Supuestos de proyección de flujo de caja - Valoración de la empresa

5 Valoración de la empresa

5.1 Supuestos de proyección de flujo de caja

3W 2W

6 V 6W v_TH, R_TH

Fig. P.1.4.1

2W 2W

15 V i_N,R_N

Fig. P.1.4.2

3W 1W

2 A v_TH, R_TH

Fig. P.1.4.3

R i

i v

(A) (B)

i v R

Fig. P.1.4.4

4W 3W

6 A i_N R_N

Fig. P.1.4.5 GATE EC BY RK Kanodia

6. vTH, RTH= ?

(A) -100 V, 75 W (B) 155 V, 55W (C) 155 V, 37W (D) 145 V, 75 W 7. RTH= ?

(A) 3W (B) 12W

8. The Thevenin impedance across the terminals ab of the network shown in fig. P.1.4.8 is

(A) 2W (B) 6W

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

(D) 8 V 5 W ⁸₅A 5 W

10. v1 = ?

(A) 6 V (B) 7 V

11. i1= ?

(A) 3 A (B) 0.75 mA

Statement for Q.12–13:

A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question..

12. As viewed from terminal x and ¢x is

(A) 8 V, 6W (B) 5 V, 6W

Page

UNIT 1 Networks

30W 25W

20W 5 A

5 V v_TH, R_TH

Fig. P.1.4.6

2 A

5 V 6W

6W R_TH

Fig. P.1.4.7

6W 2 A

8W 2 V

b a

Fig. P.1.4.8

2W 3W

2 A 4 V

R_TH

v_TH

i_N R_N

Fig. P.1.4.9

1W 1W 3W

2W 6W

8 V v₁

– + 2W

6W 18 V

Fig. P.1.4.10

4 kW i1 4 kW 20 V 6 kW

4 kW 3 kW 24 V 12 V

Fig. P.1.4.11

10W

x’

16W x

5 V 40W 8W 1 A

y’

Fig. P.1.4.12–13 GATE EC BY RK Kanodia

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13. As viewed from terminal y and y¢ is

(A) 8 V, 32W (B) 4 V, 32W

14. A practical DC current source provide 20 kW to a 50W 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

Statement for Q.15–16:

In the circuit of fig. P.1.4.15–16 when R = 0 W , the current iR equals 10 A.

15. The value of R, for which it absorbs maximum power, is

(A) 4W (B) 3W

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) 2W (B) 4W

18. iN, RN= ?

(A) 2 A, 20W (B) 2 A, -20 W (C) 0 A, 20W (D) 0 A, -20 W

19. vTH, RTH= ?

(A) 0W (B) 1.2W

20. vTH, RTH= ?

(A) 8 V, 5W (B) 8 V, 10W

21. RTH= ?

(A) 3W (B) 1.2W

22. In the circuit shown in fig. P.1.4.22 the effective resistance faced by the voltage source is

(A) 4W (B) 3W

Chap 1.4 Network Theorems

4W 2W 2W

4W R 2W 4 A

i_R

Fig. P.1.4.15–16.

10W i1

30W 20i1

R_N i_N,

Fig. P.1.4.18

6W i1

4W 3i1

R_N i_N,

Fig. P1.4.19

v₁

– +

5W 0.1v1

4 V

v_TH R_TH

Fig. P.1.4.20

v_x

– +

2W 3W

4 V vx

RTH

Fig. P.1.4.21

i i

v_s 4

Fig. P.1.4.22 GATE EC BY RK Kanodia

23. In the circuit of fig. P1.4.23 the value of RTH at terminal ab is

(A) -3 W (B) 9

8W (C) -8

3W (D) None of the above

24. RTH= ?

(A) ¥ (B) 0

125W (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.26–27:

In the circuit shown in fig. P1.4.26–27 the maximum power transfer condition is met for the load RL .

26. The value of RL will be

(A) 2W (B) 3W

27. The maximum power is

(A) 0.75 W (B) 1.5 W

28. RTH= ?

(A) 100W (B) 136.4W

29. Consider the circuits shown in fig. P.1.4.29

Page

UNIT 1 Networks

0.75v_a

v_a –

+ 4W

a 8W

9 V

b Fig. P.1.4.23

–

+ v_a

200W

50W 100W va

100

R_TH

Fig. P.1.4.24

40W 100W

200W 3i RL

6 V

Fig. P.1.4.25

i_x

3W RL

0.9 A

2W 16 V

Fig. P.1.4.26–27

100W -2ix

0.01v_x

ix 800W

100W 300W

– +

R_TH

Fig. P.1.4.28

6W 6W

2W 2W

18 V 3 A 12 V

i_b

i_a 2W

6W 6W

2W 2W

8 V 12 V

12 V

Fig. P.1.4.29a & b GATE EC BY RK Kanodia

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The relation between ia and ib is (A) ib= + 6ia (B) ib= + 2ia

(C) ib= 15. ia (D) ib=ia

30. Req= ?

(A) 18W (B) 72

13W (C) 36

13 W (D) 9W

31. In the lattice network the value of RL for the maximum power transfer to it is

(A) 6.67W (B) 9W

Statement for Q.32–33:

A circuit is shown in fig. P.1.4.32–33.

