the traces of oscilloscope the as current is leading the voltage which is the characteristic of RC circuit. Hence assuming a series RC circuit, as, Z = Rs + j Xs . Eq.1(c3) From equation 5(a3), Z = 0.707 or . Eq. 2(c3) Comparing 1(c3) & 2(c3) we have, Rs = 0.707 & Xs = . Eqs. 3(c3) Since the phase difference = 1 horizontal div. As 12 div = 2 or 1div = hence Rs = 0.707 & Xs = or Rs = 0.612 & Xs = 0.35 . Eqs. 4(c3) For capacitance as, Xs = 1/Cs, Network 7 or Cs = 1/Xs, or Cs =1/2fXs, or Cs = 1/2(1210010Exp-60.35) or Cs = 379 . Eq. 5(c3) Now if Rp & Xp be the parallel equivalent the for a parallel RC circuit the admittance is given as, Y = Rp + jXp . Eq. 6(c3) Comparing equations 2(b3) & 6(c3), Rp +j Xp = = 1.41+j = 1.22 + j0.57 Hence Rp =1.22 S & Xp = 0.57 S (Siemens) . Eq. 7(c3) For capacitance, Cp = Xp / or Cp = 0.57/2(1210010Exp-6) (after putting the values of & Xp ) or Cp = 75.6 . Eq. 8(c3) Network 8 Solution: 4 (a) Since a single source of voltage is connected to the two port network, hence the one generator z-parameter equivalent circuit for the given condition is shown below in Fig. 1(a4). Port 1 +V1 I1 z11-z12 z22-z12 (z21-z12)I1 I2 +V2 Port2 Fig. 1(4a) Now for the values of z-parameters, as, V1 & V2 in terms of I & z-parameters be given as, V1= z11I1 + z12I2 . Eq. 1(a4) V2= z21I1 + z22I2 . Eq. 2(a4) Now according to 1st given condition as an open circuit voltage is measured at port two Hence, I2 = 0, & we only have to determine the values of z11 and z21. i.e; V1= z11I1 V2= z21I1 or z11 = V1/I1I2 = 0 & z21 = V2/I1I2 = 0, Substituting the values from Table A, we have, or z11 = 10 & z21 = 2010 . Eq. 3(a4) Now according to 2nd given condition as an open circuit voltage is measured at port one. therefore, I1 = 0, & we only have, Network 9 V1= z12I2 V2= z22I2 or z12 = V1/I2I1 = 0 & z22 = V2/I2I1 = 0, Substituting the values from Table A, we have, or z12 = 10 & z22 = 1M, . Eq. 4(a4) 4(b) The y-parameter equivalent circuit for the network is given below in fig. 1(b4) with short circuit admittance parameters. Now from Table B, z = z22z11 - z21 z12, Substituting the values from Eqs. 3(a4) & 4(a4), we have, z = 980 M, Hence, y11 = z22/z, y11 =135 S, y12 = - z12/z y12 = - 1.0210Exp-6 y21 = - z21/z y21 = - 24010Exp-9 & y22 = z11/z, y22 = 1.0210Exp-6 be our required y-parameters. The -ve sign for y12 & y21 indicates that current is flowing out of network into the voltage source generator. Port 1 +V1 I1 I2 +V2 Port2 Fig. 1(b4) Network 10 4(c) To determine gain using y-parameter model as, V1=(y22I1 - y12I2)/y . Eq. 1(c4) V2=(-y21I1 + y11I2)/y . Eq. 2(c4) Where y =y11y22-y12y21 And I1 & I2 in terms of y-parameters and voltages be, I1=y11V1 + y12V2 . Eq. 3(c4) I2=y11V1 + y12V2 . Eq. 4(c4) Substituting the value of I1 from Eq. 3(c4) into Eq. 1(c4) we have, V1=(y22( y11V1 + y12V2 ) - y12I2)/y V1=(y22 y11V1 +y22 y12V2 ) - y12I2)/y y V1-y22 y11 V1 =y22 y12V2 - y12I2 (y -y22 y11) V1 =y22 y12V2 - y12I2 . Eq. 5(c4) When R = 1 M is connected at the output of two port circuit we have, I2 = -V2/R where -ve sign shows that current is flowing Outward to two port network into the resistance R. Putting the value of I2 in Eq. 5(c4), (y -y22 y11) V1 =y22 y12V2 + y12V2/R (y -y22 y11) V1 = (y22 y12V2 + y12 /R)V2 V2/ V1 = (y22 y12 + y12 /R)/ (y -y22 y11) Ay = V2/ V2 = y12 (y22 + y12 /R)/ -y12 y21 Ay = V2/ V2 = (y22 + y12 /R)/ - y21 (Asy =y11y22-y12y2 ) This is the required gain in terms of y-parameter with R =1 M. Network 11 5(a) From given CE configuration, if we look into the base emitter terminal, hie is in series with hre v0. As for CE configuration, hre is very small so it can be neglected. Therefore, rin hie .Eq. 1(a5) Now considering the given circuit, v1= hieib + ieRE, As ie = ic + ib therefor v1= hieib +(ic + ib)RE, v1= hieib + ib RE + hfeibRE (as ic = hfeib) v1=ib[hie + (1 + hfeib )RE] Hence for base, rin = v1/i1 gives, rin = v1/i1 = ib[hie + (1 + hfe )RE]/ ib (as i1= ib) = hie + (1 + hfe )RE . Eq. 2(a5) For r(stage) as R1 & R2 are parallel, i.e.; R1 R2 =R12, therefor, r(stage) = R12 rin . Eq. 3(a5) Fig. 1(a5) As magnitude of the mid-band voltage gain Gv(MF) is defined as, Gv(MF) = V2/V1 = v2/ v1 Network 12 Gv(MF) = v0/ vin . Eq. 4(a5) As from equivalent circuit v0 =ic RL & vin =ib rin or, Gv(MF) = ic RL / ib[ hie + (1 + hfe )RE] or, Gv(MF) = hfe RL /[ hie + (1 + hfe )RE] (ic=hfeib) Gv(MF) = hfe RL /[ hie + (1 + hfe )RE] Fig. 2(a5) Now if we consider the expression if (1 + hfe )RE >> hie then, Gv(MF) = RL /RE Hence, Gv(MF) = [hfe/ hie] RL /[1 + (1 + hfe )RE/ hie] Gv(MF) = [hfe/ hie] RL /[1 +G0RL] With G0RL = (1 + hfe )RE/ hie 5(b) Generally as voltage gain is given as, Gv(MF) = hfeRL/Rin Gv(MF) = hfeRL/R12 or Gv(MF) = hfeRL/R12 Substituting values from table we have, Gv(MF) = 25 For power gain as Ap, Ap = Av.Ai Where Ai = i2/i1 = hfe R1 R2(rin + R1 R2) .Solving we can have power gain. Network 13 References Edministator, J. A. (1995). Schaum's outline of Theory & problems of Electrical circuits. University of Akron: McGRAW- Hill Book Company. Saeed, M. (1995). Electrical Circuit & Devices. University of Engineering & Technology, Lahore. A-One Publisher Conpany. Theraja, B. L. & Theraja, A. K. (1995). A Text Book of Electrical Technology. New Delhi: Nerja Construction & Development Conpany. Valkemburg, M.V. (1995). Network Analysis. University of Illinois, New Jersy: PRENTICE-HALL INC. Read More