A.gate by-rk-kanodia

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MULTIPLE CHOICE QUESTION Electronics & Communication Engineering Fifth Edition R. K. Kanodia B.Tech. NODIA & COMAPNY JAIPUR GATE EC BY RK Kanodia www.gatehelp.com

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  1. 1. MULTIPLE CHOICE QUESTION Electronics&CommunicationEngineering Fifth Edition R. K. Kanodia B.Tech. NODIA & COMAPNY JAIPUR GATE EC BY RK Kanodia www.gatehelp.com
  2. 2. Price 550.00 MRP 400.00 GATE EC BY RK Kanodia www.gatehelp.com
  3. 3. GATE EC BY RK Kanodia www.gatehelp.com
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  9. 9. 1. A solid copper sphere, 10 cm in diameter is deprived of 1020 electrons by a charging scheme. The charge on the sphere is (A) 160.2 C (B) -160.2 C (C) 16.02 C (D) -16.02 C 2. A lightning bolt carrying 15,000 A lasts for 100 ms. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is (A) 13.33 mC (B) 75 C (C) 1500 mC (D) 1.5 C 3. If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is (A) 0.5 A (B) 2 A (C) 3.33 mA (D) 0.3 mA 4. The energy required to move 120 coulomb through 3 V is (A) 25 mJ (B) 360 J (C) 40 J (D) 2.78 mJ 5. i = ? (A) 1 A (B) 2 A (C) 3 A (D) 4 A 6. In the circuit of fig P1.1.6 a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v1 must be (A) 240 V (B) 120 V (C) 60 V (D) 30 V 7. In the circuit of the fig P1.1.7, the value of the voltage source E is (A) -16 V (b) 4 V (C) -6 V (D) 16 V CHAPTER 1.1 BASIC CONCEPTS Page 3 100 V 60 V v1 20 W Fig. P.1.1.6 1 A 2 A 5 A 4 A3 A i Fig. P.1.1.5 2 V 1 V 5 V 10 V 0 V 4 V E + + + + + Fig. P.1.1.7 GATE EC BY RK Kanodia www.gatehelp.com
  10. 10. 8. Consider the circuit graph shown in fig. P1.1.8. Each branch of circuit graph represent a circuit element. The value of voltage v1 is (A) -30 V (B) 25 V (C) -20 V (D) 15 V 9. For the circuit shown in fig P.1.1.9 the value of voltage vo is (A) 10 V (B) 15 V (C) 20 V (D) None of the above 10. R1 = ? (A) 25 W (B) 50 W (C) 100 W (D) 2000 W 11. Twelve 6 W resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is (A) 5 6 W (B) 6 5 W (C) 5 W (D) 6 W 12. v1 = ? (A) -11 V (B) 5 V (C) 8 V (D) 18 V 13. The voltage vo in fig. P1.1.11 is always equal to (A) 1 V (B) 5 V (C) 9 V (D) None of the above 14. Req = ? (A) 11.86 W (B) 10 W (C) 25 W (D) 11.18 W 15. vs = ? (A) 320 V (B) 280 V (C) 240 V (D) 200 V Page 4 UNIT 1 Networks 5 W 15 V 1 Avo + Fig. P.1.1.9 + 105 V 15 V + 10 V + + 35 V + 55 V + 65 V v1+ + 30 V + 30 V 100 V Fig. P.1.1.8 1 kW 2 kW 3 kW 4 kW 8 V 5 V 6 V 7 V + v1 Fig. P1.1.12 + vo 5 V 1 A 4 W Fig. P1.1.11 5 W 10 W 10 W up to Req 10 W 10 W 10 W 10 W Fig. P1.1.14 180 W 40 W 20 V + 60 W 90 W 180 W vs Fig. P.1.1.15 100 V R1 + 20 V + 70 V R2 60 W Fig. P.1.1.10 GATE EC BY RK Kanodia www.gatehelp.com
  11. 11. 24. Let i t te t ( ) = - 3 100 A and v t t e t ( ) . ( . )= - - 0 6 0 01 100 V for the network of fig. P.1.1.24. The power being absorbed by the network element at t = 5 ms is (A) 18.4 mW (B) 9.2 mW (C) 16.6 mW (D) 8.3 mW 25. In the circuit of fig. P.1.1.25 bulb A uses 36 W when lit, bulb B uses 24 W when lit, and bulb C uses 14.4 W when lit. The additional A bulbs in parallel to this circuit, that would be required to blow the fuse is (A) 4 (B) 5 (C) 6 (D) 7 26. In the circuit of fig. P.1.1.26, the power absorbed by the load RL is (A) 2 W (B) 4 W (C) 6 W (D) 8 W 27. vo = ? (A) 6 V (B) -6 V (C) -12 V (D) 12 V 28. vab = ? (A) 15.4 V (B) 2.6 V (C) -2.6 V (D) 15.4 V 29. In the circuit of fig. P.1.1.29 power is delivered by (A) dependent source of 192 W (B) dependent source of 368 W (C) independent source of 16 W (D) independent source of 40 W 30. The dependent source in fig. P.1.1.30 (A) delivers 80 W (B) delivers 40 W (C) absorbs 40 W (D) absorbs 80 W 31. In the circuit of fig. P.1.1.31 dependent source (A) supplies 16 W (B) absorbs 16 W (C) supplies 32 W (D) absorbs 32 W Page 6 UNIT 1 Networks 20 A A B C 12 V Fig. P.1.1.25 i v + N Fig. P.1.1.24 i1 RL = 2 W2i11 V 1 W Fig. P.1.1.26 0.2 A 5 W v1 + v2 + 8 W0.3v1 18 W vo + 5v2 Fig. P.1.1.27 a b 2 A 2 W 6 W 2 W 0.2i1i1 0.3i1 R 8 A Fig. P.1.1.28 500 W ix 400 W 200 W40 V 2ix Fig. P.1.1.29 5 W 20 V 5 W v1 v1 5 Fig. P.1.1.30 ix 8 V+ 2ix4 A Fig. P.1.1.31 GATE EC BY RK Kanodia www.gatehelp.com
  12. 12. 32. A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is (A) 0.75 mF (B) 1.33 mF (C) 0.6 mF (D) 1.67 mF 33. The energy required to charge a 10 mF capacitor to 100 V is (A) 0.10 J (B) 0.05 J (C) 5 10 9 - J (D) 10 10 9 - J 34. The current in a 100 mF capacitor is shown in fig. P.1.1.34. If capacitor is initially uncharged, then the waveform for the voltage across it is 35. The voltage across a 100 mF capacitor is shown in fig. P.1.1.35. The waveform for the current in the capacitor is 36. The waveform for the current in a 200 mF capacitor is shown in fig. P.1.1.36 The waveform for the capacitor voltage is 37. Ceq = ? (A) 3.5 mF (B) 1.2 mF (C) 2.4 mF (D) 2.6 mF 38. In the circuit shown in fig. P.1.1.38 i t tin ( ) sin= 300 20 mA, for t 0. Let C1 40= mF and C2 30= mF. All capacitors are initially uncharged. The v tin ( ) would be (A) -0.25cos 20t V (B) 0.25cos 20t V (C) -36cos 20t mV (D) 36cos 20t mV Chap 1.1Basic Concepts Page 7 250m 50m v t(ms) 5 4 v t(ms) 4 v t(ms) 4 v t(ms) 4 50m (C) (D) i(mA) t(ms) 5 4 Fig. P. 1.1.36 (A) (B) v 2 t(ms) 10 4 4 v 2 t(ms) 10 v 2 t(ms) 0.2 4 4 v 2 t(ms) 0.2 (C) (D) (A) (B) i(mA) 2 t(ms) Fig. P. 1.1.34 2.5 Fm 2 Fm1.5 Fm 1 FmCeq Fig. P.1.1.37 6 21 3 v t(ms) Fig. P.1.1.35 6 21 3 t(ms) i(mA) 6 2 1 3 t(ms) i(mA) 600 600 21 3 t(ms) i(mA) 2 1 3 t(ms) i(mA) (C) (D) (A) (B) iin vin + C2 60 mFC1 C2 C2 C2 C1 C1 C1 Fig. P. 1.1.38 GATE EC BY RK Kanodia www.gatehelp.com
  13. 13. SOLUTIONS 1. (C) n = 1020 , Q ne e= = =10 16 0220 . C Charge on sphere will be positive. 2. (D) D DQ i t= = 15000 100m = 15. C 3. (B) i dQ dt = = = 120 60 2 A 4. (B) W Qv= = 360 J 6. (A) 6. (A) In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise. i dQ dt = = = 600 60 10 A Applying KVL we get v1 60 100+ - = 10 20 or v1 = 240 V 7. (A) Going from 10 V to 0 V 10 5 1+ + +E = 0 or E = -16 V 8. (D) 100 65 352 2= + =v v V v v v3 2 330 65- = = V 105 65 04 3+ - - =v v v4 25= V v v4 115 55 0+ - + = v1 15= V 9. (B) Voltage is constant because of 15 V source. 10. (C) Voltage across 60 W resistor = 30 V Current = = 30 60 0 5. A Voltage across R1 is = - =70 20 50 V R1 50 0 5 100= = . W 11. (C) The current i will be distributed in the cube branches symmetrically v i i i iab = + + = 6 3 6 6 6 3 5 , R v i eq ab = = 5 W 12. (C) If we go from +side of 1 kW through 7 V, 6 V and 5 V, we get v1 7 6 5 8= + - = V 13. (D) It is not possible to determine the voltage across 1 A source. 14. (D) R R R eq eq eq = + + + + 5 10 5 10 5 ( ) + = + + +R R R Req eq eq eq 2 15 5 75 10 50 = =Req 125 1118. W Chap 1.1Basic Concepts Page 9 i a b i 3 i 3 i 6 i 6 i 3 i 3 i Fig. S. 1.1.11 1 A 2 A 5 A 4 A3 A i = 1 A 2 A 1 A 6 A Fig. S 1.1.5 + v2 + v4 + v3 + 105 V 15 V + 10 V + + 55 V + 65 V v1+ + 30 V + 30 V 100 V Fig. S 1.1.8 2 V 1 V 5 V 10 V 0 V 4 V E + + + + + Fig. S 1.1.7 5 W 10 W 5 W Req Req Fig. S 1.1.14 GATE EC BY RK Kanodia www.gatehelp.com
  14. 14. v v vo o o - + = = 20 5 5 20 5 20 V Power is P v v o= = =1 5 20 20 5 80 W 31. (D) Power P vi i i ix x x= = =2 2 2 ix = 4 A, P = 32 W (absorb) 32. (D) v v C idtt t t t 2 1 1 2 1 - = = -12 1 2 2 1 C t tm ( ) = 12 2 10C m =C 1 67. mF 33. (B) E Cv= = =-1 2 5 10 100 0 052 6 2 . J 34. (D) v c idtc = = = - - -1 10 10 100 10 2 10 0 2 0 2 3 6 3 m ( ) . V This 0.2 V increases linearly from 0 to 0.2 V. Then current is zero. So capacitor hold this voltage. 35. (D) i C dv dt = For 0 1<
  15. 15. 1. Consider the following circuits : The planner circuits are (A) 1 and 2 (B) 2 and 3 (C) 3 and 4 (D) 4 and 1 2. Consider the following graphs Non-planner graphs are (A) 1 and 3 (B) 4 only (C) 3 only (D) 3 and 4 3. A graph of an electrical network has 4 nodes and 7 branches. The number of links l, with respect to the chosen tree, would be (A) 2 (B) 3 (C) 4 (D) 5 4. For the graph shown in fig. P.1.1.4 correct set is Node Branch Twigs Link (A) 4 6 4 2 (B) 4 6 3 3 (C) 5 6 4 2 (D) 5 5 4 1 5. A tree of the graph shown in fig. P.1.2.5 is (A) a d e h (B) a c f h (C) a f h g (D) a e f g CHAPTER 1.2 GRAPH THEORY Page 12 Fig. P.1.1.4 (1) (2) (3) (4) (1) (2) (3) (4) cc 1 2 3 g 5 ed h4 b a f Fig. P.1.2.5 GATE EC BY RK Kanodia www.gatehelp.com
  16. 16. (A) 1 1 0 1 0 1 0 1 1 - - - (B) 1 0 1 1 1 0 0 1 1 - - - (C) - - - 1 1 0 0 1 1 1 0 1 (D) - - - 1 0 1 0 1 1 1 1 0 13. The incidence matrix of a graph is as given below A = - - - - - - - 1 1 1 0 0 0 0 0 1 1 1 0 0 1 0 1 0 1 1 0 0 0 1 1 The graph is 14. The incidence matrix of a graph is as given below A = - - - - - - - 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0 The graph is 15. The incidence matrix of a graph is as given below A = - - - - - - 1 1 1 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 0 0 1 1 The graph is 16. The graph of a network is shown in fig. P.1.1.16. The number of possible tree are (A) 8 (B) 12 (C) 16 (D) 20 Page 14 NetworksUNIT 1 1 2 3 4 4 4 4 1 2 3 1 2 3 1 2 3 (C) (D) (A) (B) 1 2 4 3 1 2 4 3 1 2 4 3 1 2 4 3 (C) (D) (A) (B) 1 2 4 3 5 1 2 4 3 5 (A) (B) 1 2 4 3 5 1 2 4 3 5 (C) (D) Fig. P.1.1.16 GATE EC BY RK Kanodia www.gatehelp.com
  17. 17. 22. The fundamental cut-set matrix of a graph is QF = - - - - 1 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 0 1 1 0 The oriented graph of the network is 23. A graph is shown in fig. P.1.2.23 in which twigs are solid line and links are dotted line. For this chosen tree fundamental set matrix is given below. BF = - 1 1 0 0 1 0 0 1 1 1 0 0 0 0 0 1 1 1 The oriented graph will be 24. A graph is shown in fig. P.1.2.24 in which twigs are solid line and links are dotted line. For this tree fundamental loop matrix is given as below BF = 1 1 1 0 1 0 1 1 The oriented graph will be 25. Consider the graph shown in fig. P.1.2.25 in which twigs are solid line and links are dotted line. Page 16 NetworksUNIT 1 2 3 1 5 6 4 2 3 1 5 6 4 2 3 1 5 6 4 2 3 1 5 6 4 (C) (D) (A) (B) (A) (B) 1 2 4 5 6 3 Fig. P. 1.2.23 (C) (D) 4 1 23 Fig. P.1.2.24 41 2 56 3 Fig. P. 1.2.25 (C) (D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