32. If vs1 =vs2= V then the value of v6 a is

(A) 3 V (B) 4 V

33. If vs1 = V and v6 s2 = - V then the value of v6 a is

(A) 4 V (B) -4 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) P1±P2 (B) P1 ± P2

± 2 (D) (P1 P2)

± 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 2W, the power dissipated by the bulb is

(A) 80 W (B) 1800 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) 16W (B) 8W

Chap 1.4 Network Theorems

12W 4W

6W 2W

6W 9W 18W

R_eq

Fig. P.1.4.30

7W 6 W 5 W

9W Fig. P.1.4.31

12W

1W 3W 3W 1W

1W v_s2

v_s1

– +

v_a

Fig. P.1.4.32–33

N R N R N

Fig. P.1.4.34 GATE EC BY RK Kanodia

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) 0-10V 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

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) 300W (B) -300 W

***********

SOLUTIONS

1. (B) vTH=

+ =

( )( )6 6 3 6 4 V, RTH=( || )3 6 + =2 4W 2. (A)

RN =2 4+ =2 10

|| 3 W,

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

|| 6W

4. (B) After killing all source equivalent resistance is R Open circuit voltage = v1

5. (D) The short circuit current across the terminal is

isc= ´ iN

+ = =

6 4

4 2 4 A ,

RN =6||3 2= W

6. (B) For the calculation of RTH if we kill the sources then 20W resistance is inactive because 5 A source will be open circuit

RTH=30+25=55W, vTH= + ´5 5 30=155 V

Page

UNIT 1 Networks

R_L Linear Network

– + v_ab

Fig. P.1.4.39

2 1 -1 -2 -3

-4 0

i(mA)

30 20 10

Resistive Network

– + v_ab

Fig. P.1.4.40

2W v1 2W

4W 15 V

i_sc

Fig. S.1.4.2

3W 4W

6 A

i_sc

Fig. S1.4.5 GATE EC BY RK Kanodia

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7. (C) After killing the source, RTH= 6 W

8. (B) After killing all source, RTH=3 6|| +8||8=6W

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

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

12. (B) We Thevenized the left side of xx¢ and source transformed right side of yy¢

vxx¢=vTH= +

13. (D) Thevenin equivalent seen from terminal yy¢ is vyy¢=vTH= +

GATE EC BY RK Kanodia

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

34. (C) Let Thevenin equivalent of both network

P V

sensitivity = =

k mA Solving (i) and (ii)

vTH=16

UNIT 1 Networks

v_TH

GATE EC BY RK Kanodia

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1. The natural response of an RLC circuit is described by the differential equation

d v dt

dv dt v

2 +2 + = , v( )0 0 = , dv10 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¢¢( )+3000v t¢( )+102. ´10⁸v t( )=10⁸v ts( ) (B) v t¢¢( )+1000v t¢( )+102. ´10⁸v t( )=10⁸v ts( ) (C) v t¢¢ + v t¢ + =

v t v ts

( ) ( )

. ( ) ( ) 10

10 102

8 5

(D) v t¢¢ + v t¢ + = v t v ts

( ) ( )

. ( ) ( ) 10

10 198

8 5

3.The differential equation for the circuit shown in fig.

P1.6.3 is

(A) i t¢¢L( )+1100i t¢¢L( )+11 10´ ⁸i tL( )=10⁸i ts( ) (B) i t¢¢L( )+1100i t¢¢L( )+11 10´ ⁸i tL( )=10⁸i ts( ) (C) i t¢¢ + i t¢¢ + =

i t i t

L L

L s

( ) . ( )

. ( ) ( ) 10

10 11

8 4

(D) i t¢¢ + i t¢ + = i t i t

L L

L s

( ) ( )

( ) ( ) 10

10 11

8 4

4.In the circuit of fig. P.1.6.4 vs= 0 for t > 0. The initial condition are v( )0 = V and dv6 ( )0 dt= -3000 V s. The v t( ) for t > 0 is

(A) -2e^-¹⁰⁰^t+8e^-⁴⁰⁰^t V (B) 6e^-¹⁰⁰^t+8e^-⁴⁰⁰^t V (C) 6e^-¹⁰⁰^t-8e^-⁴⁰⁰^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 = V. The v t2 ( ) for t > is

(A) 5e^-^t-7e^-³^t V (B) 7e^-^t-5e^-³^t V (C) -e^-^t+3e^-³^t V (D) 3e^-^t-e^-³^t V

CHAPTER

1.6

In document ESCUELA SUPERIOR POLITÉCNICA DEL LITORAL. Facultad de Ciencias Sociales y Humanísticas (página 45-0)