24 A 12 A4 \u2126 48 V V 5 I 5 R 5 I 4 I 7 R 5 4 A6 \u2126 24 V V 7 I 7 R 7 8 A2 \u2126 16 V d

24 a 12 a4 ω 48 v v 5 i 5 r 5 i 4 i 7 r 5 4 a6 ω

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= (24 A 12 A)4 = 48 V V 5 = I 5 R 5 = ( I 4 I 7 ) R 5 = (4 A)6 = 24 V V 7 = I 7 R 7 = (8 A)2 = 16 V d. P = 2 7 7 I R = (8 A) 2 2 = 128 W P = EI = (240 V)(24 A) = 5760 W 26. a. R T = R 4 || ( R 6 + R 7 + R 8 ) = 2 || 7 = 1.56 R T = R 2 || ( R 3 + R 5 + R T ) = 2 || (4 + 1 + 1.56 ) = 1.53 R T = R 1 + R T = 4 + 1.53 = 5.53 b. I = 2 V/5.53 = 361.66 mA c. I 3 = 2 ( ) 2 (361.66 mA) 2 6.56 2 6.56 I Ω Ω = Ω + Ω Ω + Ω = 84.50 mA I 8 = 2 (84.5 mA) 2 + 7 Ω Ω Ω = 18.78 mA 27. The 12 resistors are in parallel. Network redrawn: R T = 12 I s = 24 V 12 T E = R Ω = 2 A I 2 = 2 A 2 2 s I = = 1 A I 12 = 2 24 ( ) 2 A 24 12 3 I Ω Ω = Ω + Ω P 10 = ( I 10 ) 2 10 = 2 2 A 10 3 Ω = 4.44 W 28. a. R 10 + R 11 || R 12 = 1 + 2 || 2 = 2 R 4 || ( R 5 + R 6 ) = 10 || 10 = 5 R 1 + R 2 || ( R 3 + 5 ) = 3 + 6 || 6 = 6 R T = 2 || 3 || 6 = 2 || 2 = 1 I = 12 V/1 = 12 A b. I 1 = 12 V/6 = 2 A I 3 = 6 (2 A) 6 + 6 Ω Ω Ω = 1 A I 4 = 1 A 2 = 0.5 A c. I 6 = I 4 = 0.5 A d. I 10 = 12 A 2 = 6 A
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Chapter 7 61 29. a. E = (40 mA)(1.6 k Ω ) = 64 V b. 2 48 V 12 mA L R = = 4 k Ω 3 24 V 8 mA L R = = 3 k Ω c. 1 R I = 72 mA 40 mA = 32 mA 2 R I = 32 mA 12 mA = 20 mA 3 R I = 20 mA 8 mA = 12 mA R 1 = 1 1 64 V 48 V 16 V = 32 mA 32 mA R R V = I = 0.5 k Ω R 2 = 2 2 48 V 24 V 24 V = 20 mA 20 mA R R V = I = 1.2 k Ω R 3 = 3 3 24 V 12 mA R R V = I = 2 k Ω 30. 1 R I = 40 mA 2 R I = 40 mA 10 mA = 30 mA 3 R I = 30 mA 20 mA = 10 mA 5 R I = 40 mA 4 R I = 40 mA 4 mA = 36 mA R 1 = 1 1 120 V 100 V 20 V = 40 mA 40 mA R R V = I = 0.5 k Ω R 2 = 2 2 100 V 40 V 60 V = 30 mA 30 mA R R V = I = 2 k Ω R 3 = 3 3 40 V 10 mA R R V = I = 4 k Ω R 4 = 4 4 36 V 36 mA R R V = I = 1 k Ω R 5 = 5 5 60 V 36 V 24 V = 40 mA 40 mA R R V = I = 0.6 k Ω
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Chapter 7 62 P 1 = 2 1 1 I R = (40 mA) 2 0.5 k Ω = 0.8 W (1 watt resistor) P 2 = 2 2 2 I R = (30 mA) 2 2 k Ω = 1.8 W (2 watt resistor) P 3 = 2 3 3 I R = (10 mA) 2 4 k Ω = 0.4 W (1/2 watt or 1 watt resistor) P 4 = 2 4 4 I R = (36 mA) 2 1 k Ω = 1.3 W (2 watt resistor) P 5 = 2 5 5 I R = (40 mA) 2 0.6 k Ω = 0.96 W (1 watt resistor) All power levels less than 2 W . Four less than 1 W . 31. a. yes, R L ± R max (potentiometer) b. VDR: 2 R V = 3 V = 2 2 1 2 (12 V) (12 V) = 1 k R R R R + Ω R 2 = 3 V(1 k ) 12 V Ω = 0.25 k Ω = 250 Ω R 1 = 1 k Ω 0.25 k Ω = 0.75 k Ω = 750 Ω c. 1 R V = E V L = 12 V 3 V = 9 V 1 R V = 9 V = 1 1 2 (12 V) ( ) L R R R R + & (Chose 1 R V rather than 2 L R R V & since numerator of VDR equation "cleaner") 9 R 1 + 9( R 2 || R L ) = 12 R 1 1 2 1 2 3( ) 2 eq. 2 unk( = 10 k ) 1 k L L R R R R R R = Ω + = Ω & R 1 = 2 2 3 L L R R R R + 2 2 3 10 k 10 k R R Ω + Ω and R 1 ( R 2 + 10 k Ω ) = 30 k Ω R 2 R 1 R 2 + 10 k Ω R 1 = 30 k Ω R 2 R 1 + R 2 = 1 k Ω : (1 k Ω R 2 ) R 2 + 10 k Ω (1 k Ω R 2 ) = 30 k Ω R 2 2 2 R + 39 k Ω R 2 10 k Ω 2 = 0 R 2 = 0.255 k Ω , 39.255 k Ω R 2 = 255 Ω R 1 = 1 k Ω R 2 = 745 Ω 32. a. V ab = 80 (40 V) 100 Ω Ω = 32 V V bc = 40 V 32 V = 8 V b. 80 Ω || 1 k Ω = 74.07 Ω 20 Ω || 10 k Ω = 19.96 Ω V ab = 74.07 (40 V) 74.07 + 19.96 Ω Ω Ω = 31.51 V V bc = 40 V 31.51 V = 8.49 V
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Chapter 7 63 c. P = 2 2 (31.51 V (8.49 V ) ) + 80 20 Ω Ω = 12.411 W + 3.604 W = 16.02 W d. P = 2 2 (32 V (8 V ) ) + 80 20 Ω Ω = 12.8 W + 3.2 W = 16 W The applied loads dissipate less than 20 mW of power. 33. a. I CS = 1 mA b. R shunt = max (100 )(1 mA) = 20 A 1 mA m CS CS R I I I Ω 0.1 20 = 5 m 34. 25 mA: R shunt = (1 k )(50 A) 25 mA 0.05 mA μ Ω 2 50 mA: R shunt = (1 k )(50 A) 50 mA 0.05 mA μ Ω = 1 100 mA: R shunt 0.5 35. a. R s = max 15 V (50 A)(1 k ) = 50 A VS CS V V I μ μ Ω = 300 k b. /V = 1/ I CS = 1/50 μ A = 20,000 36. 5 V: R s = 5 V (1 mA)(100 ) 1 mA
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