  18. 18. A fundamental loop matrix for this tree is given as below BF = - - - 1 0 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 The oriented graph will be 26. In the graph shown in fig. P.1.2.26 solid lines are twigs and dotted line are link. The fundamental loop matrix is (A) 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 - - - - - - (B) - - - - - - 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 (C) 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 - - - (D) 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 - - 27. Branch current and loop current relation are expressed in matrix form as i i i i i i i i 1 2 3 4 5 6 7 8 0 1 1 0 0 0 1 = - - 1 1 0 0 1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 - - I I I I 1 2 3 4 where ij represent branch current and Ik loop current. The number of independent node equation are (A) 4 (B) 5 (C) 6 (D) 7 28. If the number of branch in a network is b, the number of nodes is n and the number of dependent loop is l, then the number of independent node equations will be (A) n l+ - 1 (B) b - 1 (C) b n- + 1 (D) n - 1 Statement for Q.2930: Branch current and loop current relation are expressed in matrix form as i i i i i i i i 1 2 3 4 5 6 7 8 0 0 1 0 1 1 = - - -1 0 0 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 0 0 1 - - - I I I I 1 2 3 4 where ij represent branch current and Ik loop current. 29. The rank of incidence matrix is (A) 4 (B) 5 (C) 6 (D) 8 Chap 1.2 Page 17 Graph Theory g bh a c i e d f Fig. P.1.2.26 (C) (D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
  19. 19. 30. The directed graph will be 31. A network has 8 nodes and 5 independent loops. The number of branches in the network is (A) 11 (B) 12 (C) 8 (D) 6 32. A branch has 6 node and 9 branch. The independent loops are (A) 3 (B) 4 (C) 5 (D) 6 Statement for Q.3334: For a network branch voltage and node voltage relation are expressed in matrix form as follows: v v v v v v v v 1 2 3 4 5 6 7 8 1 0 0 1 0 1 0 0 0 = 0 1 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 - - - - V V V V 1 2 3 4 where vi is the branch voltage and Vk is the node voltage with respect to datum node. 33. The independent mesh equation for this network are (A) 4 (B) 5 (C) 6 (D 7 33. The oriented graph for this network is ************ Page 18 NetworksUNIT 1 5 3 421 5 3 421 5 3 421 5 3 421 (C) (D) 5 3 4 2 1 7 6 8 5 3 4 2 1 7 6 8 5 3 4 2 1 7 6 8 5 3 4 2 1 7 6 8 (C) (D) GATE EC BY RK Kanodia www.gatehelp.com
  20. 20. SOLUTIONS 1. (A) The circuit 1 and 2 are redrawn as below. 3 and 4 can not be redrawn on a plane without crossing other branch. Fig. S1.2.1 2. (B) Other three circuits can be drawn on plane without crossing Fig. S1.2.1 3. (C) l b n= - - =( )1 4. 4. (B) There are 4 node and 6 branches. t n= - =1 3, l b n= - + =1 3 5. (C) From fig. it can be seen that a f h g is a tree of given graph 6. (B) From fig. it can be seen that a d f is a tree. 7. (D) D is not a tree Fig. S .1.2.7 8. (D) it is obvious from the following figure that 1, 3, and 4 are tree Fig. S. 1.2.8 Chap 1.2 Page 19 Graph Theory (1) (2) (3) cc ged h b a f Fig. S 1.2.5 (1) (2) c e d fa b Fig. S. 1.2.6 2 2 a a 1 1 3 3 c c d d e e b b 4 4 f f 2 a 1 3 c d e b 4 f 2 a 1 3 c d e b 4 f 2 a 1 3 c d e b 4 f (5) (1) (2) (3) (4) (C) (D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
  21. 21. This in similar to matrix in (A). Only place of rows has been changed. 27. (A) Number of branch =8 Number of link =4 Number of twigs =8 4 4- = Number of twigs =number of independent node equation. 28. (D) The number of independent node equation are n - 1. 29. (A) Number of branch b = 8 Number of link l = 4 Number of twigs t b l= - = 4 rank of matrix = - = =n t1 4 30. (B) We know the branch current and loop current are related as [ ] [ ][ ]i B Ib T L= So fundamental loop matrix is Bf = - - - - - - 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 1 f-loop 1 include branch (2, 4, 6, 7) and direction of branch2 is opposite to other (B only). 31. (B) Independent loops =link l b n= - -( )1 = -5 7b , b = 12 32. (B) Independent loop =link l b n= - - =( )1 4 33. (A) There are 8 branches and 4 1 5+ = node Number of link = - + =8 5 1 4 So independent mesh equation =Number of link. 34. (D) We know that [ ] [ ]v A Vb r T n= So reduced incidence matrix is Ar = - - - - 1 0 0 0 1 0 0 1 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 0 1 0 At node-1, three branch leaves so the only option is (D). *********** Page 22 NetworksUNIT 1 g bh a c i e d fl3 l4 l1 l2 Fig. S 1.2.26 GATE EC BY RK Kanodia www.gatehelp.com
  22. 22. 1. v1 = ? (A) 0.4vs (B) 1.5vs (C) 0.67vs (D) 2.5vs 2. va = ? (A) -11 V (B) 11 V (C) 3 V (D) -3 V 3. v1 = ? (A) 120 V (B) -120 V (C) 90 V (D) -90 V 4. va = ? (A) 4.33 V (B) 4.09 V (C) 8.67 V (D) 8.18 V 5. v2 = ? (A) 0.5 V (B) 1.0 V (C) 1.5 V (D) 2.0 V 6. ib = ? (A) 0.6 A (B) 0.5 A (C) 0.4 A (D) 0.3 A CHAPTER 1.3 Page 23 METHODS OF ANALYSIS 6R 3R 4vs vsv16R + Fig. P1.3.1 3 A 2 W va 3 W 1 A Fig. P1.3.2 12 V 10 W 4 W 4 A1 W 2 W 10 V va Fig. P1.3.4 v2 60 W 30 W 0.5 A 10 V 30 W 20 W + Fig. P1.3.5 10 W 30 V 3 A 20 W 30 W v1 + 6 A 60 W9 A60 W Fig. P1.3.3 10 V 36 W 69 W 0.5 A 37 W64 W ib Fig. P1.3.6 GATE EC BY RK Kanodia www.gatehelp.com
  23. 23. 7. i1 = ? (A) 3.3 A (B) 2.1 A (C) 1.7 A (D) 1.1 A 8. i1 = ? (A) 1 mA (B) 1.5 mA (C) 2 mA (D) 2.5 mA 9. i1 = ? (A) 4 A (B) 3 A (C) 6 A (D) 5 A 10. i1 = ? (A) 20 mA (B) 15 mA (C) 10 mA (D) 5 mA 11. i1 = ? (A) 0.01 A (B) -0.01 A (C) 0.03 A (D) 0.02 A 12. The value of the current measured by the ammeter in Fig. P1.3.12 is (A) 2 3 A (B) 5 3 A (C) - 5 6 A (D) 2 9 A 13. i1 = ? (A) 10 mA (B) -10 mA (C) 0.4 mA (D) -0.4 mA Page 24 UNIT 1 Networks 2 A 4 A 3 W 2 W i1 4 W 3 V Fig. P1.3.9 i145 V 500 W 2 kW 15 mA Fig. P1.3.10 7.5mA 90 kW 10 kW 90 kW 10 kW i2 i1 75 V Fig. P1.3.8 6.6 V i1 100 W 40 W 0.1 A 0.06 A 50 W 60W 0.1A Fig. P1.3.11 75 V 4 W 3 W 5 W 10 W 2 W8 W 6 A i1 Fig. P1.3.7 2 A 3 A 6 W 4 W7 W 5 W 2 W Ammeter Fig. P1.3.12 40 mA 100 W i1 200 W 50 W 10 mA Fig. 1.3.13 GATE EC BY RK Kanodia www.gatehelp.com
  24. 24. 14. The values of node voltage are va = 12 V, vb = 9.88 V and vc = 5.29 V. The power supplied by the voltage source is (A) 19.8 W (B) 27.3 W (C) 46.9 W (D) 54.6 W 15. i i i1 2 3, , = ? (A) 3 A, 2 A, and 4 A (B) 3 A, 3 A, and 8 A (C) 1 A, 3 A, and 4 A (D) 1 A, 2 A, and 8 A 16. vo = ? (A) 6 5 V (B) 8 5 V (C) 6 7 V (D) 5 7 V 17. The mesh current equation for the circuit in Fig. P1.3.17 are (A) 4 2 0 2 8 2 0 2 5 12 8 20 1 2 3 - - - - = - i i i (B) 6 2 0 2 12 2 0 2 7 12 8 20 1 2 3 - - - = i i i (C) 6 2 0 2 12 2 0 2 7 12 8 20 1 2 3 - - - - = i i i (D) 4 2 0 2 8 2 0 2 5 12 8 20 1 2 3 - - - = i i i 18. For the circuit shown in Fig. P1.3.18 the mesh equation are (A) 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k k k k k - - - - - - i i i = - 6 0 5 (B) 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k k k k k - - - - i i i = - 6 0 5 (C) - - - 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k k k k k i i i = - 6 0 5 (D) - - - - - 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k k k k k i i i = - 6 0 5 Chap 1.3Methods of Analysis Page 25 15 V 3 W 9 W 2 W i2 i3 21 V6 W i1 Fig. P1.3.15 i2 i3i1 4 W 8 W 5 W 12 V 2 W 2 W 8 V 20 V Fig. 1.3.17 4 mA vo2 kW1 mA 2 kW 1 kW 2 mA 1 kW + 1 kW Fig. P1.3.16 va 6 W 4 W 3 W 2 W1 A12 V vb vc Fig. 1.3.14 6 kW 5 mA6 V 6 kW 6 kW 6 kW i1 i2 i3 Fig. 1.3.18 GATE EC BY RK Kanodia www.gatehelp.com
  25. 25. The value of R4 is (A) 40 (B) 15 (C) 5 (D) 20 26. va = ? (A) 26 V (B) 19 V (C) 13 V (D) 18 V 27. v = ? (A) 60 V (B) -60 V (C) 30 V (D) -30 V 28. i1 = ? (A) 66.67 mA (B) 46.24 mA (C) 23.12 mA (D) 33.33 mA 29. va = ? (A) 342 V (B) 171 V (C) 198 V (D) 396 V 30. ia = ? (A) 14 mA (B) -6 5. mA (C) 7 mA (D) -21 mA 31. v2 = ? (A) 5 V (B) 75 V (C) 3 V (D) 10 V 32. i1 = ? Chap 1.3Methods of Analysis Page 27 2.5 Wk 10 kW 10 Wk va 10 Wk 5 Wk 4 mA20 V Fig. P1.3.26 2 A 10 W 20 Wv 15 W 5 W 4 A Fig. P.3.1.27 40 V 300 W 0.4i1 i1 500 W Fig. P1.3.28 10 A 10 W 4 A 200 W 5 A 100 W 20 W 20 A 50 W 40 Wva Fig. P1.3.29 50 W 150 W 4 V 8 V2 V 100 W 200 W225 W 75 W 50 W ia Fig. P1.3.30 v2 + 50 W 100 W10 V 0.04v2 Fig. P1.3.31 2 W 4 A8 V i1 0.5i1 4 W 6 V Fig. P1.3.32 v1 R4 v2 25i2 R2 R3 R1 i1 i2 i3 Fig. P1.3.25 GATE EC BY RK Kanodia www.gatehelp.com
  26. 26. (A) -1.636 A (B) -3 273. A (C) -2.314 A (D) -4 628. A 33. vx = ? (A) 32 V (B) -32 V (C) 12 V (D) -12 V 34. ib = ? (A) 4 mA (B) -4 mA (C) 12 mA (D) -12 mA 35. vb = ? (A) 1 V (B) 1.5 V (C) 4 V (D) 6 V 36. vx = ? (A) -3 V (B) 3 V (C) 10 V (D) -10 V 37. va = ? (A) 25.91 V (B) -25.91 V (C) 51.82 V (D) -51.82 V 38. For the circuit of Fig. P1.3.38 the value of vs , that will result in v1 = 0, is (A) 28 V (B) -28 V (C) 14 V (D) -14 V 39. i i1 2, = ? (A) 2.6 A, 1.4 A (B) 2.6 A, -1.4 A (C) 1.6 A, 1.35 A (D) 1.2 A, -1.35 A 40. v1 = ? Page 28 UNIT 1 Networks 1 kW 4va6 V 3 kW ib 2 kW va Fig. P1.3.34 ia 4 kW 2 V 2 kW 5ia vb Fig. P1.3.35 iy 2 A 100 W vx50 W + 25iy 0.2vx 50 W Fig. P1.3.36 1.6 A 100 W 0.02vx 50 W vx + Fig. P1.3.33 0.8va 10 A5 W2.5 W va 16 A 2 W Fig. P1.3.37 3 A 2 A 0.1v1 10 W 20 W 40 Wvs 48 V + v1 Fig. P1.3.38 4 W 2 W 18 Vi115 V i2 2ix 6 W ix Fig. P1.3.39 3 W 14 V 2vy + v1 7 A 6 W 2 A + vy 3 W 2 W Fig. P1.3.40 GATE EC BY RK Kanodia www.gatehelp.com
  27. 27. (A) 10 V (B) -10 V (C) 7 V (D) -7 V 41. vx = ? (A) 9 V (B) -9 V (C) 10 V (D) -10 V 42. The power being dissipated in the 2 W resistor in the circuit of Fig. P1.3.42 is (A) 76.4 W (B) 305.6 W (C) 52.5 W (D) 210.0 W 43. i1 = ? (A) 0.12 A (B) 0.24 A (C) 0.36 A (D) 0.48 A ***************** SOLUTIONS 1. (B) Applying the nodal analysis v v R v R R R R v s s s1 4 6 3 1 6 1 3 1 6 15= + + + = . 2. (C) va = + + =2 3 1 3 1 11( ) ( ) V 3. (D) - + - + = v v1 1 60 60 6 9 = -v1 90 V 4. (C) v va a- + = 10 4 2 4 =va 8 67. V 5. (D) v v2 2 20 10 30 0 5+ + = . =v2 2 V 6. (B) Using Thevenin equivalent and source transform va = + + + + = 25 2 60 15 3 14 1 3 1 15 15 23 8 3 . V i1 25 15 23 14 3 2 09= - = . . A 7. (A) ib = + + = 10 64 36 0 5 0 6. . A 8. (B) 75 90 10 7 51 1= + -k k mi i( . ) 150 100 151 1= =ki i . mA 9. (B) 3 2 3 4 31 1 1= + - =i i i( ) A 10. (B) 45 2 500 151 1= + +k mi i( ) =i1 15 mA 11. (D) 6 6 50 100 0 1 40 0 06 60 0 11 1 1 1. ( . ) ( . ) ( . )= + + + - + -i i i i i1 0 02= . A Chap 1.3Methods of Analysis Page 29 +vx 0.6 A 600 W 900 W 0.3 A 500 W 0.5vx 500 W Fig. P1.3.41 2.5 A 2 W3 W 30 V 5 W 6ia4 W 2 A ia Fig. P1.3.42 500 W 100 W 100 W 400 W + + vy vx 0.005vy 0.6 A 0.001vy 180 V Fig. P1.3.43 25 V W 2 W 10 W 60 V 5 W3 W i1 va 8 3 Fig. S.1.3.6 GATE EC BY RK Kanodia www.gatehelp.com
