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Problems Section 2-2 Engineering and Linear Models P2.2-1 The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P2.2-2 (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and the line passes through the origin so the equation of the line is 0.12v i= . The element is indeed linear. (b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV (c) When v = 4 V, 4 33.3 A 0.12 i = = P2.2-3 (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is 256.5v i= . The element is indeed linear. (b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V (c) When v = 12 V, 12 0.04678 256.5 i = = A = 46.78 mA. P2.2-4 Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence, the property of homogeneity is not satisfied. The element is not linear. P2.2-5 (a) 0.4 3.2 V 10 40 8 v v v v= + = ⇒ = 0.08 A 40 vi = = (b) 2 20.4 0.8 0 10 2 5 v v vv= + ⇒ + − = Using the quadratic formula 0.2 1.8 0.8, 1.0 V 2 v − ±= = − When v = 0.8 V then 20.8 0.32 A 2 i = = . When v = -1.0 V then ( )21 0.5 A 2 i −= = . (c) 2 20.4 0.8 0.8 0 10 2 5 v v vv= + + ⇒ + + = Using the quadratic formula 0.2 0.04 3.2 2 v − ± −= So there is no real solution to the equation. Section 2-4 Resistors P2.4-1 3 A and = 7 3 = 21 V and adhere to the passive convention 21 3 = 63 W is the power absorbed by the resistor. si i v Ri v i P v i = = = × ∴ = = × P2.4-2 3 3 3 mA and 24 V 24 8000 8 k .003 = (3 10 ) 24 = 72 10 72 mW si i v vR i P − − = = = = = = = Ω × × × = P2.4-3 =10 V and 5 10 2 A 5 and adhere to the passive convention 2 10 20 W is the power absorbed by the resistor sv v R vi R v i p v i = = Ω = = = ∴ = = ⋅ = P2.4-4 24 V and 2 A 24 12 2 24 2 = 48 W sv v i vR i p vi = = = = = = Ω = = ⋅ P2.4-5 1 2 1 2 1 1 1 1 1 150 V; 50 ; 25 and adhere to the passive convention so 150 3 A 50 sv v v R R v i vi R = = = = Ω = Ω = = = 2 2 2 2 2 150 and do not adhere to the passive convention so 6 A 25 vv i i R = − = − = − 1 1 1 1The power absorbed by is 150 3 450 WR P v i= = ⋅ = 2 2 2 2The power absorbed by is 150( 6) 900 WR P v i= − = − − = P2.4-6 1 2 1 2 1 1 1 1 1 1 1 1 1 2 A ; =4 and 8 and do not adhere to the passive convention so 4 2 8 V. The power absorbed by is ( 8)(2) 16 W. si i i R R v i v R i R P v i = = = Ω = Ω =− =− ⋅ =− =− =− − = 2 2 2 2 2 2 2 2 2 and do adhere to the passive convention so 8 2 16 V . The power absorbed by 16 2 32 W. v i v R i R is P v i = = ⋅ = = = ⋅ = P2.4-7 2 2 2 2 2 Model the heater as a resistor, then (250)with a 250 V source: 62.5 1000 (210)with a 210 V source: 705.6 W 62.5 v vP R R P vP R = ⇒ = = = = = = Ω P2.4-8 2 2 2 5000 125The current required by the mine lights is: A 120 3 Power loss in the wire is : Thus the maximum resistance of the copper wire allowed is 0.05 0.05 5000 0.144 (125/3) now Pi v i R PR i = = = ×= = = Ω 6 6 2 since the length of the wire is 2 100 200 m 20,000 cm thus / with = 1.7 10 cm from Table 2.5 1 1.7 