[a] The 5 kω and 7 kω resistors are in series. The simplified circuit is shown below:

Transcription

1 P 3.3 [a] The 5 kω and 7 kω resistors are in series. The simplified circuit is shown below: [b] The 800Ω and 1200Ω resistors are in series, as are the 300Ω and 200Ω resistors. The simplified circuit is shown below:

2 [c] The 35 Ω, 15 Ω, and 25 Ω resistors are in series. as are the 10 Ω and 40 Ω resistors. The simplified circuit is shown below: [d] The 50Ω and 90Ω resistors are in series, as are the 80Ω and 70Ω resistors. The simplified circuit is shown below: P 3.4 [a] The 36 Ω and 18 Ω resistors are in parallel. The simplified circuit is shown below: [b] The 200Ω and 120Ω resistors are in parallel, as are the 210Ω and 280Ω resistors. The simplified circuit is shown below: [c] The 100 kω, 150 kω, and 60 kω resistors are in parallel, as are the 75 kω and 50 kω resistors. The simplified circuit is shown below:

3 [d] The 750Ω and 500Ω resistors are in parallel, as are the 1.5 kω and 3 kω resistors. The simplified circuit is shown below:

4 P 3.7 [a] Circuit in Fig. P3.7(a): R eq = ([(15 60) + (30 45) + 20] 50) = [( ) 50] = (50 50) = = 60Ω Circuit in Fig. P3.7(b) begin by simplifying the 75Ω resistor and all resistors to its right: [( ) ] 75 = ( ) 75 = ( ) 75 = = 30Ω Now simplify the remainder of the circuit: R eq = ([( ) 50] + (20 60)) 40 = [(50 50) + 15] 40 = ( ) 40 = = 20Ω Circuit in Fig. P3.7(c) begin by simplifying the left and right sides of the circuit: R left = [( ) 2000] = ( ) = = 1500Ω R right = [( ) 1000] = ( ) = = 1500Ω Now find the equivalent resistance seen by the source: R eq = (R left R right ) = ( ) = = 4000 = 4 kω Circuit in Fig. P3.7(d): R eq = ([( ) 1000] + 100) ([( ) 500] + 300) = [( ) + 100] [( ) + 300] = ( ) ( ) = = 300Ω [b] Note that in every case, the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit. For the circuit in Fig. P3.7(a): P = V s 2 = 302 R eq 60 = 15 W For the circuit in Fig. P3.7(b): P = I 2 s (R eq) = (0.08) 2 (20) = = 128 mw For the circuit in Fig. P3.7(c): P = V s 2 = 202 R eq 4000 = 0.1 = 100 mw For the circuit in Fig. P3.7(d): P = I 2 s(r eq ) = (0.05) 2 (300) = 0.75 = 750 mw

5 P 3.12 [a] v o = 160(3300) ( ) = 66 V [b] i = 160/8000 = 20 ma P R1 = ( )( ) = 1.88 W P R2 = ( )( ) = 1.32 W [c] Since R 1 and R 2 carry the same current and R 1 > R 2 to satisfy the voltage requirement, first pick R 1 to meet the 0.5 W specification ( ) i R1 =, Therefore, R R 1 Thus, R R 1 or R 1 17,672Ω Now use the voltage specification: R 2 (160) = 66 R ,672 Thus, R 2 = 12,408Ω P = 20R 2 R so R 2 = 10Ω 3 = 20R e 40 + R e so R e = Ω Thus, = 10R L 10 + R L so R L = 24Ω

6 P 3.16 R eq = 10 [6 + 5 (8 + 12)] = 10 ( ) = 10 (6 + 4) = 5Ω v 10A = v 10Ω = (10 A)(5Ω) = 50 V Using voltage division: v 5Ω = 5 (8 + 12) (8 + 12) (50) = 4 (50) = 20 V Thus, p 5Ω = v2 5Ω 5 = = 80 W

7 P 3.20 [a] 20 kω + 40 kω = 60 kω 30 kω 60 kω = 20 kω v o1 = 20,000 (180) = 120 V (10, ,000) v o = 40,000 60,000 (v o1) = 80 V

8 [b] i = ,000 30,000i = 135 V = 4.5 ma v o = 40,000 (135) = 90 V 60,000 [c] It removes the loading effect of the second voltage divider on the first voltage divider. Observe that the open circuit voltage of the first divider is v o1 = 30,000 (180) = 135 V 40,000 Now note this is the input voltage to the second voltage divider when the current-controlled voltage source is used.