  28. 28. 38. (D) If v1 0= , the dependent source is a short circuit v v v vs1 1 1 40 10 48 20 2 3+ - + - = - v1 0= - - = - vs 10 48 20 1 vs = - 14 V 39. (D) i i ix = -1 2 15 4 2 61 1 2 1 2= - - + -i i i i i( ) ( ) 8 4 151 2i i- = K(i) - = + -18 2 62 2 1i i i( ) 3 4 91 2i i- = K(ii) i1 12= . A, i2 1 35= - . A 40. (B) 14 3 6 2 7 2 2 71 1 1= + + - - + + -i v i v iy y( ) ( ) v iy = -3 21( ) 14 3 9 2 6 9 2 71 1 1 1= + - + - + -i i i i( ) ( ) ( ) 14 20 18 54 141= - - -i i1 5= A v1 6 5 2 7 2 3 5 2 2 5 7 10= - - + - + - = -( ) ( ) ( ) V 41. (D) Let i1 and i2 be two loop current 0 5 500 5001 1 2. ( )v i i ix = + - , v ix = -500 1 5 2 01 2i i- = K(i) 500 900 0 3 600 0 6 02 1 2 2( ) ( . ) ( . )i i i i- + + + - = - + =5 20 0 91 2i i . K(ii) i1 20= mA, vx = - = -500 20 10m V 42. (C) 30 5 3 2 5 4 2 5 2= + - + - +i i ia a a( . ) ( . ) ia = + + = 30 7 5 2 12 329 . . A 6 19 75ia = . V voltage across 2 W resistor 30 19 75 10 25- =. . V, P = = ( . ) . 10 25 2 52 53 2 W 43. (A) v ix = 500 1 v i vy x= -400 0 0011( . ) = - =400 0 5 2001 1 1( . )i i i 180 500 100 0 6 200 100 0 0051 1 1 1= + - + + +i i i i vy( . ) ( . ) 180 900 60 100 0 005 2001 1= - + i i. i1 0 12= . A ************ Page 32 UNIT 1 Networks 3 A 2 A 10 W 20 W 40 Wvs 48 V + v1 Fig. S1.3.38 3 W 14 V 2vy + v1 i1 7 A 6 W 2 A + vy 3 W 2 W Fig. S1.3.40 GATE EC BY RK Kanodia www.gatehelp.com
  29. 29. 1. v RTH TH, = ? (A) 2 V, 4 W (B) 4 V, 4 W (C) 4 V, 5 W (D) 2 V, 5 W 2. i RN N, = ? (A) 3 A, 10 3 W (B) 10 A, 4 W (C) 1,5 A, 6 W (D) 1.5 A, 4 W 3. v RTH TH, = ? (A) -2 V, 6 5 W (B) 2 V, 5 6 W (C) 1 V, 5 6 W (D) -1 V, 6 5 W 4. A simple equivalent circuit of the 2 terminal network shown in fig. P1.4.4 is 5. i RN N, = ? (A) 4 A, 3 W (B) 2 A, 6 W (C) 2 A, 9 W (D) 4 A, 2 W CHAPTER 1.4 Page 33 NETWORKS THEOREM 3 W 2 W 6 V 6 W RTHvTH, Fig. P.1.4.1 2 W 2 W 4 W15 V RN iN, Fig. P.1.4.2 2 W 3 W 1 W2 A RTHvTH, Fig. P.1.4.3 R R R R i i v (C) (D) (A) (B) vi R Fig. P.1.4.4 2 W 4 W 3 W6 A iN RN Fig. P.1.4.5 GATE EC BY RK Kanodia www.gatehelp.com
  30. 30. 6. v RTH TH, = ? (A) -100 V, 75 W (B) 155 V, 55 W (C) 155 V, 37 W (D) 145 V, 75 W 7. RTH = ? (A) 3 W (B) 12 W (C) 6 W (D) 8. The Thevenin impedance across the terminals ab of the network shown in fig. P.1.4.8 is (A) 2 W (B) 6 W (C) 6 16. W (D) 4 3 W 9. For In the the circuit shown in fig. P.1.4.9 a network and its Thevenin and Norton equivalent are given The value of the parameter are vTH RTH iN RN (A) 4 V 2 W 2 A 2 W (B) 4 V 2 W 2 A 3 W (C) 8 V 1.2 W 30 3 A 1.2 W (D) 8 V 5 W 8 5 A 5 W 10. v1 = ? (A) 6 V (B) 7 V (C) 8 V (D) 10 V 11. i1 = ? (A) 3 A (B) 0.75 mA (C) 2 mA (D) 1.75 mA Statement for Q.1213: A circuit is given in fig. P.1.4.1213. Find the Thevenin equivalent as given in question.. 12. As viewed from terminal x and x is (A) 8 V, 6 W (B) 5 V, 6 W (C) 5 V, 32 W (D) 8 V, 32 W Page 34 UNIT 1 Networks 30 W 25 W 20 W 5 A 5 V RTHvTH, Fig. P.1.4.6 2 A 5 V 6 W 6 W RTH Fig. P.1.4.7 3 W 6 W2 A 8 W 8 W 2 V b a Fig. P.1.4.8 2 W 3 W 2 A4 V RTH vTH iN RN Fig. P.1.4.9 1 W 1 W 3 W 2 W 6 W8 V v1 + 2 W 6 W 18 V Fig. P.1.4.10 4 kW 4 kWi1 6 kW 20 V 24 V4 kW 3 kW12 V Fig. P.1.4.11 10 W x 16 Wx 5 V 1 A40 W 8 W y y Fig. P.1.4.1213 GATE EC BY RK Kanodia www.gatehelp.com
  31. 31. 13. As viewed from terminal y and y is (A) 8 V, 32 W (B) 4 V, 32 W (C) 5 V, 6 W (D) 7 V, 6 W 14. A practical DC current source provide 20 kW to a 50 W load and 20 kW to a 200 W load. The maximum power, that can drawn from it, is (A) 22.5 kW (B) 45 kW (C) 30.3 kW (D) 40 kW Statement for Q.1516: In the circuit of fig. P.1.4.1516 when R = 0 W , the current iR equals 10 A. 15. The value of R, for which it absorbs maximum power, is (A) 4 W (B) 3 W (C) 2 W (D) None of the above 16. The maximum power will be (A) 50 W (B) 100 W (C) 200 W (D) value of E is required 17. Consider a 24 V battery of internal resistance r = 4 W connected to a variable resistance RL . The rate of heat dissipated in the resistor is maximum when the current drawn from the battery is i . The current drawn form the battery will be i 2 when RL is equal to (A) 2 W (B) 4 W (C) 8 W (D) 12 W 18. i RN N, = ? (A) 2 A, 20 W (B) 2 A, -20 W (C) 0 A, 20 W (D) 0 A, -20 W 19. v RTH TH, = ? (A) 0 W (B) 1.2 W (C) 2.4 W (D) 3.6 W 20. v RTH TH, = ? (A) 8 V, 5 W (B) 8 V, 10 W (C) 4 V, 5 W (D) 4 V, 10 W 21. RTH = ? (A) 3 W (B) 1.2 W (C) 5 W (D) 10 W 22. In the circuit shown in fig. P.1.4.22 the effective resistance faced by the voltage source is (A) 4 W (B) 3 W (C) 2 W (D) 1 W Chap 1.4Network Theorems Page 35 4 W 2 W 2 W 4 W R 2 W 4 AE iR Fig. P.1.4.1516. 10 W i1 5 W 30 W20i1 RNiN, Fig. P.1.4.18 6 W i1 4 W3i1 RNiN, Fig. P1.4.19 v1 + 5 W0.1v1 4 V vTH RTH Fig. P.1.4.20 vx + 2 W 3 W 4 V vx 4 RTH Fig. P.1.4.21 4 W i i vs 4 Fig. P.1.4.22 GATE EC BY RK Kanodia www.gatehelp.com
  32. 32. 23. In the circuit of fig. P1.4.23 the value of RTH at terminal ab is (A) -3 W (B) 9 8 W (C) - 8 3 W (D) None of the above 24. RTH = ? (A) (B) 0 (C) 3 125 W (D) 125 3 W 25. In the circuit of fig. P.1.4.25, the RL will absorb maximum power if RL is equal to (A) 400 3 W (B) 2 9 kW (C) 800 3 W (D) 4 9 kW Statement for Q.2627: In the circuit shown in fig. P1.4.2627 the maximum power transfer condition is met for the load RL . 26. The value of RL will be (A) 2 W (B) 3 W (C) 1 W (D) None of the above 27. The maximum power is (A) 0.75 W (B) 1.5 W (C) 2.25 W (D) 1.125 W 28. RTH = ? (A) 100 W (B) 136.4 W (C) 200 W (D) 272.8 W 29. Consider the circuits shown in fig. P.1.4.29 Page 36 UNIT 1 Networks 0.75va va + 4 W a 8 W 9 V b Fig. P.1.4.23 + va 200 W 50 W100 W va 100 RTH Fig. P.1.4.24 40 W 100 W 200 W 3i RL 6 V i Fig. P.1.4.25 ix 3 W RL0.9 A 2 W 16 V Fig. P.1.4.2627 100 W -2ix 0.01vx ix 800 W 100 W 300 W + vx RTH Fig. P.1.4.28 2 W 6 W 6 W 6 W 2 W 2 W 12 V18 V 3 A ib ia 2 W 6 W 6 W 6 W 2 W 2 W 8 V 12 V 12 V Fig. P.1.4.29a & b GATE EC BY RK Kanodia www.gatehelp.com
  33. 33. The relation between ia and ib is (A) i ib a= + 6 (B) i ib a= + 2 (C) i ib a= 15. (D) i ib a= 30. Req = ? (A) 18 W (B) 72 13 W (C) 36 13 W (D) 9 W 31. In the lattice network the value of RL for the maximum power transfer to it is (A) 6.67 W (B) 9 W (C) 6.52 W (D) 8 W Statement for Q.3233: A circuit is shown in fig. P.1.4.3233. 32. If v vs s1 2 6= = V then the value of va is (A) 3 V (B) 4 V (C) 6 V (D) 5 V 33. If vs1 6= V and vs2 6= - V then the value of va is (A) 4 V (B) -4 V (C) 6 V (D) -6 V 34. A network N feeds a resistance R as shown in fig. P1.4.34. Let the power consumed by R be P. If an identical network is added as shown in figure, the power consumed by R will be (A) equal to P (B) less than P (C) between P and 4P (D) more than 4P 35. A certain network consists of a large number of ideal linear resistors, one of which is R and two constant ideal source. The power consumed by R is P1 when only the first source is active, and P2 when only the second source is active. If both sources are active simultaneously, then the power consumed by R is (A) P P1 2 (B) P P1 2 (C) ( )P P1 2 2 (D) ( )P P1 2 2 36. A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 W, the power dissipated by the bulb is (A) 80 W (B) 1800 W (C) 112.5 W (D) 228 W 37. The following results were obtained from measurements taken between the two terminal of a resistive network Terminal voltage 12 V 0 V Terminal current 0 A 1.5 A The Thevenin resistance of the network is (A) 16 W (B) 8 W (C) 0 (D) Chap 1.4Network Theorems Page 37 12 W 4 W 6 W 2 W 6 W 9 W 18 W Req Fig. P.1.4.30 RL 7 W 6 W 5 W 9 W Fig. P.1.4.31 12 W 1 W 3 W 3 W 1W 1 W vs2vs1 + va Fig. P.1.4.3233 N R N R N Fig. P.1.4.34 GATE EC BY RK Kanodia www.gatehelp.com
  34. 34. 38. A DC voltmeter with a sensitivity of 20 kW/V is used to find the Thevenin equivalent of a linear network. Reading on two scales are as follows (a) 0 10- V scale : 4 V (b) 0 -15 V scale : 5 V The Thevenin voltage and the Thevenin resistance of the network is (A) 16 3 V, 200 3 kW (B) 32 3 V, 1 15 MW (C) 18 V, 2 15 MW (D) 36 V, 200 3 kW 39. Consider the network shown in fig. P.1.4.39. The power absorbed by load resistance RL is shown in table : RL 10 kW 30 kW P 3.6 MW 4.8 MW The value of RL , that would absorb maximum power, is (A) 60 kW (B) 100 W (C) 300 W (D) 30 kW 40. Measurement made on terminal ab of a circuit of fig.P.1.4.40 yield the current-voltage characteristics shown in fig. P.1.4.40. The Thevenin resistance is (A) 300 W (B) -300 W (C) 100 W (D) -100 W *********** SOLUTIONS 1. (B) vTH = + = ( )( )6 6 3 6 4 V, RTH = + =( || )3 6 2 4 W 2. (A) RN = + =2 4 2 10 3 || W, v1 15 2 1 2 1 2 1 4 6= + + = V i i v sc N= = =1 2 3 A 3. (C) vTH = + = ( )( )( )2 3 1 3 3 1 V, RTH = =1 5 5 6 || W 4. (B) After killing all source equivalent resistance is R Open circuit voltage = v1 5. (D) The short circuit current across the terminal is i isc N= + = = 6 4 4 2 4 A , RN = =6 3 2|| W 6. (B) For the calculation of RTH if we kill the sources then 20 W resistance is inactive because 5 A source will be open circuit RTH = + =30 25 55 W, vTH = + =5 5 30 155 V Page 38 UNIT 1 Networks RL Linear Network + vab Fig. P.1.4.39 21-1-2-3-4 0 i(mA) 30 20 10 v Resistive Network + vab a b Fig. P.1.4.40 2 W 2 Wv1 4 W15 V isc Fig. S.1.4.2 2 W 3 W4 W6 A isc Fig. S1.4.5 GATE EC BY RK Kanodia www.gatehelp.com