10 20,000 0.236 cm 0.144 L R L A LA R ρ ρ ρ − − = × = = = × Ω⋅ × ×= = = − *P2.4-9 380 4200.7884 0.8108 102 380 98 420 gain= ≤ ≤ =+ + 0.7884 0.8108 0.7996 2 nominal gain += = 0.7996 0.7884 0.8108 0.7996 100 100 1.40% 0.7996 0.7996 gain tolerance − −= × = × = So 0.7996 1.40%gain = ± P2.4-10 Label the current i as shown. That current is the element current in both resistors. First a 40 v i = Next a b b a 40 40 v v v R i R R v = = ⇒ = For example, 7.0540 24 11.75 R = = Ω Section 2-5 Independent Sources P2.5-1 (a) ( )2215 3 A and 5 3 45 W 5 svi P R i R = = = = = = (b) and do not depend on .si P i The values of and are 3 A and 45 W, both when 3 A and when 5 A.s si P i i= = P2.5-2 (a) 2 2 10 5 2 10 V and 20 W 5s vv R i P R = = ⋅ = = = = (b) and do not depend on . sv P v The values of and are 10V and 20 W both when 10 V and when 5 Vs sv P v v= = P2.5-3 Consider the current source: and do not adhere to the passive convention, so 3 12 36 W is the power supplied by the current source. s s cs s s i v P i v= = ⋅ = Consider the voltage source: and do adhere to the passive convention, so 3 12 36 W is the power absorbed by the voltage source. The voltage source supplies 36 W. s s vs s s i v P i v= = ⋅ = ∴ − P2.5-4 Consider the current source: and adhere to the passive convention so 3 12 36 W is the power absorbed by the current source. Current source supplies 36 W. s s cs s s i v P i v= = ⋅ = − Consider the voltage source: and do not adhere to the passive convention so 3 12 36 W is the power supplied by the voltage source. s s vs s s i v P i v= = ⋅ = P2.5-5 (a) 2(2 cos ) (10 cos ) 20 cos mWP v i t t t= = = (b) 1 1 1 2 0 0 0 1 1 20 cos = 20 sin 2 10 5 sin 2 mJ 2 4 w P dt t dt t t⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∫ ∫ P2.5-6 (a) capacity 800 mAhtime to discharge 16 hours current 50 mA = = = Section 2-6 Voltmeters and Ammeters P2.6-1 (a) 5 10 0.5 vR i = = = Ω (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source adhere to the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers -6 W. P2.6-2 The voltmeter current is zero so the ammeter current is equal to the current source current except for the reference direction: i = -2 A The voltage v is the voltage of the current source. The power supplied by the current source is 40 W so 40 2 20 Vv v= ⇒ = P2.6-3 (a) m 900 12 10.8 V 900 100 v ⎛ ⎞= =⎜ ⎟+⎝ ⎠ 12 10.8 0.1 10% 12 − = = (b) We require m m m m m 12 12 100 0.02 0.98 4900 12 100 R R R R R ⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒+ ≥ Ω (checked: LNAP 6/16/04) P2.6-4 (a) m 1000 2 1.98 A 1000 10 i ⎛ ⎞= =⎜ ⎟+⎝ ⎠ 2 1.98% error 100 0.99% 2 −= × = (b) m m m 10002 2 1000 10000.05 0.95 52.63 2 1000 R R R ⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒+ ≤ Ω (checked: LNAP 6/17/04) P2.6-5 a.) ( )R R25 25 2 50 Vv i= = − = − ( )m R12 12 50 62 Vv v= − = − − = b.) Element Power supplied voltage source ( ) ( )s12 12 2 24 