Il . Fig. 2. (1) I (2) 2 Rl (3) 3 (4) 4 tttTt I;tItf; { How many mesh equations afe needed to solve for mesh currents shown in Fis. 34? $(s) 5 c(6) 6 any one of the (7) 7 (8) 8 € ) 34 How many node equatigns are needed to solve for the node voltages in the network of Fig. 34? Hint: Convert all voltage sources to current sources. (Hint: Node Analysis, page 23) (Choose your answer from the choices given for Question l.) Which of the following methods of analysis would be most appropriate to solve for the voltage across Rr in the network of Fig. 34? (l) Thevinin’s theorem (4) Superposition (2) Branch current (5) Nodal analysis ,K:l Mesh analysis 4. In the network of Fig. 35 the mesh equation for mesh 2 is (l) (R r -jXc, + jXL)It + jX t,I2 * jXs,lt : -E , (2) -jXt,/r + (Rz * jXc,+ jXr,)lz -jXc,lt –E2 (3) jXc,lt -jXc,lz + (R: + jX r,-jXr, -jXc)lt –E2 ‘#4) -jXL,l1 * (R, -jXc” * jX r,)1, * jXsrls –82 (5) (Rt* jXg,+ jXL)It-jXt,Iz-j X c , l t : * Er (6) none ofthe above. . — 5. In the network of Fig. 35 the mesh equation for mesh I is l(t) (Choosey our answersfr om the choicesg ivenf or Question4 .)
I –II R5 (1) (4) IIlIIII(IITIIIIII Xt ^ + 45 Iv)E2 cr 🙂 determinant for the solution of .I3 is l l ^v ‘.1c3r1 -y’ ncz c /+ l -ilD€II; R2 Fig. 35 6. The numerator (2) (3) Rr * /Xc, * jXr, -jXr, -jXr, Rz * jXc” * jXr, -jXc, -jXc” Rr * jXc, * jXt, -jX;, -jXt, Rz * jXc, * jXr, -jXr, -jXc, Rt -jXc, * jXt, *jXr, -jXt, Rz -jXc” * jXt, jX,, -jXr, Rt -jXc, * jXL, iXr, jXt, Rz -jXc, * jXr, -jXc, jXr” F . _tr^ 0 -: v I-^ a -jXc, f i : + j X r , * j X c ,F . u l E2 R r * j X L , -j X c ,E1 p-” 0 , ‘ 1 -J.I C, jXc” 7. Thei mpedancdee terminanat z for then etworko f Fig.3 5i s (choosey our answefrr om thec hoicegs ivenf or euestion6 .) 8. In then etworko f Fig.3 5 all resistorasr e 5 o and eacho f the reactive elementhsa sa valueo f l0 o (+r0 for thei nductora nd -10 for the capacitor).I f the two voltages ourcesa rei n phasew ith a voltageo f 5 v each, what is the value of It?…