  35. 35. 7. (C) After killing the source, RTH = 6 W 8. (B) After killing all source, RTH = + =3 6 8 8 6|| || W 9. (D) v voc TH= + = =2 2 4 8 V R RTH N= + = =2 3 5 W , i v R N TH TH = = 8 5 A 10. (A) If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent v1 4 1 1 12 1 2 1 1 1 1 6 1 1 2 6= + + + + + + + = V 11. (B) If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent i1 20 6 8 2 4 2 0 75= - - + + = k k k . mA 12. (B) We Thevenized the left side of xx and source transformed right side of yy v vxx TH = = + + = 4 8 8 24 1 8 1 24 5 V, RTH = + =8 16 8 6||( ) W 13. (D) Thevenin equivalent seen from terminal yy is v vyy TH = = + + = 4 24 8 8 1 24 1 8 7 V, RTH = + =( )||8 16 8 6 W 14. (A) ir r + = 50 50 20 2 k, ir r + = 200 200 20 2 k ( ) ( )r r+ = +200 4 502 2 r = 100 W i = 30 A, Pmax = = ( ) . 30 100 4 22 5 2 kW 15. (C) Thevenized the circuit across R, RTH = 2 W 16. (A) isc = 10 A, RTH = 2 W, Pmax = = 10 2 2 50 2 W Chap 1.4Network Theorems Page 39 4 V 12 V 1 W 1 W v1 + 2 W 6 W 1 W Fig. S1.4.10 2 kW 4 kWi1 2 kW 20 V 6 V 8 V Fig. S1.4.11 8 W x 16 Wx 4 V 8 V y y 8 W Fig. S1.4.12 i RLr Fig. S1.4.14 4 W 2 W 2 W 4 W 2 W Fig. S1.4.15 6 W 6 W RTH Fig. S.1.4.7 GATE EC BY RK Kanodia www.gatehelp.com
  36. 36. Now in this circuit all straight-through connection have been cut as shown in fig. S1.4.32b va = + + + = 6 2 3 2 3 1 5 ( ) V 33. (B) Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.33 va = - + = - 6 6 3 2 1 4 ( || ) V 34. (C) Let Thevenin equivalent of both network P V R R RTH TH = + 2 P V R R R V R R RTH TH TH TH = + = + 2 4 2 2 2 Thus P P P< < 4 35. (C) i P R 1 1 = and i P R 2 2 = using superposition i i i= + =1 2 P R P R 1 2 i R P P2 1 2 2 = ( ) 36. (C) r v i oc sc = = 1 2. W P = + = 24 1 2 2 2 112 5 2 2 ( . ) . W 37. (B) R v i TH oc sc = = = 12 15 8 . W 38. (A) Let 1 1 20 50 sensitivity = = k mA For 0 -10 V scale Rm = =10 20 200k kW For 0 -50 V scale Rm = =50 20 1k MW For 4 V reading i = = 4 10 50 20 mA v R RTH TH TH= + = +20 20 200 4 20m m mk ...(i) For 5 V reading i = = 5 50 50 5m mA v R RTH TH TH= + = +5 5 1 5 5m m mM ...(ii) Solving (i) and (ii) vTH = 16 3 V, RTH = 200 3 kW 39. (D) v10 10 3 6 6k k m= =. v30 30 4 8 12k k m= =. V 6 10 10 = + R v TH TH 10 6 60v RTH TH= + 12 30 30 = + v R TH TH 5 2 60v RTH TH= + RTH = 30 kW 40. (D) At v = 0 , isc = 30 mA At i = 0, voc = - 3 V R v i TH oc sc = = - = - 3 30 100 m W ************ Page 42 UNIT 1 Networks vTHvTHvTH RTH R R RTH RTH Fig. S1.4.34 6 V + va 2 W 6 W 1 W 3 W Fig. S1.4.33 6 V + va 2 W 6 W 1 W 3 W Fig. S.1.4.32b GATE EC BY RK Kanodia www.gatehelp.com
  37. 37. 1. The natural response of an RLC circuit is described by the differential equation d v dt dv dt v 2 2 2 0+ + = , v( )0 10= , dv dt ( )0 0= . The v t( ) is (A) 10 1( )+ - t e t V (B) 10 1( )- - t e t V (C) 10e t- V (D) 10te t- V 2. The differential equation for the circuit shown in fig. P1.6.2. is (A) + + =v t v t v t v ts( ) ( ) . ( ) ( )3000 102 10 108 8 (B) + + =v t v t v t v ts( ) ( ) . ( ) ( )1000 102 10 108 8 (C) + + = v t v t v t v ts ( ) ( ) . ( ) ( ) 10 2 10 1028 5 (D) + + = v t v t v t v ts ( ) ( ) . ( ) ( ) 10 2 10 1988 5 3. The differential equation for the circuit shown in fig. P1.6.3 is (A) + + =i t i t i t i tL L L s( ) ( ) ( ) ( )1100 11 10 108 8 (B) + + =i t i t i t i tL L L s( ) ( ) ( ) ( )1100 11 10 108 8 (C) + + = i t i t i t i tL L L s ( ) . ( ) . ( ) ( ) 10 11 10 118 4 (D) + + = i t i t i t i tL L L s ( ) ( ) ( ) ( ) 10 11 10 118 4 4. In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial condition are v( )0 6= V and dv dt( )0 3000= - V s. The v t( ) for t > 0 is (A) - +- - 2 8100 400 e et t V (B) 6 8100 400 e et t- - + V (C) 6 8100 400 e et t- - - V (D) None of the above 5. The circuit shown in fig. P1.6.5 has been open for a long time before closing at t = 0. The initial condition is v( )0 2= V. The v t( ) for t > is (A) 5 7 3 e et t- - - V (B) 7 5 3 e et t- - - V (C) - +- - e et t 3 3 V (D) 3 3 e et t- - - V CHAPTER 1.6 THE RLC CIRCUITS Page 54 2 W vs 10 Fm100 W 1 mH v Fig. P1.6.2 10 Fm 10 W 100 W iL is Fig. P.1.6.3 vs 25 Fm 1 H 80 W vC + Fig. P1.6.4 t=0 W1 H vC + F4 3 3 1 Fig. P.1.6.5 GATE EC BY RK Kanodia www.gatehelp.com
  38. 38. Statement for Q.67: Circuit is shown in fig. P.1.6. Initial conditions are i i1 20 0 11( ) ( )= = A 6. i1 1( )s = ? (A) 0.78 A (B) 1.46 A (C) 2.56 A (D) 3.62 A 7. i2 1( )s = ? (A) 0.78 A (B) 1.46 A (C) 2.56 A (D) 3.62 A 8. v tC( ) = ? for t > 0 (A) 4 1000 2000 e et t- - - V (B) ( )3 6000 2000 + - t e t V (C) 2 1000 2000 e et t- - + V (D) ( )3 6000 2000 - - t e t V 9. The circuit shown in fig. P1.6.9 is in steady state with switch open. At t = 0 the switch is closed. The output voltage v tC( ) for t > 0 is (A) - +- - 9 12400 300 e et t (B) e t tt- +400 3 300 4 300[ cos sin ] (C) e t tt- +300 3 400 4 300[ cos sin ] (D) e t tt- +300 3 400 2 25 300[ cos . sin ] 10. The switch of the circuit shown in fig. P1.6.10 is opened at t = 0 after long time. The v t( ) , for t > 0 is (A) 4 22 e tt- sin V (B) - - 4 22 e tt sin V (C) 4 22 e tt- cos V (D) - - 4 22 e tt cos V 11. In the circuit of fig. P1.6.23 the switch is opened at t = 0 after long time. The current i tL ( ) for t > 0 is (A) e t tt- +2 2 4( cos sin ) A (B) e t tt- -2 3 4( sin cos ) A (C) e t tt- - +2 4 2( sin cos ) A (D)e t tt- -2 2 4( sin cos ) A Statement for Q.1214: In the circuit shown in fig. P1.6.1214 all initial condition are zero. 12. If i ts( ) = 1 A, then the inductor current i tL ( ) is (A) 1 A (B) t A (C) t + 1 A (D) 0 A 13. If i t ts( ) .= 0 5 A, then i tL ( ) is (A) 0 5 3 25 10 3 . .t + - A (B) 2 3250t - A (C) 0 5 0 25 10 3 . .t - - A (D) 2 3250t + A 14. If i t es t ( ) = - 2 250 A then i tL ( ) is (A) 4000 3 250 te t- A (B) 4000 3 250 e t- A (C) 200 7 250 e t- A (D) 200 7 250 te t- A Chap 1.6The RLC Circuits Page 55 3 H1 W 2 H 2 W i1 i2 Fig. P1.6.67 30 (- ) mAu t 10 Fm 25 mH 100 W vC + Fig. P1.6.8 0.8 H 500 W 9 V 250 W 5 Fm vC + t=0 Fig. P1.6.9 3 W 1 W H6 V t=0 F2 1 4 1 vC + Fig. P1.6.10 2 W t=0 4 H 7 A4 W 8 W iL F4 1 Fig. P1.6.11 W 100 65 i u ts ( ) A 1 mF iL vL + 10 mH Fig. P1.5.12-14 GATE EC BY RK Kanodia www.gatehelp.com
  39. 39. 15. The forced response for the capacitor voltage v tf ( ) is (A) 0 2 117 10 3 . .t + - V (B) 0 2 117 10 3 . .t - - V (C) 117 10 0 23 . . -- t V (D) 117 10 0 23 . . +- t V 16. For a RLC series circuit R L= =20 0 6W , . H, the value of C will be [CD =critically damped, OD =over damped, UD =under damped]. CD OD UD (A) C = 6 mF C >6 mF C 6 mF (C) C >6 mF C = 6 mF C < 6 mF (D) C 6 mF 17. The circuit shown in fig. P1.6.17 is critically damped. The value of R is (A) 40 W (B) 60 W (C) 120 W (D) 180 W 18. The step response of an RLC series circuit is given by d i t dt di t dt i t 2 2 5 10 ( ) ( ) ( )+ + = , i( )0 2+ = , di dt ( )0 4 + = . The i t( ) is (A) 1 4+ - e tt cos A (B) 4 2 4- - e tt cos A (C) 2 4+ - e tt sin A (D) 10 4+ - e tt sin A 19. In the circuit shown in fig. P 1.5.19 v t( ) for t > 0 is (A) 50 46 5 3 62 3 4 - + - ( . sin cos )t t e t V (B) 50 46 5 3 62 3 4 + + - ( . sin cos )t t e t V (C) 50 62 4 46 5 4 3 + + - ( cos . sin )t t e t V (D) 50 62 4 46 5 4 3 - + - ( cos . sin )t t e t V 20. In the circuit of fig. P1.6.20 the switch is closed at t = 0 after long time. The current i t( ) for t > 0 is (A) -10 8sin t A (B) 10 8sin t A (C) -10 8cos t A (D) 10 8cos t A 21. In the circuit of fig. P1.6.21 switch is moved from 8 V to 12 V at t = 0. The voltage v t( ) for t > 0 is (A) 12 4 2 2 2- + - ( cos sin )t t e t V (B) 12 4 2 8 2- + - ( cos sin )t t e t V (C) 12 4 2 8 2+ + - ( cos sin )t t e t V (D) 12 4 2 2 2+ + - ( cos sin )t t e t V Page 56 UNIT 1 Networks 120 WR 10 mF 4 H Fig. P1.6.17 2 (- ) Au t 1 H 0.04 F 4 W 2 W 50 ( ) Vu t vC + Fig. P1.6.19 5 W iL 20 V t=0 H 1 4 F 1 16 vC + Fig. P1.6.20 2 W 12 V 1 H 8 V F 1 6 vC + t=0 Fig. P1.6.21 vx + iL 20 mH50 Wavx 100 W Fig. P1.6.15 GATE EC BY RK Kanodia www.gatehelp.com
  40. 40. 22. In the circuit of fig. P1.5.22 the voltage v t( ) is (A) 40 20 0 6 15 0 6 0 8 - + - ( cos . sin . ) . t t e t V (B) 35 15 0 6 20 0 6 0 8 + + - ( cos . sin . ) . t t e t V (C) 35 15 0 6 20 0 6 0 8 - + - ( cos . sin . ) . t t e t V (D) 35 15 0 6 0 8 - - cos . . t e t V 23. In the circuit of fig. P1.6.23 the switch is opened at t = 0 after long time. The current i t( ) for t > 0 is (A) e et t- - +2 306 0 869. . A (B) - +- - e et t2 306 0 869 2. . A (C) e et t- - +4 431 0 903. . A (D) 2 4 431 0 903 e et t- - -. . A 24. In the circuit of fig. P1.6.24 switch is moved from position a to b at t = 0. The i tL ( ) for t > 0 is (A) ( )4 6 4 - t e t A (B) ( )3 6 4 - - t e t A (C) ( )3 9 5 - - t e t A (D) ( )3 8 5 - - t e t A 25. In the circuit shown in fig. P1.6.25 a steady state has been established before switch closed. The i t( ) for t > 0 is (A) 0 73 4 582 . sin .e tt- A (B) 0 89 6 382 . sin .e tt- A (C) 0 73 4 584 . sin .e tt- A (D) 0 89 6 384 . sin .e tt- A 26. The switch is closed after long time in the circuit of fig. P1.6.26. The v t( ) for t > 0 is (A) - + - 8 6 43 e tt sin V (B) - + - 12 4 43 e tt cos V (C) - + + - 12 4 4 3 4 3 ( cos sin )t t e t V (D) - + + - 12 4 4 6 4 3 ( cos sin )t t e t V 27. i t( ) = ? (A) 6 6 500 6 5000 50 - + - ( cos sin )t t e t mA (B) 8 8 500 0 06 5000 50 - + - ( cos . sin )t t e t mA (C) 6 6 5000 0 06 5000 50 - + - ( cos . sin )t t e t mA (D) 6 500050 e tt- sin mA Chap 1.6The RLC Circuits Page 57 10 W 5 W 10 W 2 A H 3 4 F 1 3 t=0 Fig. P1.6.23 14 W 0.02 F t=0 b a 2 W 12 V 4 A 6 W iL 2 H Fig. P1.6.24 5 W 20 W 1 H 5 W i 100 V F 1 25 t=0 Fig. P1.6.25 1 H 6 W 1 W 4 V t=0 2 A F 1 25 vC + Fig. P1.6.26 2 kW 5 Fm i 12 ( ) Vu t 8 mH Fig. P1.6.27 0.2 F 1 W5 H 5 W3 ( ) Au t 2 W 20 V vC + Fig. P1.6.22 GATE EC BY RK Kanodia www.gatehelp.com
