Solucionario Van Valkenburg, Ejercicios de Teoría de Circuitos

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Dedicated to: Prof. Dr. sohail aftab qureshi Conventions for Describing Networks 2 - 1. For the controlled (monitored) source shown in the figure, prepare a plot similar to that given in Fig. 2-8(b). v 2 v 1 = Vb  Vb v 1 = Va  Va i 2 Fig. 2-8 (b) Solution: Open your book & see the figure (P/46) It is voltage controlled current source. i 2 +Ve axis v 2

• Ve axis gv 1 i 2 gv 1 + v 2 current source

Dedicated to: Prof. Dr. sohail aftab qureshi Slope: y = mx + c (x 1 , y 1 ) = (0, V/Rb) (x 2 , y 2 ) = (V, 0) m = (y 2 – y 1 )/(x 2 – x 1 ) = (0 – V/Rb)/(V - 0) = (-V/Rb)/V = (-V/Rb)(1/V) = - 1/Rb y-intercept = V/Rb x-intercept = V Slope y-intercept x-intercept

• 1/Rb V/Rb V 2 - 4. The magnetic system shown in the figure has three windings marked 1- 1’, 2 - 2’, and 3- 3’. Using three different forms of dots, establish polarity markings for these windings. Solution: Open your book & see the figure (P/46) Lets assume current in coil 1- 1’ has direction up at 1 (increasing). It produces flux  (increasing) in that core in clockwise direction. 1 1’ 2 2’ 3 3’ According to the Lenz’s law current produced in coil 2 - 2’ is in such a direction that it opposes the increasing flux . So direction of current in 2- 2’ is down at 2’. Hence ends 1 & 2’ are of same polarity at any instant. Hence are marked with. Similarly assuming the direction of current in coil 2- 2’, we can show at any instant 2 & 3’ have same polarities and also 1 & 3 have same polarities. 2 - 5. Place three windings on the core shown for Prob. 2-4 with winding senses selected such that the following terminals have the same mark: (a) 1 and 2, 2 and 3, 3 and 1, (b) 1’ and 2’, 2’ and 3’, 3’ and 1’.

Dedicated to: Prof. Dr. sohail aftab qureshi Solution: Open your book & see the figure (P/47) 1 1’ 2 2’ 3 3’ (a) 1 1’ 2 2’ 3 3’ (b) 2 - 6. The figure shows four windings on a magnetic flux-conducting core. Using different shaped dots, establish polarity markings for the windings. Solution: Open your book & see the figure (P/47) i 1 i 3 i 4  1 (Follow Fleming’s right hand rule)

i 2 Coil 1 Coil 3 Coil 2 Coil 4

Dedicated to: Prof. Dr. sohail aftab qureshi V = self induced e.m.f. (1) + self induced e.m.f. (2) + mutually induced e.m.f. (1) + mutually induced e.m.f. (2) V = L 1 di/dt + L 2 di/dt + M di/dt + M di/dt Let Leq be the equivalent inductance then V = Leq di/dt Leq di/dt = (L 1 + L 2 + M + M) di/dt  Leq = L 1 + L 2 + M + M Leq = L 1 + L 2 + 2M M L 1 L 2 i V (b) This is the case of subtractive flux.  V = L 1 di/dt + L 2 di/dt - M di/dt - M di/dt Let Leq be the equivalent inductance then V = Leq di/dt Leq di/dt = (L 1 + L 2 - M - M) di/dt  Leq = L 1 + L 2 - M - M Leq = L 1 + L 2 - 2M 2 - 9. A transformer has 100 turns on the primary (terminals 1- 1’) and 200 turns on the secondary (terminals 2- 2’). A current in the primary causes a magnetic flux, which links all turns of both the primary and the secondary. The flux decreases according to the law  = e

• t Weber, when t  0. Find: (a) the flux linkages of the primary and secondary, (b) the voltage induced in the secondary. Solution: N 1 = 100 N 2 = 200  = e
• t (t  0) Primary flux linkage  1 = N 1  = 100 e-t Secondary flux linkage  2 = N 2  = 200 e
• t Magnitude of voltage induced in secondary v 2 = d  2 /dt = d/dt(200 e
• t ) v 2 = - 200 e-t Hence secondary induced voltage has magnitude v 2 = 200 e
• t 2 - 10. In (a) of the figure is shown a resistive network. In (b) and (c) are shown graphs with two of the four nodes identified. For these two graphs, assign resistors