Wi− = − = − current source ( )62 2 124 W= resistor ( ) ( )R R 50 2 100 Wv i− × = − − − = − total 0 P2.6-6 a.) R R 12 0.48 A 25 25 v i = = = m R 2 0.482 1.52 Ai i= − = − = − b.) Element Power supplied voltage source ( ) ( )m12 12 1.52 18.24 Wi = − = − current source ( ) ( )s 2 12 2 =24 Wv = resistor ( )( )R R 12 0.48 5.76 Wv i− × = − = − total 0 Section 2-7 Dependent Sources P2.7-1 8 4 2 b a vr i = = = Ω P2.7-2 2 A8 V ; 2 A ; 0.25 8 V a b b a b iv g v i g v = = = = = = P2.7-3 32 A 8 A ; 32A ; 4 8 A a b b a b ii d i i d i = = = = = = P2.7-4 8 V2 V ; 8 V ; 4 2 V b a a b a vv b v v b v = = = = = = P2.7-5 4 2 2 R = − = Ω− and 2 V4 0.5 A A = = −− (checked: LNAP 6/6/04) P2.7-6 2 V, 4 8 A and 2.2 Vc d c dv i v v= − = = − = id and vd adhere to the passive convention so (2.2) ( 8) 17.6 Wd dP v i= = − = − is the power received by the dependent source. The power supplied by the dependent source is 17.6 W. P2.7-7 1.25 A, 2 2.5 V and 1.75 Ac d c di v i i= = = = id and vd adhere to the passive convention so (2.5) (1.75) 4.375 Wd dP v i= = = is the power received by the dependent source. Section 2-8 Transducers P2.8-1 360 = , = 360 (360)(23V)= = 75.27 (100 k )(1.1 mA) m p va R I θ θ θ °Ω P2.8-2 AAD590 : =1 , K =20 V (voltage condition satisfied) k v μ ° 4 A < < 13 A 4 K< <13 K i TiT k μ μ ° ° ⎫⎪ ⇒⎬= ⎪⎭ Section 2-9 Switches P2.9-1 At t = 1 s the left switch is open and the right switch is closed so the voltage across the resistor is 10 V. 3 10= = 2 mA 5 10 vi R =× At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 15 V. 3 15= = 3 mA 5 10 vi R =× P2.9-2 At t = 1 s the current in the resistor is 3 mA so v = 15 V. At t = 4 s the current in the resistor is 0 A so v = 0 V. P2.9-3 (a) v = 12 V (b) 100 12 11.43 V 105 v ⎛ ⎞= =⎜ ⎟⎝ ⎠ (c) v = 0 V (d) 100 12 0.1188 0.12 V 10100 v ⎛ ⎞= =⎜ ⎟⎝ ⎠ � Section 2-10 How Can We Check…? P2.10-1 =40 V and = ( 2) 2 A. (Notice that the ammeter measures rather than .) 40 V So 20 2 A Your lab partner is wrong. o s s s o s v i i v i − − = − = = i P2.10-2 12We expect the resistor current to be = 0.48 A. The power absorbed by 25 this resistor will be = (0.48) (12) = 5.76 W. A half watt resistor can't absorb this much power. You should n s s vi R P i v = = = ot try another resistor. CH2sec2 Problems Section 2-2 Engineering and Linear Models P2.2-1 P2.2-2 P2.2-3 P2.2-4 P2.2-5 CH2sec4 Section 2-4 Resistors P2.4-1 P2.4-2 P2.4-3 P2.4-4 P2.4-5 P2.4-6 P2.4-7 P2.4-8 *P2.4-9 P2.4-10 CH2sec5 Section 2-5 Independent Sources P2.5-1 P2.5-2 P2.5-3 P2.5-4 P2.5-5 P2.5-6 CH2sec6 Section 2-6 Voltmeters and Ammeters P2.6-1 P2.6-2 P2.6-3 P2.6-4 P2.6-5 P2.6-6 CH2sec7 Section 2-7 Dependent Sources P2.7-1 P2.7-2 P2.7-3 P2.7-4 P2.7-5 P2.7-6 P2.7-7 CH2sec8 Section 2-8 Transducers P2.8-1 P2.8-2 CH2sec9 Section 2-9 Switches P2.9-1 P2.9-2 P2.9-3 CH2sec10 Section 2-10 How Can We Check…? P2.10-1 P2.10-2
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