  41. 41. 28. In the circuit of fig. P1.6.28 i( )0 1= A and v( )0 0= . The current i t( ) for t > 0 is (A) 4 6 38 0 5 + - . . e At (B) 4 6 38 0 5 - - . . e t A (C) 4 3 1 32 113 1 32 0 5 + + - ( cos . . sin . ) . t t e t A (D) 4 3 1 32 113 1 32 0 5 - + - ( cos . . sin . ) . t t e t A 29. In the circuit of fig. P1.6.29 a steady state has been established before switch closed. The v to( ) for t > 0 is (A) 100 10 te t- V (B) 200 10 te t- V (C) 400 50 te t- V (D) 800 50 te t- V 30. In the circuit of fig. P1.6.30 a steady state has been established before switch closed. The i t( ) for t > 0 is (A) 2 22 e tt- sin A (B) - - e tt2 2sin A (C) - - - 2 1 2 ( )t e t A (D) 2 1 2 ( )- - t e t A 31. In the circuit of fig. P1.6.31 a steady state has been established. The i t( ) for t > 0 is (A) 9 2 810 2 5 + -- - e et t. A (B) 9 8 210 2 5 - + - e et t. A (C) 9 2 10 10 2 5 + + - ( cos sin ) . t t e t A (D) 9 10 2 10 2 5 + + - (cos sin ) . t t e t A *************** SOLUTIONS 1. (A) s s2 2 1 0+ + = s = - -1 1, , v t A A t e t ( ) ( )= + - 1 2 v( )0 10= V, dv dt A A ( )0 0 1 1 2= = - + A A1 2 10= = 2. (A) i v dv dt L = + - 100 10 10 6 v i di dt vs L L = + +- 2 10 3 = + + + - - - - 2 100 10 10 10 1 100 10 106 3 6 2 v dv dt dv dt d vt dt v2 + 10 3000 1028 v t v t v t v ts( ) ( ) ( ) . ( )= + + 3. (C) i v i dv dt s C L C = + + 100 10m v i di dt C L L = + - 10 10 3 i i di dt i d dt i di dt s L L L L L = + + + +- - - 0 1 10 10 10 105 5 3 . ( ) = + + + +- - - 0 1 10 10 105 4 8 2 2 . i di dt i di dt d i dt L L L L L + + = i t i t i t i tL L L s ( ) . ( ) . ( ) ( ) 10 11 10 118 4 4. (A) v dv dt v v dts 80 25 0+ + - =m ( ) d v dt dv dt 2 2 500 40000 0+ + = s s2 500 40000 0+ + = s = - -100 400, , v t Ae Bet t ( ) = +- -100 400 A B+ = 6, - - = -100 400 3000A B B = 8, A = -2 5. (C) The characteristic equation is s s RC LC 2 1 0+ + = After putting the values, s s2 4 3 0+ + = v t Ae Bet t ( ) = +- -3 , Page 58 UNIT 1 Networks 10 W 10 mF5 W3 A vo + 1 H t=0 Fig. P1.6.29 i 1 H 2 W 6 V 1 W F 1 4 t=0 Fig. P1.6.30 40 W 4 H10 W 6 ( ) Au t3 A i 10 mF Fig. P1.6.31 0.5 F2 W1 H i 4 ( ) Au t Fig. P1.6.28 iL 2 W vs 10 Fm100 W 1 mH v Fig. S1.6.2 GATE EC BY RK Kanodia www.gatehelp.com
  42. 42. v( )0 2+ = V A B+ = 2 iL ( )0 0+ = iR( )0 2 3 4 8 3 = = , - = + C dv dt ( )0 8 3 dv dt ( )0 8 + = - , - - = -A B3 8, B A= = -3 1, 6. (D) i di dt di dt 1 1 2 5 3 0+ - = , 2 3 3 02 2 1 i di dt di dt + - = ( )1 5 3 01 2+ - =s i si , - + + =3 2 3 01 2si s i( ) ( ) ( )( ) 1 5 3 3 2 3 01 1 + - + =s i s s i s 6 13 2 02 s s+ + = s = - - 1 6 2, i A e Be t t 1 1 6 2 = + - - , i A B( )0 11= + = In differential equation putting t = 0 and solving di dt di dt 1 20 33 2 0 143 6 ( ) , ( )+ + = - = - - - = - A B 6 2 33 2 , A B= =3 8, , i e e t t 1 6 2 3 8= + - - , i e e1 1 6 2 1 3 8 3 62( ) .s = + = - - A 7. (A) i Ce De t t 2 6 2 = + - - i C D2 0 11( ) = = + , di dt C D2 0 143 6 6 2 ( ) = - = - - C = -1 and D = 12 i e e t t 2 6 2 12= - + - - A, i e e2 1 6 2 1 12 0 78( ) .s = + = - - A 8. (B) vC( )0 30 100 3+ = =m V C dv dt i i C dv dt C L L C( ) ( ) ( ) ( )0 0 0 0 0- - + + = = = = s s2 3 3 6 100 25 10 1 25 10 10 10 + + - - - s = -2000, -2000 v t A A t eC t ( ) ( )= + - 1 2 2000 dv t dt A e A A t eC t t( ) ( ) ( )= + + -- - 2 2000 1 2 2000 2000 v AC( )0 31 + = = , dv dt AC( )0 2000 3 02= - = A2 6000= 9. (B) v iC L( ) , ( )0 3 0 12+ + = = -V mA v i dv dt C L C 250 5 10 06 + + =- 3 250 12 5 10 0 06 - + =- + m dv dt C( ) = + dv dt C( )0 0 s s2 6 6 250 5 10 1 0 8 5 10 0+ + =- - . + + =s s2 4 800 25 10 0 = - s j400 300 v t e A t A tC t ( ) ( cos sin )= +-400 1 2300 300 A1 3= , dv dt A A AC( ) , 0 400 300 41 2 2= - + = 10. (B) v( )0 0+ = , iL ( )0 2+ = A, 1 4 0 2 dv xdt C( )+ = - s s2 4 8 0+ + = s j= - 2 2 v t e A t A tC t ( ) ( cos sin )= +-2 1 22 2 A1 0= , dv dt AC ( ) ( ) ( ) 0 8 2 0 0 0 2 2 + = - = - + + + , A2 4= - 11. (D) iL ( )0 4+ = - , vC( )0 8+ = V 4 0 8 4 8 di dt L ( ) ( ) + = - - di dt L ( )0 10 + = s v v iC C L 4 1 2 0+ + = , v si iC L L= +4 8 s i si s jL L 2 4 5 0 2+ + = = - , i t e A t A tL t ( ) ( cos sin )= +-2 1 2 A1 4= - , di dt A AL ( ) ( ) 0 10 2 01 2 + = = - + + , A2 2= 12. (A) i v dv dt is L= + +- 100 65 10 3 , v di dt L = - 10 10 3 i di dt d i dt is L L L= + + =- - -65 100 10 10 10 10 10 03 3 3 2 ( ) ( ) d i dt di dt i iL L L s 2 5 5 650 10 10+ + = Trying i t BL ( ) = 0 0 10 105 5 + + =B , B = 1, iL = 1 A 13. (A) Trying i t At BL ( ) = + , 0 650 10 10 0 55 5 + + + =A At B t( ) ( . ), A = 0 5. Chap 1.6The RLC Circuits Page 59 8 W2 W iL 4 H F4 1 vC + Fig. S1.6.11 GATE EC BY RK Kanodia www.gatehelp.com
  43. 43. di dt A B ( ) . . 0 16 3 4 431 0 903 + = - = - - A B= =1 1, 24. (C) vC( )0 0= , iL ( )0 4 6 6 2 3= + = 0 02 0 0 3. ( ) ( ) dv dt iC L= = dv dt C( )0 150= a = + = 6 14 2 2 5, wo = = 1 2 0 02 5 . a wo= critically damped v t A Bt e t ( ) ( )= + + - 12 5 0 12 150 5= + = - +A A B, A B= - =12 90, v t t e t ( ) ( )= + - - 12 90 12 5 i t e t e t eL t t t ( ) . ( ) ( ) . ( ) ( )= - - + = -- - - 0 02 5 90 12 0 02 90 3 95 5 5 25. (A) v( )0 100 5 5 5 20 50 3 + = + + = , iL ( )0 0+ = if = 0 A di dt L ( )0 20 50 3 10 3 + = - = a = = 4 2 1 2, wo = = 1 1 1 25 5 s j= - - = - 2 4 25 2 4 58. i t A t B t e t ( ) ( cos . sin . )= + - 4 58 4 58 2 26.(A) iL ( )0 0+ = , vL ( )0 4 12 8+ = - = - 1 25 0 0 0 dv dt iL L ( ) ( ) + + = = a = = 6 2 3, Wo = = 1 1 1 25 5 / b = - - = - 3 9 25 3 4j v t A t B t e t 1 3 12 4 4( ) ( cos sin )= - + + - v AL ( )0 8 12= - = + , =A 4 dv dt A BL ( )0 0 3 4= = - + , =B 3 27. (C) a = = = 1 2 1 2 2 54 50 RC k W LC m o = = = 1 1 8 5 5000 m a < Wo, underdamped response. s j= - - = - 50 50 5000 50 50002 2 i t A t B t e t ( ) ( cos sin )= + + - 6 5000 5000 50 mA i A( )0 6 6= = + , = -A 6 di dt A B ( )0 50 5000 0= - + = , B = -0 06. 28.(D) i( )0 1+ = A, v Ldi dt ( ) ( ) 0 0+ + = a = = 1 2 2 0 5 0 5 . . , Wo = - = 1 1 0 5 2 . s j= - - = 0 5 0 5 2 0 5 1 3232 . . . . i t A t B t e t( ) ( cos . sin . ) . = + + - 4 1 32 1 32 0 5 1 4= + A, = -A 3 di dt A B ( ) . . 0 0 0 5 1 32= = + , B = -113. 29. (B) Vo( )0 0+ = ,iL ( )0 1+ = A di dt vL ( ) ( ) 0 0 01 + = = a = = 1 2 5 0 01 10 . , Wo = = 1 1 0 01 10 . a = Wo, so critically damped response s = - -10 10, i t A Bt e t ( ) ( )= + - 3 10 , i A( )0 1 3= = + di dt A B ( )0 10 + = - + i t t eL t ( ) ( )= - + - 3 2 20 10 , v Ldi t dt teo L t = = -( ) 200 10 30. (C) i( )0 6 1 2 2+ = - + = - A, v di dt c( ) ( ) 0 2 1 2 0+ + = = = a = = = 1 2 1 2 1 0 25 2 RC . , W LC o = = 1 2 a = Wo, critically damped response s = - -2 2, i t A Bt e t ( ) ( )= + -2 , A = -2 di t dt Bt e B et t( ) ( ) ( ) ( )= - + - + + - 2 2 02 2 At t = 0, = -B 2 31. (A) i( )0 3+ = A, vC( )0 0+ = V = + 4 0di dt ( ) is = 9 A, R = =10 40 8|| W a = = = 1 2 1 2 8 0 01 6 25 RC . . W LC o = = = 1 1 4 0 01 5 . a > Wo, so overdamped response s = - - = - -6 25 6 25 25 10 2 52 . . , . i t Ae Bet t ( ) . = + +- - 9 10 2 5 3 9= + +A B, 0 10 2 5= - -A B. On solving, A = 2, B = -8 ************ Chap 1.6The RLC Circuits Page 61 GATE EC BY RK Kanodia www.gatehelp.com
  44. 44. 1. i t( ) = ? (A) 20 300 68 2cos ( . )t + A (B) 20 300 68 2cos( . )t - A (C) 2 48 300 68 2. cos( . )t + A (D) 2 48 300 68 2. cos( . )t - A 2. v tC( ) ?= (A) 0 89 10 63 433 . cos ( . )t - V (B) 0 89 10 63 433 . cos ( . )t + V (C) 0 45 10 26 573 . cos ( . )t + V (D) 0 45 10 26 573 . cos ( . )t - V 3. v tC( ) = ? (A) 1 2 2 45cos ( )t - V (B) 1 2 2 45cos ( )t + V (C) 1 2 2 45sin ( )t - V (D) 1 2 2 45sin ( )t + V 4. v tC( ) ?= (A) 2 25 5 150. cos ( )t + V (B) 2 25 5 150. cos ( )t - V (C) 2 25 5 140 71. cos ( . )t + V (D) 2 25 5 140 71. cos ( . )t - V 5. i t( ) = ? (A) 2 2 5 77sin ( . )t + A (B) cos ( . )2 84 23t - A (C) 2 2 5 77sin ( . )t - A (D) cos ( . )2 84 23t + A CHAPTER 1.7 SINUSOIDAL STEADY STATE ANALYSIS Page 62 i 25 mH20cos 300 Vt 3 W ~ Fig. P1.7.1 2 Wcos 10 A 3 t 1 mF~ vC + Fig. P1.7.2 5 W 0.1 F vC + cos 2 Vt ~ Fig. P1.7.3 3 H 50 mF 9 W vC + 8cos 5 Vt ~ Fig. P1.7.4 1 W 4 W 0.25 F 4 H i 10cos 2 Vt ~ Fig. P1.7.5 GATE EC BY RK Kanodia www.gatehelp.com
  45. 45. 13. In the bridge shown in fig. P1.7.13, Z1 300= W, Z j2 300 600= -( ) W, Z j3 200 100= +( )W. The Z4 at balance is (A) 400 300+ j W (B) 400 300- j W (C) j100 W (D) - j900 W 14. In a two element series circuit, the applied voltage and the resulting current are v t t( ) sin ( )= +60 66 103 V, i t t( ) . sin ( . )= + 2 3 10 68 33 A. The nature of the elements would be (A) R C- (B) L C- (C) R L- (D) R R- 15. Vo = ? (A) 223 56 - V (B) 223 56 V (C) 124 154 - V (D) 124 154 V 16. v to( ) = ? (A) 315 112. cos ( )t + V (B) 43 2 23. cos ( )t + V (C) 315 112. cos ( )t - V (D) 43 2 23. cos ( )t - V Statement for Q.17-18: The circuit is as shown in fig. P1.7.17-18 17. i t1( ) = ? (A) 2 36 4 4107. cos ( . )t - A (B) 2 36 4 4107. cos ( . )t + A (C) 1 37 4 4107. cos ( . )t - A (D) 2 36 4 4107. cos ( . )t + A 18. i t2( ) = ? (A) 2 04 4 92 13. sin ( . )t + A (B) - + 2 04 4 2 13. sin ( . )t A (C) 2 04 4 2 13. cos ( . )t + A (D) - + 2 04 4 92 13. cos ( . )t A 19. Ix = ? (A) 394 46 28. . A (B) 4 62 97 38. . A (C) 7 42 92 49. . A (D) 6 78 49 27. . A 20. Vx = ? (A) 29 11 166. V (B) 29 11 166. - V (C) 43 24 124. V (D) 43 24 124. - V Page 64 UNIT 1 Networks 4 W 0.5Ix Ix j3 W10 30 V o - 2j W ~ Fig. P1.7.19 10sin ( +30 ) Vt o 1 H 3 W 1 F 20cos ( -45 ) Vt o ~ ~vo + Fig. P1.7.16 1 H 1 W 1 W 1 H 5cos 4 Vt 10cos (4 -30 ) Vt o i1 i2 1 W 1 F ~ ~ Fig. P1.7.1718 j20 40 W 50 W120 -15 V o - 30j 6 30 A o ~ ~ Vo Fig. P1.7.15 3 0 A o Vx + 4Vx 20 W j10 W20 W ~ Fig. P1.7.20 Z 1 Z 2 Z 4 Z 3 ~ Fig. P1.7.13 GATE EC BY RK Kanodia www.gatehelp.com
  46. 46. Statement for Q.2732: Determine the complex power for hte given values in question. 27. P = 269 W, Q = 150 VAR (capacitive) (A) 150 269- j VA (B) 150 269+ j VA (C) 269 150- j VA (D) 269 150+ j VA 28. Q = 2000 VAR, pf = 0 9. (leading) (A) 4129 8 2000. + j VA (B) 2000 4129 8+ j . VA (C) 2000 4129 8- j . . VA (D) 4129 8 2000. - j VA 29. S = 60 VA, Q = 45 VAR (inductive) (A) 39 69 45. + j VA (B) 39 69 45. - j VA (C) 45 39 69+ j . VA (D) 45 39 69- j . VA 30. Vrms = 220 V, P = 1 kW, | |Z = 40 W (inductive) (A) 1000 68125- j . VA (B) 1000 68125+ j . VA (C) 68125 1000. + j VA (D) 68125 1000. - j VA 31. Vrms = 21 20 V, Vrms = 21 20 V, Irms = - 8 5 50. A (A) 154 6 89 3. .+ j VA (B) 154 6 89 3. .- j VA (C) 61 167 7+ j . VA (D) 61 167 7- j . VA 32. Vrms = 120 30 V, Z j= +40 80 W (A) 72 144+ j VA (B) 72 144- j VA (C) 144 72+ j VA (D) 144 72- j VA 33. Vo = ? (A) 7 1 32 29. . kV (B) 42 59 32 29. . kV (C) 38 49 24 39. . kV (D) 38 49 32 29. . kV 34. A relay coil is connected to a 210 V, 50 Hz supply. If it has resistance of 30 W and an inductance of 0.5 H, the apparent power is (A) 30 VA (B) 275.6 VA (C) 157 VA (D) 187 VA 35. In the circuit shown in fig. P1.7.35 power factor is (A) 56.31 (leading) (B) 56.31 (lagging) (C) 0.555 (lagging) (D) 0.555 (leading) 36. The power factor seen by the voltage source is (A) 0.8 (leading) (B) 0.8 (lagging) (C) 36.9 (leading) (D) 39.6 (lagging) 37. The average power supplied by the dependent source is (A) 96 W (B) -96 W (C) 92 W (D) -192 W 38. In the circuit of fig. P1.7.38 the maximum power absorbed by ZL is (A) 180 W (B) 90 W (C) 140 W (D) 700 W Page 66 UNIT 1 Networks v1+ 1 W4 W v110cos 2 Vt ~ F3 1 4 3 Fig. P1.7.36 j1.92Ix 4.8 W 8 W1.6Ix2 90 A o ~ Fig. P1.7.37 6 0 A o VO + ~ 20 kW 0.8 pf lagging 16 kW 0.9 pf lagging Fig. P1.7.33 4 W - 2j - 2j j5 Fig. P1.7.35 10 W j15 - 10j120 0 V o ~ ZL Fig. P1.7.38 GATE EC BY RK Kanodia www.gatehelp.com