Dedicated to: Prof. Dr. sohail aftab qureshi to the branches and identify the two remaining nodes such that the resulting networks are topologically identical to that shown in (a). Solution: Open your book & see the figure (P/48) b R 2 R 3 R 1 a c R 5 R 4 d R 4 d c R 5 R 1 R 3 b R 2 a

Dedicated to: Prof. Dr. sohail aftab qureshi Maximum number of branches = 7 2 - 16. Display five different trees for the graph shown in the figure. Show branches with solid lines and chords with dotted lines. (b) Repeat (a) for the graph of (c) in Prob. 2-11. Solution: Open your book & see the figure (P/49)

Dedicated to: Prof. Dr. sohail aftab qureshi b):

2 - 17. Determine all trees of the graphs shown in (a) of Prob. 2-11 and (b) of Prob. 2-

10. Use solid lines for tree branches and dotted lines for chords. Solution: Open your book & see the figure (P/49) All trees:

Dedicated to: Prof. Dr. sohail aftab qureshi

All trees of Solution:

Dedicated to: Prof. Dr. sohail aftab qureshi Before solving exercise following terms should be kept in mind:

1. Node
2. Branch
3. Tree
4. Transformer theory
5. Slope
6. Straight line equation
7. Intercept
8. Self induction
9. Mutual induction
10. Current controlled voltage source
11. Voltage controlled current source
12. Coordinate system ALLAH MUHAMMAD (P.B.U.H)

Dedicated to: Prof. Dr. sohail aftab qureshi Diagram (b) reduces to

C 1 + v Ceq ’

From result obtained by (a) 1/Ceq = (1/C 1 + 1/Ceq ’) 1/Ceq = (1/C 1 + 1/C 2 + C 3 ) 3 - 2. What must be the relationship between Leq and L 1 , L 2 and M for the networks of (a) and of (b) to be equivalent to that of (c)? Solution: Open your book & see the figure (P/87) In network (a) applying KVL v = L 1 di/dt + L 2 di/dt + Mdi/dt + Mdi/dt v = (L 1 + L 2 + M + M)di/dt v = (L 1 + L 2 + 2M)di/dt In network (c) v = Leqdi/dt If (a) & (c) are equivalent (L 1 + L 2 + 2M)di/dt = Leqdi/dt (L 1 + L 2 + 2M) = Leq In network (b) applying KVL v = L 1 di/dt + L 2 di/dt - Mdi/dt - Mdi/dt v = (L 1 + L 2 - M – M)di/dt v = (L 1 + L 2 - 2M)di/dt In network (c) v = Leqdi/dt If (b) & (c) are equivalent (L 1 + L 2 - 2M)di/dt = Leqdi/dt (L 1 + L 2 - 2M) = Leq 3 - 3. Repeat Prob. 3-2 for the three networks shown in the accompanying figure. Solution: Open your book & see the figure (P/87)

Dedicated to: Prof. Dr. sohail aftab qureshi

M v i 1 L 1 L 2 loop 1 loop 2 i 2

Applying KVL in loop 1 v = L 1 d(i 1 – i 2 )/dt + Mdi 2 /dt v = L 1 di 1 /dt - L 1 di 2 /dt + Mdi 2 /dt v = L 1 di 1 /dt + Mdi 2 /dt - L 1 di 2 /dt v = L 1 di 1 /dt + (M - L 1 )di 2 /dt Applying KVL in loop 2 0 = L 2 di 2 /dt + L 1 d(i 2 – i 1 )/dt + {-Mdi 2 /dt} + {-Md(i 2 – i 1 )/dt} 0 = L 2 di 2 /dt + L 1 di 2 /dt - L 1 di 1 /dt - Mdi 2 /dt - Md(i 2 – i 1 )/dt 0 = L 2 di 2 /dt + L 1 di 2 /dt - L 1 di 1 /dt - Mdi 2 /dt - Mdi 2 /dt + Mdi 1 /dt 0 = L 2 di 2 /dt + L 1 di 2 /dt - L 1 di 1 /dt - 2Mdi 2 /dt + Mdi 1 /dt 0 = (M – L 1 ) di 1 /dt + (L 1 + L 2 – 2M) di 2 /dt Writing in matrix form L 1 M – L 1 di 1 /dt v = M – L 1 L 1 + L 2 – 2M di 2 /dt 0 v M – L 1 0 L 1 + L 2 – 2M di 1 /dt = L 1 M – L 1 M – L 1 L 1 + L 2 – 2M