9 P 3.23 [a] The equivalent resistance of the 6 kω resistor and the resistors to its right is 6 k (5 k + 7 k) = 6 k 12 k = 4 kω

10 Using voltage division, v 6k = [b] v 5k = (18) = 6 V 5000 (6) = 2.5 V P 3.24 [a] The equivalent resistance of the 100Ω resistor and the resistors to its right is 100 ( ) = = 60Ω Using current division, i 50 = [b] v 70 = ( ) (0.03) = 120 (0.03) = = 18 ma 200 ( ) 100 (0.018) = 60 (0.018) = = 7.2 ma

11 P 3.28 [a] v 6k = 6 (18) = 13.5 V v 3k = 3 (18) = 4.5 V v x = v 6k v 3k = = 9 V [b] v 6k = 6 8 (V s) = 0.75V s v 3k = 3 12 (V s) = 0.25V s v x = (0.75V s ) (0.25V s ) = 0.5V s

12 P 3.32 Use current division to find the current in the 8Ω resistor. Begin by finding the equivalent resistance of the 8Ω resistor and all resistors to its right: R eq = ([(20 80) + 4] 30) + 8 = 20Ω i 8 = 60 R eq (0.25) = (0.25) = = ma R eq 20 Use current division to find i 1 from i 8 : i 1 = 30 [4 + (80 20)] 30 (i 8 ) = (0.1875) = = 75 ma 30

13 Use current division to find i 4Ω from i 8 : i 4Ω = 30 [4 + (80 20)] 4 + (80 20) (i 8 ) = (0.1875) = = ma 20 Finally, use current division to find i 2 from i 4Ω : i 2 = (i 4Ω) = (0.1125) = 0.09 = 90 ma 20

14 P 3.58 [a] Use the figure below to transform the Y to an equivalent : R a = R b = R c = (25)(30) + (25)(50) + (30)(50) 30 (25)(30) + (25)(50) + (30)(50) 50 (25)(30) + (25)(50) + (30)(50) 25 = = Ω = = 70Ω = = 140Ω

15 Replace the Y with its equivalent in the circuit to get the figure below: Find the equivalent resistance to the right of the 13Ω and 7Ω resistors: 70 [( ) + (20 140)] = 30Ω Thus, the equivalent resistance seen from the terminals a-b is: R ab = = 50Ω [b] Use the figure below to transform the to an equivalent Y: R 1 = (50)(20) = 10Ω R 2 = (50)(30) = 15Ω R 3 = (20)(30) = 6Ω

16 Replace the with its equivalent Y in the circuit to get the figure below: Find the equivalent resistance to the right of the 13Ω and 7Ω resistors: ( ) ( ) + 6 = 30Ω Thus, the equivalent resistance seen from the terminals a-b is: R ab = = 50Ω [c] Convert the delta connection R 1 R 2 R 3 to its equivalent wye. Convert the wye connection R 1 R 3 R 4 to its equivalent delta.

17 P 3.61 [a] After the 20 Ω 100 Ω 50 Ω wye is replaced by its equivalent delta, the circuit reduces to

18 Now the circuit can be reduced to i = 96 (1000) = 240 ma 400 i o = 400 (240) = 96 ma 1000 [b] i 1 = 80 (240) = 48 ma 400 [c] Now that i o and i 1 are known return to the original circuit v 2 = (50)(0.048) + (600)(0.096) = 60 V i 2 = v = = 600 ma [d] v g = v ( ) = = V p g = (v g )(1) = W Thus the current source delivers W.

19 P 3.67 [a] After making the Y-to- transformation, the circuit reduces to

20 Combining the parallel resistors reduces the circuit to Now note: 0.75R + 3RR L = 2.25R RR L 3R + R L 3R + R L Therefore ( 2.25R 2 ) RR L 3R 3R + R L R ab = ( 2.25R 2 ) = 3R(3R + 5R L) RR L 15R + 9R L 3R + 3R + R L If R = R L, we have R ab = 3R L(8R L ) 24R L Therefore R ab = R L = R L [b] When R = R L, the circuit reduces to i o = i i(3r L ) = 1 4.5R L 1.5 i i = v i R L, v o = 0.75R L i o = 1 2 v i, Therefore v o v i = 0.5