  47. 47. 39. The value of the load impedance, that would absorbs the maximum average power is (A) 12 8 49 6. .- j W (B) 12 8 49 6. .+ j W (C) 339 86 3. .- j W (D) 339 86 3. .+ j W Statement for Q.4041: In a balanced Yconnected three phase generator Vab = 400 Vrms 40. If phase sequence is abc then phase voltage V V Va b c, , and are respectively (A) 231 0 231 120 231 240 , , (B) 231 30 231 150 231 90 - - , , (C) 231 30 231 150 231 90 - , , (D) 231 60 231 180 231 60 - , , 41. If phase sequence is acb then phase voltage are (A) 231 0 231 120 231 240 , , (B) 231 30 231 150 231 90 - - , , (C) 231 30 231 150 231 90 - , , (D) 231 60 231 180 231 60 - , , 42. A balanced three-phase Y-connected load has one phase voltage Vc = 277 45 V. The phase sequence is abc. The line to line voltage VAB is (A) 480 45 V (B) 480 45 - V (C) 339 45 V (D) 339 45 - V 43. A three-phase circuit has two parallel balanced D loads, one of the 6 W resistor and one of 12 W resistors. The magnitude of the total line current, when the line-to-line voltage is 480 Vrms , is (A) 120 Arms (B) 360 Arms (C) 208 Arms (D) 470 Arms 44. In a balanced three-phase system, the source has an abc phase sequence and is connected in delta. There are two parallel Y-connected load. The phase impedance of load 1 and load 2 is 4 4+ j W and 10 4+ j W respectively. The line impedance connecting the source to load is 0 3 0 2. .+ j W. If the current in a phase of load 1 is I = 10 20 Arms, the current in source in ab branch is (A) 15 122 - Arms (B) 8 67 122. - Arms (C) 15 27 9 . Arms (D) 8 67 57 9. . - Arms 45. An abc phase sequence 3-phase balanced Y-connected source supplies power to a balanced D connected load. The impedance per phase in the load is 10 8+ j W. If the line current in a phase is IaA = - 28 10 28 66. . Arms and the line impedance is zero, the load voltage VAB is (A) 207 8 140. - Vrms (B) 148 3 40. Vrms (C) 148 3 40. - Vrms (D) 207 8 40. Vrms 46. The magnitude of the complex power supplied by a 3-phase balanced Y-Y system is 3600 VA. The line voltage is 208 Vrms . If the line impedance is negligible and the power factor angle of the load is 25, the load impedance is (A) 5 07 10 88. .+ j W (B) 10 88 5 07. .+ j W (C) 432 14 6. .+ j W (D) 14 6 432. .+ j W *********** Chap 1.7Sinusoidal Steady State Analysis Page 67 j100 3 20 A o 80 W - 40j ~ ZL Fig. P1.7.39 GATE EC BY RK Kanodia www.gatehelp.com
  48. 48. SOLUTIONS 1. (D) Z j j= + = + = 3 25 300 3 7 5 8 08 68 2( )( ) . . .m W I = = - 20 0 8 08 68 2 2 48 68 2 . . . . A i t t( ) . cos ( . )= - 2 48 300 68 2 A 2. (A) Y j j= + = + = 1 2 1 10 0 5 112 63 433 ( )( ) . . .m VC = = - ( ) . . . . 1 0 112 63 43 0 89 63 43 V v t tC( ) . cos ( . )= - 0 89 10 63 433 V 3. (A) Z j j= + - = - = - 5 0 1 2 5 5 5 5 45 ( . )( ) VC = - - = - ( ) ( )1 0 5 90 5 2 45 1 2 45 V v tC( ) = 1 2 2 45cos ( )t - V 4. (D) Z j j j= + + - = +9 3 5 50 5 9 11( )( ) ( ) ( )m = Z 14 21 50 71. . W VC = = ( )( ) . . . . 8 0 4 90 14 21 50 71 2 25 140 71 - V v t tC( ) . cos ( . )= - 2 25 5 140 71 V 5. (B) V j j j a = + - + + = + 10 0 1 1 1 1 2 1 4 8 10 0 105 0 4. . V I V j j a = + = + = - 4 8 10 0 1 10 1 84 23. A i t t( ) cos ( . )= - 2 84 23 A 6. (D) w p p= = 2 10 10 2 103 4 Y j j = + - +( )( ) ( )( ) 1 2 10 160 2 10 1 36 4 4 m p m p = -0 0278 0 0366. .j S Z Y j= = + 1 1316 17 33. . W 7. (C) Z j j= - + w m w ( ) ||( ( ) ) 22 6 27m = - + + - -j j j 10 22 6 27 10 6 27 10 22 6 3 6 w w w w ( ) ( )m = - + - 27 10 22 6 10 22 6 27 10 22 3 6 6 2 j j w w w m - - - = j j36 10 22 27 10 22 27 10 22 0 6 3 6 2 w w w m =w 1278 f = w p2 Hz = 1278 2 203 p = Hz 8. (C) Vs = 477 68. V, V2 7 51= 35. V V Vs1 2 7 68 7 51 159= - = 47 - 35 = 125. . . 9. (B) vin = + - =3 14 10 52 2 ( ) 10. (C) I1 744 118= - mA, I2 540 100= mA I I I= + = - + 1 2 744 118 540 5 100. = - 460 164 i t t( ) cos ( )= - 460 3 164 mA 11. (A) 2 4 20 5 10 45 = - + - 0 + V j V j C C ( )( )( ) ( )1 4 10 5 10 5 4 8+ - + = + - +j j j V j j jC - = +60 100 10j V jC( ) VC = - 64.711 6. 12. (D) X X XL C= + = 0 So reactive power drawn from the source is zero. 13. (B) Z Z Z Z1 4 3 2= 300 300 600 200 1004Z j j= - +( )( ) = -Z j4 400 300 14. (A) R C- causes a positive phase shift in voltage Z Z= | | q , - < 0 is (A) 9 t V (B) 9e t- V (C) 9 V (D) 0 V 15. A unit step current of 1 A is applied to a network whose driving point impedance is Z s V s I s s s ( ) ( ) ( ) ( ) ( ) = = + + 3 2 2 The steady state and initial values of the voltage developed across the source would be respectively (A) 3 4 V, I V (B) 1 4 V, 3 4 V (C) 3 4 V, 0 V (D) 1 V, 3 4 V 16. In the circuit of Fig. P1.8.16 i( )0 1= A, vC( )0 8= V and v e u tt 1 2 10 2 4 = - ( ). The i t( ) is (A) 1 15 10 2 10 4 10 10 3 22 4 4 4 [ ] ( )e e e u tt t t- - - - - A (B) 1 15 10 2 10 4 10 10 3 22 4 4 4 [ ] ( )- + +- - - e e e u tt t t A (C) 1 3 10 2 10 4 10 10 3 22 4 4 4 [ ] ( )e e e u tt t t- - - + + A (D) 1 3 10 2 10 4 10 10 3 22 4 4 4 [ ] ( )- + -- - - e e e u tt t t A 17. In the circuit shown in Fig. P1.8.18 v( )0 8- = V and i t tin ( ) ( )= 4d . The v tC( ) for t 0 is (A) 164e t- V (B) 208e t- V (C) 208 1 3 ( )- - e t V (D) 164 3 e t- V Page 74 UNIT 1 Networks A 2 W Vo( )s + - 2Vo( )s a b s 1 ( +1)s 4 Fig. P1.8.12 1 F vC2 + - 1 FvC1 + - t=0 Fig. P1.8.13 10 V S1 5 V Va 1 V S2 5 V2 F + - 6 V +- 3 F 4 F + - Fig. P1.8.14 50 W 1m H v1 2.5 Fm i vC + Fig. P1.8.16 iin 20 mF50 W vC + Fig. P1.8.17 GATE EC BY RK Kanodia www.gatehelp.com
  49. 55. 18. The driving point impedance Z s( ) of a network has the pole zero location as shown in Fig. P1.8.18. If Z( )0 3= , the Z s( ) is (A) 4 3 12 ( )s s s + + + (B) 2 3 2 22 ( )s s s + + + (C) 2 3 2 22 ( )s s s + + + (D) 4 3 22 ( )s s s + + + Statement for Q.19-21: The circuit is as shown in the fig. P1.8.1921. All initial conditions are zero. 19. I s I s o in ( ) ( ) = ? (A) ( )s s + 1 2 (B) 2 1 1 s s( )+ - (C) ( )s s+ - 1 1 (D) s s( )+ - 1 1 20. If i t tin ( ) ( )= 4d then i to( ) will be (A) 4d( ) ( )t e u tt - - A (B) 4 4d( ) ( )t e u tt - - A (C) 4 4e u t tt- -( ) ( )d A (D) e u t tt- -( ) ( )d A 21. If i t tu tin ( ) ( )= then i to( ) will be (A) e u tt- ( ) A (B) ( ) ( )1 - - e u tt A (C) u t( ) A (D) ( ) ( )2 - - e u tt A 22. The voltage across 200 mF capacitor is given by V s s s s C( ) ( ) = + + 2 6 3 The steady state voltage across capacitor is (A) 6 V (B) 0 V (C) (D) 2 V 23. The transformed voltage across the 60 mF capacitor is given by V s s s s C( ) ( )( ) = + + + 20 6 10 3 4 The initial current through capacitor is (A) 0.12 mA (B) - 0 12. mA (C) 0.48 mA (D) - 0 48. mA 24. The current through an 4 H inductor is given by I s s s L ( ) ( ) = + 10 2 The initial voltage across inductor is (A) 40 V (B) 20 V (C) 10 V (D) 5 V 25. The amplifier network shown in fig. P1.8.25 is stable if (A) K 3 (B) K 3 (C) K 1 3 (D) K 1 3 26. The network shown in fig. P1.8.26 is stable if (A) K 5 2 (B) K 5 2 (C) K 2 5 (D) K 2 5 Chap 1.8Circuit Analysis in the s-Domain Page 75 iin io 1 H 1 W 1 F 1 W Fig. P1.8.1921 4 W 1 H1 F 2 W v1 v2 + - Amplifier gain=K + - Fig.P1.8.25 Kv2 1 W v2 + - 2 F1 W 1 F Fig.P1.8.26 jw 1 -1 -1-3 s Fig. P1.8.18 GATE EC BY RK Kanodia www.gatehelp.com
  50. 56. 27. A circuit has a transfer function with a pole s = - 4 and a zero which may be adjusted in position as s a= - The response of this system to a step input has a term of form Ke t-4 . The K will be (H= scale factor) (A) H a 1 4 - (B) H a 1 4 + (C) H a 4 4 - (D) H a 4 4 + 28. A circuit has input v t t u tin ( ) cos ( )= 2 V and output i t t u to( ) sin ( )= 2 2 A. The circuit had no internal stored energy at t = 0. The admittance transfer function is (A) 2 s (B) s 2 (C) s (D) 1 s 29. A two terminal network consists of a coil having an inductance L and resistance R shunted by a capacitor C. The poles of the driving point impedance function Z of this network are at - 1 2 3 2 j and zero at -1. If Z( )0 1= the value of R, L, C are (A) 3 W, 3 H, 1 3 F (B) 2 W, 2 H, 1 2 F (C) 1 W, 2 H, 1 2 F (D) 1 W, 1 H, 1 F 30. The current response of a network to a unit step input is I s s s s o = + + + 10 2 11 302 ( ) ( ) The response is (A) Under damped (B) Over damped (C) Critically damped (D) None of the above Statement for Q.31-33: The circuit is shown in fig. P1.8.31-33. 31. The current ratio transfer function I I o s is (A) s s s s ( )+ + + 4 3 42 (B) s s s s ( ) ( )( ) + + + 4 1 3 (C) s s s s 2 3 4 4 + + +( ) (D) ( )( ) ( ) s s s s + + + 1 3 4 32. The response is (A) Over damped (B) Under damped (C) Critically damped (D) cant be determined 33. If input is is 2u t( ) A, the output current io is (A) ( ) ( )2 3 3 e te u tt t- - - A (B) ( ) ( )3 3 te e u tt t- - - A (C) ( ) ( )3 3 e e u tt t- - - A (D) ( ) ( )e e u tt t- - -3 3 A 34. In the network of Fig. P1.8.34, all initial condition are zero. The damping exhibited by the network is (A) Over damped (B) Under damped (C) Critically damped (D) value of voltage is requires 35. The voltage response of a network to a unit step input is V s s s s o( ) ( ) = + + 10 8 162 The response is (A) under damped (B) over damped (C) critically damped (D) cant be determined 36. The response of an initially relaxed circuit to a signal vs is e u tt-2 ( ). If the signal is changed to ( )vs dv dt s + 2 , the response would be (A) 5 2 e u tt- ( ) (B) - - 3 2 e u tt ( ) (C) 4 2 e u tt- ( ) (D) - - 4 2 e u tt ( ) Page 76 UNIT 1 Networks is 4 W 1 H io 3 1 F Fig. P1.8.31-33 vs 2 H 2 W vo + - 1 4 F Fig. P1.8.34 GATE EC BY RK Kanodia www.gatehelp.com