Dedicated to: Prof. Dr. sohail aftab qureshi

Applying KVL in loop 1 v = L 1 d(i 1 – i 2 )/dt - Mdi 2 /dt v = L 1 di 1 /dt - L 1 di 2 /dt - Mdi 2 /dt v = L 1 di 1 /dt + Mdi 2 /dt - L 1 di 2 /dt v = L 1 di 1 /dt - (L 1 + M)di 2 /dt Applying KVL in loop 2 0 = L 2 di 2 /dt + L 1 d(i 2 – i 1 )/dt + Mdi 2 /dt + Md(i 2 – i 1 )/dt 0 = L 2 di 2 /dt + L 1 di 2 /dt - L 1 di 1 /dt + Mdi 2 /dt + Md(i 2 – i 1 )/dt 0 = L 2 di 2 /dt + L 1 di 2 /dt - L 1 di 1 /dt + Mdi 2 /dt + Mdi 2 /dt - Mdi 1 /dt 0 = L 2 di 2 /dt + L 1 di 2 /dt - L 1 di 1 /dt + 2Mdi 2 /dt - Mdi 1 /dt 0 = - (L 1 + M) di 1 /dt + (L 1 + L 2 + 2M) di 2 /dt Writing in matrix form L 1 - (L 1 + M) di 1 /dt v =

• (L 1 + M) L 1 + L 2 + 2M di 2 /dt 0 v - (L 1 + M) 0 L 1 + L 2 + 2M di 1 /dt = L 1 - (L 1 + M)
• (L 1 + M) L 1 + L 2 + 2M v - (L 1 + M) 0 L 1 + L 2 + 2M = (v)( L 1 + L 2 + 2M) – 0 = (v)(L 1 + L 2 + 2M) L 1 - (L 1 + M)
• (L 1 + M) L 1 + L 2 + 2M = (L 1 )(L 1 + L 2 + 2M) - (L 1 + M)(L 1 + M)

Dedicated to: Prof. Dr. sohail aftab qureshi = (L 1 )(L 1 + L 2 + 2M) - (L 1 + M)^2 = (L 1 2

• L 1 L 2 + 2L 1 M) - M 2

• L 1 2 - 2ML 1 = L 1 2

• L 1 L 2 + 2L 1 M - M 2

• L 1 2
• 2ML 1 = L 1 L 2 – M^2 di 1 /dt = (v)(L 1 + L 2 + 2M)/L 1 L 2 – M^2 di 1 /dt {(L 1 L 2 – M^2 )/(L 1 + L 2 + 2M)} = v In network (c) v i 1 Leq v = Leqdi 1 /dt For (a) & (c) to be equal di 1 /dt {(L 1 L 2 – M 2 )/(L 1 + L 2 + 2M)} = Leqdi 1 /dt (L 1 L 2 – M^2 )/(L 1 + L 2 + 2M) = Leq 3 - 4. The network of inductors shown in the figure is composed of a 1-H inductor on each edge of a cube with the inductors connected to the vertices of the cube as shown. Show that, with respect to vertices a and b, the network is equivalent to that in (b) of the figure when Leq = 5/6 H. Make use of symmetry in working this problem, rather than writing kirchhoff laws. Solution: 1 - H Open your book & see the figure (P/88) 1 - H 1 - H 1 - H 1 - H 1 - H 1 - H 1 1’ 1 - H 1 - H 1 - H 1 - H