  51. 57. 37. Consider the following statements in the circuit shown in fig. P1.8.37 1. It is a first order circuit with steady state value of vC = 10 3 , i = 5 3 A 2. It is a second order circuit with steady state of vC = 2 V , i = 2 A 3. The network function V s I s ( ) ( ) has one pole. 4. The network function V s I s ( ) ( ) has two poles. The true statements are (A) 1 and 3 (B) 1 and 4 (C) 2 and 3 (D) 2 and 4 38. The network function s s s s 2 2 10 24 8 15 + + + + represent a (A) RC admittance (B) RL impedance (C) LC impedance (D) None of the above 39. The network function s s s s s ( ) ( )( )( ) + + + + 4 1 2 3 represents an (A) RC impedance (B) RL impedance (C) LC impedance (D) None of these 40. The network function s s s s ( ) ( )( ) 3 8 1 3 + + + represents an (A) RL admittance (B) RC impedance (C) RC admittance (D) None of the above 41. The network function ( )( ) ( )( ) s s s s s + + + + 1 4 2 5 is a (A) RL impedance function (B) RC impedance function (C) LC impedance function (D) Above all 42. The network function s s s 2 7 6 2 + + + is a (A) RL impedance function (B) RL admittance (C) LC impedance function (D) LC admittance 43. A valid immittance function is (A) ( )( ) ( )( ) s s s s + + + - 4 8 2 5 (B) s s s s ( ) ( )( ) + + + 1 2 5 (C) s s s s s ( )( ) ( )( ) + + + + 2 3 1 4 (D) s s s s s ( )( ) ( )( ) + + + + 2 6 1 4 44. The network function s s s s 2 2 8 15 6 8 + + + + is a (A) RLadmittance (B) RC admittance (C) LC admittance (D) Above all 45. A impedance function is given as Z s s s s s ( ) ( )( ) ( ) = + + + 3 2 4 3 The network for this function is ************ Chap 1.8Circuit Analysis in the s-Domain Page 77 i 4 W2 H 10 V 1 W vC + 2 1 F Fig. P1.8.37 8 F 1 F 3 W8 1 F 3 1 W 1 F 3 W 3 1 W 3 W 1 F 3 1 W 8 1 F 3 W 1 H 3 1 W 8 1 H (C) (D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
  52. 58. 18. (B) Z s K s s j s j K s s s ( ) ( ) ( ( ))( ( )) ( ) = + - - + - - = + + + 3 1 1 3 2 22 Z K K( )0 3 2 3 2= = = 19. (D) I s I s s s s s s s s o in ( ) ( ) = + + + + = + 1 1 1 1 1 20. (B) I sin ( ) = 4 I s s s s o( ) = + = - + 4 1 4 4 1 i t t e u to t ( ) ( ) ( )= - - 4 4d 21. (B) I s s in ( ) = 1 2 , I s s s s s o( ) ( ) = + = - + 1 1 1 1 1 i t u t e u to t ( ) ( ) ( )= - - = - - ( ) ( )1 e u tt 22. (D) v sV s s s C s C s ( ) lim ( ) lim = = + + = 0 0 2 6 3 2 V 23. (D) v sV sC s C( ) lim ( )0+ = = + + + = s s s s ( ) ( )( ) 20 6 10 3 4 2 V i Cdv dt C C = I s C sV s vC C C( ) [ ( ) ( )]= - + 0 = + + + - - 60 10 20 6 10 3 4 26 s s s s ( ) ( )( ) = - + + + - 480 10 10 3 10 43 12 6 2 ( )s s s i sI sC s C( ) lim ( ) .0 480 10 0 486+ - = = - = - mA 24. (A) v L d i dt V s L sI s iL L L L L= = - + ( ) [ ( ) ( )]0 i sI s s L s L( ) lim ( )0 10 2 0+ = = + = V s s s s s L ( ) ( ) = + = + 40 2 40 2 v sV s s s L s L( ) lim ( )0 40 2 40+ = = + = 25. (A) V s KV s2 1( ) ( )= V s V s KV s s s 1 1 1 2 4 1 0 ( ) ( ) ( ) + - + + = 4 1 2 2 0+ + + - =s s K s K s2 6 2 1 0+ - + =( ) ( )6 2 0- >K K < 3 26. (B) Let v1 be the node voltage of middle node V s KV s sV s s s 1 2 22 1 2 ( ) ( ) ( ) = + + + + = +( ) ( ) ( ) ( )3 1 21 2s V s s K V s V s sV s s 2 12 2 1 ( ) ( ) = + + =( ) ( ) ( )2 1 22 1s V s sV s + + = +( )( ) ( )3 1 2 1 2 2s s s s K 2 5 2 1 02 s K s+ - + =( ) , 5 2 0- >K , K < 5 2 27. (A) H s H s a s ( ) ( ) = + + 4 R s H s a s s Ha s H a s ( ) ( ) ( ) = + + = + - +4 4 1 4 4 r t Ha u t H a e t ( ) ( )= + - - 4 1 4 4 28. (A) V s s s in ( ) = +2 1 , I s s o( ) = + 2 12 , I s V s s o in ( ) ( ) = 2 29. (D) Z s sL R sC sL R sC ( ) ( ) = + + + 1 1 = + + + 1 12 C s R L s R L LC Z s K s s j s j ( ) ( ) = + + + + - 1 1 2 3 2 1 2 3 2 = + + + K s s s ( ) ( ) 1 12 Since Z K( ) ,0 1 1= =thus 1 1 1 1 1 C R L LC = = =, , C L R= = =1 1 1, , 30. (B) The characteristic equation is s s s2 2 11 30 0( )+ + = s s s2 6 5( ) ( )+ + =0 s = - -6 5, , Being real and unequal, it is overdamped. Page 80 UNIT 1 Networks sL R Cs 1Z s( ) Fig. S1.8.29 GATE EC BY RK Kanodia www.gatehelp.com
  53. 59. 31. (B) I I s s s s s s s o s = + + + = + + + 4 4 3 4 1 3 ( ) ( )( ) 32. (A) The characteristic equation is ( ) ( )s s+ + =1 3 0. Being real and unequal root, it is overdamped response. 33. (C) i u ts = 2 ( ) I s s s( ) = 2 I s s s s s s o ( ) ( ) ( )( ) = + + + = + - + 2 4 1 3 3 1 1 3 i e e u to t t = -- - ( ) ( )3 3 A 34. (B) V s V s s s s s o s ( ) ( ) = + + = + + 2 4 2 2 1 22 The roots are imaginary so network is underdamped. 35. (C) The characteristic equation is s s s( )2 8 16 0+ + = , ( )s + =4 02 , s = - -4 4, Being real and repeated root, it is critically damped. 36. (B) v e u to t = -2 ( ) V s H s V s s o s( ) ( ) ( )= = + 1 2 = +v v dv dt s s s2 = +V s s V ss s( ) ( ) ( )1 2 = = +V s H s V s s V s H so s s( ) ( ) ( ) ( ) ( ) ( )1 2 = + + = - + V s s s s o( ) 1 2 2 2 3 2 = - - v s e u to t 2 3 2 d( ) ( ) 37. (C) It is a second order circuit. In steady state i = + = 10 4 1 2 A , v = =2 1 2 V I s s s s s ( ) ( ) ( ) = + + + = + + + 10 2 4 1 1 1 2 5 2 2 12 V s s s s s ( ) ( ) ( ) = + + + + = + + 10 1 1 2 2 4 1 1 1 2 10 2 12 V s I s s ( ) ( ) = + 2 2 , It has one pole at s = -2 38. (D) s s s s s s s s 2 2 10 24 8 15 4 6 3 5 + + + + = + + + + ( )( ) ( )( ) The singularity near to origin is pole. So it may be RC impedance or RL admittance function. 39. (D) Poles and zero does not interlace on negative real axis so it is not a immittance function. 40. (C) The singularity nearest to origin is a zero. So it may be RL impedance or RC admittance function. Because of (D) option it is required to check that it is a valid RC admittance function. The poles and zeros interlace along the negative real axis. The residues of Y s s RC ( ) are real and positive. 41. (B) The singularity nearest to origin is a pole. So it may be RC impedance or RL admittance function. 42. (A) s s s s s s 2 7 6 2 1 6 2 + + + = + + + ( )( ) ( ) The singularity nearest to origin is at zero. So it may be RC admittance or RL impedance function. 43. (D) (A) pole lie on positive real axis (B) poles and zero does not interlace on axis. (C) poles and zero does not interlace on axis. (D) is a valid immittance function. 44. (A) s s s s s s s s 2 2 8 15 6 8 3 5 2 4 + + + + = + + + + ( ) ( ) ( ) ( ) The singularity nearest to origin is a pole. So it may be a RL admittance or RC impedance function. 45. (A) The singularity nearest to origin is a pole. So this is RC impedance function. Z s s s s s ( ) = + + + = + + + 3 8 1 3 3 8 1 3 1 3 ************** Chap 1.8Circuit Analysis in the s-Domain Page 81 GATE EC BY RK Kanodia www.gatehelp.com
  54. 60. Statement for Q.1-2: In the circuit of fig. P1.9.1-2 i t1 4 2= sin A, and i2 0= . 1. v1 = ? (A) -16 2cos t V (B) 16 2cos t V (C) 4 2cos t V (D) -4 2cos t V 2. v2 = ? (A) 2 2cos t V (B) -2 2cos t V (C) 8 2cos t V (D) -8 2cos t V Statement for Q.3-4: Consider the circuit shown in Fig. P1.9.3-4 3. If i1 0= and i t2 2 4= sin A, the voltage v1 is (A) 24 4cos t V (B) -24 4cos t V (C) 15 4. cos t V (D) -15 4. cos t V 4. If i e t 1 2 = - V and i2 0= , the voltage v2 is (A) - - 6 2 e t V (B) 6 2 e t- V (C) 15 2 . e t- V (D) - - 15 2 . e t V Statement for Q.5-6: Consider the circuit shown in fig. P19.5-6 5. If current i t1 3 4= cos A and i2 0= , then voltage v1 and v2 are (A) v t1 24 4= - sin V, v t2 24 4= - sin V (B) v t1 24 4= sin V, v t2 36 4= - sin V (C) v t1 15 4= . sin V, v t2 4= sin V (D) v t1 15 4= - . sin V, v t2 4= -sin V 6. If current i1 0= and i t2 4 3= sin A, then voltage v1 and v2 are (A) v t1 24 3= cos V, v t2 36 3= cos V (B) v t1 24 3= cos V, v t2 36 3= - cos V (C) v t1 24 3= - cos V, v t2 36 3= cos V (D) v t1 24 3= - cos V, v t2 36 3= - cos V CHAPTER 1.9 MAGNETICALLY COUPLED CIRCUITS Page 82 i1 i2 3 H - + v2 - + v1 3 H 4 H Fig. P1.9.5-6 i1 i2 2 H - + v2 - + v1 2 H 3 H Fig. P1.9.5-6 i1 i2 1 H - + v2 - + v1 2 H 1 H Fig. P1.9.1-2 GATE EC BY RK Kanodia www.gatehelp.com
  55. 61. Statement for Q.7-8: In the circuit shown in fig. P1.9.7-8, i t1 3 3= cos A and i t2 4 3= sin A. 7. v1 = ? (A) 6 2 3( cos sin )- +t t V (B) 6 2 3( cos sin )t t+ V (C) - +6 2 3( cos sin )t t V (D) 6 2 3( cos sin )t t- V 8. v2 = ? (A) 3 8 3 3( cos sin )t t- V (B) 6 2 3( cos sin )t t+ V (C) 3 8 3 3 3( cos sin )t t+ V (D) 6 2 3( cos sin )t t- V Statement for Q.9-10: In the circuit shown in fig. P1.9.9-10, i t1 5 3= sin A and i t2 3 3= cos A 9. v1 =? (A) 9 5 3 3 3( cos sin )t t+ V (B) 9 5 3 3 3( cos sin )t t- V (C) 9 4 3 5 3( cos sin )t t+ V (D) 9 5 3 3 3( cos sin )t t- V 10. v2 = ? (A) 9 4 3 5 3( sin cos )- +t t V (B) 9 4 3 5 3( sin cos )t t- V (C) 9 4 3 5 3( sin cos )- -t t V (D) 9 4 3 5 3( sin cos )t t+ V 11. In the circuit shown in fig. P1.9.11 if current i t1 5 500 20= - cos ( ) mA and i t2 20 500 20= - cos ( ) mA, the total energy stored in system at t = 0 is (A) 151.14 mJ (B) 45.24 mJ (C) 249.44 mJ (D) 143.46 mJ 12. Leq = ? (A) 4 H (B) 6 H (C) 7 H (D) 0 H 13. Leq = ? (A) 2 H (B) 4 H (C) 6 H (D) 8 H 14. Leq = ? (A) 8 H (B) 6 H (C) 4 H (D) 2 H 15. Leq = ? (A) 0.4 H (B) 2 H (C) 1.2 H (D) 6 H 16. The equivalent inductance of a pair of a coupled inductor in various configuration are (a) 7 H after series adding connection (b) 1.8 H after series opposing connection (c) 0.5 H after parallel connection with dotted terminal connected together. Chap 1.9Magnetically Coupled Circuits Page 83 i1 i2 3 H - + v2 - + v1 3 H 4 H Fig. P1.9.9-10 i1 i2 k=0.6 - + v2 - + v1 0.4 H2.5 H Fig. P1.9.11 i1 i2 1 H - + v2 - + v1 2 H 2 H Fig. P1.9.7-8 3.6 H 1 H 1.4 H Leq Fig. P1.9.12 4 H 2 H 2 H Leq Fig. P1.9.13 Leq 4 H 4 H 6 H Fig. P1.9.14 Leq 2 H 4 H 2 H Fig. P1.9.15 GATE EC BY RK Kanodia www.gatehelp.com
  56. 62. The value of L L1 2, and M are (A) 3 H, 1.6 H, 1.2 H (B) 1.6 H, 3 H, 1.4 H (C) 3.7 H, 0.7 H, 1.3 H (D) 2 H, 3 H, 3 H 17. Leq = ? (A) 0.2 H (B) 1 H (C) 0.4 H (D) 2 H 18. Leq = ? (A) 1 H (B) 2 H (C) 3 H (D) 4 H 19. In the network of fig. P1.9.19 following terminal are connected together (i) none (ii) A to B (iii) B to C (iv) A to C The correct match for equivalent induction seen at terminal a b- is (i) (ii) (iii) (iv) (A) 1 H 0.875 H 0.6 H 0.75 H (B) 13 H 0.875 H 0.6 H 0.75 H (C) 13 H 7.375 H 6.6 H 2.4375 H (D) 1 H 7.375 H 6.6 H 2.4375 H 20. Leq = ? (A) 1 H (B) 2 H (C) 3 H (D) 4 H 21. Leq = ? (A) 41 5 H (B) 49 5 H (C) 51 5 H (D) 39 5 H Statement for Q.22-24: Consider the circuit shown in fig. P1.9.2224. 22. The voltage VAG of terminal AD is (A) 60 V (B) -60 V (C) 180 V (D) 240 V 23. The voltage vBG of terminal BD is (A) 45 V (B) 33 V (C) 69 V (D) 105 V 24. The voltage vCG of terminal CD is (A) 30 V (B) 0 V (C) -36 V (D) 36 V Page 84 UNIT 1 Networks 3 H 5 H 3 H Leq Fig. P1.9.18 a b A C B 2 H 4 H 1 H 3 H 2 H 5 H Fig. P1.9.19 1 H 1 H 2 H 2 H 3 H Leq Fig. P1.9.20 3 H 2 H 4 H Leq 2 H 3 H Fig. P1.9.21 6 At 20 H 15 At A C B 3 H 5 H D 6 H 4 H Fig. P1.9.2224 2 H 4 H 2 H Leq Fig. P1.9.17 GATE EC BY RK Kanodia www.gatehelp.com
  57. 63. 33. In the circuit of fig. P1.9.33 the w = 2 rad s. The resonance occurs when C is (A) 1 F (B) 1 2 F (C) 1 3 F (D) 1 6 F 34. In the circuit of fig. P1.9.34, the voltage gain is zero at w = 333.33 rad s. The value of C is (A) 100 mF (B) 75 mF (C) 50 mF (D) 25 mF 35. In the circuit of fig. P1.9.35 at w = 333.33 rad s, the voltage gain v vout in is zero. The value of C is (A) 3.33 mF (B) 33.33 mF (C) 3.33 mF (D) 33.33 mF 36. The Thevenin equivalent at terminal ab for the network shown in fig. P1.9.36 is (A) 6 V, 10 W (B) 6 V, 4 W (C) 0 V, 4 W (D) 0 V, 10 W 37. In the circuit of fig. P1.9.37 the maximum power delivered to RL is (A) 250 W (B) 200 W (C) 150 W (D) 100 W 38. The average power delivered to the 8 W load in the circuit of fig. P1.9.38 is (A) 8 W (B) 1.25 kW (C) 625 kW (D) 2.50 kW 39. In the circuit of fig. P1.9.39 the ideal source supplies 1000 W, half of which is delivered to the 100 W load. The value of a and b are (A) 6, 0.47 (B) 5, 0.89 (C) 0.89, 5 (D) 0.47, 6 40. I2 = ? (A) 1.65 Arms (B) 0.18 Arms (C) 0.66 Arms (D) 5.90 Arms Page 86 UNIT 1 Networks C 2 HZin 4 W2 H 2 H Fig. P1.9.33 0.09 H vin C 2 F - + vout 0.12 H 0.27 H 20 W 40 W ~ Fig. P1.9.34 C k=0.5 0.12 H 0.27 Hvin - + vout 20 W 40 W ~ 20 W Fig. P1.9.35 a b 20 W 20Ix Ix 60 W 1 : 4 Fig. P1.9.36 100Vrms 10 W 1 : 4 RL~ Fig. P1.9.37 -0.04V2 8 W - + V1 I1 I2 50Vrms 300 W 5 : 1 ~ - + V2 Fig. P1.9.38 100 W100Vrms 4 W 1 : a ~ 1 : b 25 W Fig. P1.9.39 3 W I2 50Vrms 3 : 1 ~ 4 : 3 25 W 2 W Fig. P1.9.40 GATE EC BY RK Kanodia www.gatehelp.com
  58. 64. 41. V2 = ? (A) -12.31 V (B) 12.31 V (C) -9.231 V (D) 9.231 V 42. The power being dissipated in 400 W resistor is (A) 3 W (B) 6 W (C) 9 W (D) 12 W 43. Ix = ? (A) 1921 57 4. . A (B) 2 931 59 4. . A (C) 1 68 43 6. . A (D) 179 43 6. . A 44. Zin = ? (A) 46 3 6 8. .+ j W (B) 432 1 0 96. .+ j W (C) 10 8 9 6. .+ j W (D) 615 4 0 38. .+ j W ******************** SOLUTIONS 1. (B) v di dt di dt di dt 1 1 2 1 2 1 2= + = = 16 2cos t V 2. (C) v di dt di dt di dt 2 2 1 1 1 1= + =( ) ( ) = 8 2cos t V 3. (B) v di dt di dt di dt 1 1 2 2 3 3 3= - = - = - 24 4cos t V 4. (C) v di dt di dt 2 2 1 4 3= - = - = - 3 61 2di dt e t V 5. (A) v di dt di dt di dt 1 1 2 1 2 2 2= - = = -24 4sin t V v di dt di dt di dt t2 2 1 1 3 2 2 24 4= - + = = - sin V 6. (D) v di dt di dt di dt 1 1 2 2 2 2 2= - = - = - 24 3cos t V v di dt di dt di dt t2 2 1 2 3 2 3 36 3= + = - = - cos V 7. (A) v di dt di dt 1 1 2 2 1= + = - + = -18 12 6 2 3sin cos ( cos sin )t t t t V 8. (A) v di dt di dt 2 2 1 2 1= + = - = -24 3 9 3 3 8 3 3 3cos sin ( cos sin )t t t t V 9. (A) v di dt di dt 1 1 2 3 3= - = +45 3 27 3cos sint t = +9 5 3 3 3( cos sin )t t V 10. (D) v di dt di dt 2 2 1 4 3= - + = +36 3 45 3sin cost t = +9 4 3 5 3( sin cos )t t V 11. (A) W L i L i Mi i= + + 1 2 1 2 1 1 2 2 2 2 1 2 At t = 0, i1 4 20 4 7= - =cos ( ) . mA i2 20 20 18 8= - =cos ( ) . mA , M = =0 6 2 5 0 4 0 6. . . . W = + + 1 2 2 5 4 7 1 2 0 4 18 8 0 6 4 7 18 82 2 ( . )( . ) ( . )( . ) . ( . )( . ) = 151 3. mJ 12. (C) L L L Meq = + + =1 2 2 7 H Chap 1.9Magnetically Coupled Circuits Page 87 48 W 400 W10Vrms 1 : 2 ~ 1 : 5 1 W 4 W Fig. P1.9.42 - 4j Ix j6100 0 V o 8 W 10 W ~ 2 : 1 Fig. P1.9.43 j16 6 W Zin 1 : 5 4 : 1 6 W 24 W - 10j Fig. P1.9.44 50 W 10 W - + V2 80Vrms 40 W 5 : 2 ~ Fig. P1.9.41 GATE EC BY RK Kanodia www.gatehelp.com
  59. 65. 13. (A) L L L Meq = + -1 2 2 = + - =4 2 2 2 2 H 14. (C) L L L M L L M eq = - + - 1 2 2 1 2 2 = - + - = 24 16 6 4 8 4 H 15. (A) L L L M L L M eq = - + + 1 2 2 1 2 2 = - + = 8 4 6 4 0 4. H 16. (C) L L M1 2 2 7+ + = , L L M1 2 2 1 8+ - = . + =L L1 2 4 4. , M = 1 3. L L M L L M 1 2 2 1 2 2 0 5 - + - = . , L L1 2 2 1 3 0 5 1 8- = . . . L L1 2 2 59= . , ( ) . .L L1 2 2 2 4 4 4 2 59 9- = - = L L1 2 3- = , L1 37= . , L2 0 7= . 17. (D) L L M L eq = -1 2 2 = - =4 4 2 2 H 18. (B) L L M L eq = -1 2 2 = - =5 9 3 2 H 19. (A) 20. (D) V sI sI sIL1 1 1 2= + = V sI sI sI sIL2 2 1 2= + - = , V sI sI sIL3 3 2= - = V V V V sIL L L L= + + =1 2 3 4 Leq = 4 H 21. (B) Let I1 be the current through 4 H inductor and I2 and I3 be the current through 3 H, and 2 H inductor respectively I I I1 2 3= + , V V2 3= 3 3 2 22 1 3 1sI sI sI sI+ = + + =3 22 1 3I I I 4 2 3I I= = =I I I I2 1 3 1 5 4 5 , V sI sI sI sI s I= + + + +4 3 2 3 31 2 3 2 1 = + + 7 6 5 2 4 5 1 1 3sI s I s I V sI= 49 5 1 , Leq = 49 5 H 22. (C) v d t dt d t dt AG = + =20 6 4 15 180 ( ) ( ) V 23. (B) v d t dt d t dt d t dt BG = + - =3 15 4 6 6 6 33 ( ) ( ) ( ) V 24. (C) v d t dt CG = - = -6 6 36 ( ) V 25. (B) Z Z M Z = +11 2 2 22 w = + + + 4 50 1 10 50 1 5 5 50 1 2 2 2 j j ( ) ( ) ( ) = +4 77 115. .j W 26. (B) V j js = - ( . ) ( . ) ( . )( )( )0 8 10 12 0 0 2 10 2 0 + + + [ ( . )( )]( . )3 0 5 10 12 0 2 0j = + = 9 6 21 6 26 64 66 04. . . .j V 27. (A) [ ( ) ( ) ] ( )( . ) ( )j I j100 2 10 100 0 4 2 0 02p p+ + = = - -I j2 0 4 0 0064. . , V I jo = = - -10 4 0 0642 . = - 4 179 1. = - v to 4 100 179 1cos ( . )p V 28. (B) 30 30 6 8 4 12 4 10 = - + - + - +I j j j j j( ) = + = -I j j 30 30 10 6 2 57 0 043 ( ) . . V I j jo = - +( )12 4 10 = - +( . . )( )2 57 0 043 10 8j j = +26 067 20 14. .j = 32 9 37 7. . V 29. (A) ( )- + + - = -j j I jI j2 4 31 2 ( )j I jI4 2 12 302 1+ - = - V I j1 1 45 0 56= - -. . , V I jx = - = +2 2 9 1121 . . = 311 2112. . V Page 88 UNIT 1 Networks 3 H 2 H 5 H -1 H 2 H Fig. S.1.9.19 I1 I2 3 -90 A o +- Vx -j j4 j 12 30 V o 2 W 2 W j4 ~~ Fig. S1.9.29 GATE EC BY RK Kanodia www.gatehelp.com
  60. 66. 30. (D) Z j j j j j j j eq = + + + - + + - 10 8 14 10 2 6 14 10 2 6 ( ) ( ) = +112 112. .j W 31. (C) Z j Zin o= -( )||( )6 Z j j j j o = + + - + 20 12 30 5 2 4 2 ( ) ( ) = +0 52 15 7. .j Z j j j j in = - + - + + ( )( . . ) ( . . ) 6 0 52 15 7 6 0 52 15 7 = -0 20 9 7. .j W 32. (D) M k L L= 1 2 , M 2 12 160 10= - Z j L M Z j L in L = + + w w w 1 2 2 2 = + + + - - j j j 250 10 2 10 250 10 160 10 2 10 250 1 3 6 3 2 12 ( ) 0 80 103 6 - Z jin = +0 02 0 17. . W 33. (D) V jI C j I j I1 1 1 2 2 4 2 2= - + + 0 4 4 2 22 1= + +( )j I j I I j I j 2 12 2 1 = - +( ) V I j C j j 1 1 2 4 2 1 = - + + + = - + + -j j C C j C C 8 2 2 2 Z j j C C j C C in = - + + -8 2 2 2 Im ( )Zin = 0 - + - =j j C j C8 2 0 C = 1 6 34. (A) j j C 30 3 1000 0- = , C = 100 mF 35. (D) The p equivalent circuit of coupled coil is shown in fig. S1.9.35 L L M M L L k k 1 2 2 1 2 2 1- = -( ) = = 0 12 0 27 10 5 0 5 0 27 2 . . ( . . ) . . Output is zero if - + = j jC 0 27 0 . w w C = = 1 0 27 33 332 . . w mF 36. (C) Applying 1 V test source at ab terminal, Vab = 1 V, Ix = = 1 20 0 05. A, V2 4= V , 4 60 20 0 052= + I . I2 0 05= . A I I I I Iin x x= + = + =1 24 0 25. A R I TH in = = 1 4 W , VTH = 0 37. (A) Impedance seen by RL = =10 4 1602 W For maximum power RL = 160 W, Zo = 10 W PLmax = + = 100 10 10 10 250 2 W 38. (B) I V 2 2 8 = , I I V 1 2 2 5 40 = = , V V1 25= Chap 1.9Magnetically Coupled Circuits Page 89 I1 I2 j4V1 4 W~ j4 2j2 2C -j Fig. S.1.9.33 0.03 H 0.18 H 2 F - + Vout - 3j Vin 1000C 20 W j30 ~ 40 W 0.09 H Fig. P1.9.34 M L L M1 2 - 2 L L M1 2 - 2 L L M1 2 - 2 M L1 L2 L M2 - L M1 - Fig. S1.9.35 20Ix Ix 60 W 1 V 20 W 1 : 4 Fig. S1.9.36 j10 j18 j20 j10 j10j8 Fig. S.1.9.30 GATE EC BY RK Kanodia www.gatehelp.com
  61. 67. Statement for Q.1-4: The circuit is given in fig. P.1.10.14 1. [ ] ?z = (A) - - - 1 2 3 2 17 6 1 2 (B) 1 2 3 2 17 6 1 2 (C) - - - 17 6 1 2 1 6 3 2 (D) 17 6 1 2 1 6 3 2 2. [ ] ?y = (A) 3 8 1 8 1 8 17 24 (B) 3 8 1 8 1 8 17 24 - - (C) 17 6 1 2 1 2 3 2 (D) 17 6 1 2 1 8 3 2 - - 3. [ ] ?h = (A) 6 17 3 17 3 17 24 17 - (B) 8 3 1 3 1 3 2 3 - (C) 6 17 3 17 3 17 24 17 - (D) 8 3 1 3 1 3 2 3 - 4. [ ]T = ? (A) 17 3 8 2 3 (B) 17 3 8 2 3 - - (C) - - - 17 3 8 2 3 (D) 17 3 8 2 3 - - 5. [ ]z = ? (A) 21 16 1 8 1 8 7 12 (B) 7 9 1 6 1 6 7 4 (C) 21 16 1 8 1 8 7 12 - - (D) 7 9 1 3 1 3 7 4 CHAPTER 1.10 Page 91 TWO PORT NETWORK I1 2 W 1 W 2 W I2 V1 + - 3 W V2 + - Fig. P.1.10.14 I1 2 W 1 W 2 W I2 V1 + - 3 W V2 + - 2 W Fig. P.1.10.5 GATE EC BY RK Kanodia www.gatehelp.com
  62. 68. 6. [ ]y = ? (A) 11 41 2 41 2 41 19 41 (B) 11 41 2 41 2 41 19 41 - - (C) 19 41 2 41 2 41 11 41 (D) 19 41 2 41 2 41 11 41 - - Statement for Q.7-10: A two port is described by V I V1 1 22= + , I I V2 1 22 0 4= - + . 7. [ ] ?z = (A) 11 5 5 2 5 - - . (B) 11 5 5 2 5. (C) 1 2 5 0 4 - . (D) 1 2 2 0 4- . 8. [ ]y = ? (A) 11 5 5 2 5. (B) 1 2 2 4 4 - - . (C) - - 2 4 4 4 2 . (D) 11 5 5 2 5 - - . 9. [ ] ?h = (A) 3 6 4 4 - - (B) 4 2 2 4 4 - - . (C) 1 2 2 0 4- . (D) 11 5 5 2 5. 10. [ ] ?T = (A) 2 2 0 5 0 2 0 5 . . . . (B) 2 2 0 5 0 2 0 5 . . . . - - (C) 1 2 2 0 4- . (D) 1 2 2 0 4 - - - . 11. [ ]y = ? (A) 1 2 1 3 2 1- (B) 3 2 1 1 2 1 - (C) 1 2 1 2 1 4 3 4 - (D) - 1 4 3 4 1 2 1 2 12. [ ]z = ? (A) 4 3 2 3 2 3 2 3 - (B) 1 2 1 2 1 2 1 - (C) - 2 3 2 3 4 3 2 3 (D) 1 2 1 1 2 1 2 - 13. [ ]y = ? (A) 7 4 1 4 1 2 5 4 - (B) 7 4 1 4 3 4 5 4 - (C) 10 19 2 19 6 19 14 19 (D) 6 19 14 19 10 19 2 19 Page 92 UNIT 1 Networks I1 2 W 1 W I2 V1 + - V2 + - 1 W I1 Fig. P.1.10.11 I1 2 W 1 W 2 W I2 V1 + - 3 W V2 + - 1 W Fig. P.1.10.6 1 W 2 W 2 W 2V1I1 V1 + - I2 V2 + - Fig. P.1.10.12 2 W 2 W 1 W2 W 2V1 I1 V1 + - I2 V2 + - Fig. P.1.10.13 GATE EC BY RK Kanodia www.gatehelp.com
  63. 69. 14. [ ]z = ? (A) 2 3 3 3 (B) - - 3 2 3 3 (C) 3 3 3 2 (D) 3 3 3 2- - 15. [ ] ?z = (A) 2 2 3 2 2 (B) - - 2 3 2 2 2 (C) 2 3 2 2 2 (D) 2 2 3 2 2 - - 16. [ ]y = ? (A) - - - 1 1 1 2 (B) 1 1 1 2 - - (C) - - - 2 3 1 3 1 3 1 3 (D) - - - 2 3 1 3 1 3 1 3 17. [ ]z = ? (A) 3 2 6 1 7 (B) 6 1 7 3 2 (C) 7 4 1 1 2 3 (D) 1 2 3 7 4 1 18. [ ]T = ? (A) 0 35 1 2 3 33 . . - - (B) 2 3 33 0 35 1 - - . . (C) 2 3 33 0 35 1 . . (D) 0 35 1 2 3 33 . . 19. [ ]h = ? (A) 4 3 2 2 1 2 - (B) - 2 1 2 4 3 2 (C) 4 3 2 2 1 2 - (D) 2 1 2 4 3 2 - Chap 1.10Two Port Networks Page 93 2 W 2 W W 2V1 I2 V2 + - I1 V1 + - 2 3V2 3 2 Fig. P.1.1.15 2 W 1 W 3 W I2 V2 + - I1 V1 + - V2 Fig. P.1.10.16 2 W 2 W2 W 2 W I2 V2 + - I1 V1 + - I1 Fig. P.1.10.17 4 W 4 W I1 I2 V1 + - V2 + - 10 1 V2 5 1V1 Fig. P.1.10.18 2 W 4 W I1 I2 V1 + - V2 + - V2 2 1 I2 Fig. P.1.10.19 2 W 4V3 1 W 2I2 1 W I1 I2 V1 + - V2 + - V3 + - Fig. P.1.10.14 GATE EC BY RK Kanodia www.gatehelp.com

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