chapter 02 - Chapter 2 - Circuit Elements Exercises Ex....

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Unformatted text preview: Chapter 2 - Circuit Elements Exercises Ex. 2.3-1 m ( i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied Therefore the element is linear. Ex. 2.3-2 m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied Therefore the element is not linear. Ex. 2.5-1 v 2 (10 ) P= = =1 W R 10 0 2 Ex. 2.5-2 P= v 2 (10 cos t ) 2 = = 10 cos 2 t W R 10 Ex. 2.8-1 ic = − 1 .2 A , vd = 2 4 id = 4 ( − 1 .2) = − 4 .8 V A id and vd adhere to the passive convention so P = vd id = (24) (−4.8) = −115.2 W is the power received by the dependent source 2-1 Ex. 2.8-2 vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V id and vd adhere to the passive convention so P = vd id = (2.2) (−8) = −17.6 W is the power received by the dependent source. The power supplied by the dependent source is 17.6 W. Ex. 2.8-3 ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A id and vd adhere to the passive convention so P = vd id = (2.5) (1.75) = 4.375 W is the power received by the dependent source. 2-2 Ex. 2.9-1 θ = 4 5 ° , I = 2 m A, R p = 2 0 k Ω a= θ 45 (20 kΩ) = 2.5 kΩ ⇒ aR = p 360 360 vm = (2 ×10−3 )(2.5 ×103 ) = 5 V Ex. 2.9-2 v = 10 V, i = 280 µA, k = 1 µA °K for AD590 °K i = 280° K i = kT ⇒ T = = (280µA)1 µA k Ex. 2.10-1 At t = 4 s both switches are open, so i = 0 A. Ex. 2.10.2 At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V . At t = 6 s the switch is in the down position, so v = 0 V. Problems Section 2-3 Engineering and Linear Models P2.3-1 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. P2.3-2 (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 v = 0.12 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 4 (c) When v = 4 V, i = = 33 A = 33 A. 0.12 2-3 P2.3-3 (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 v = 256.5 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 12 (c) When v = 12 V, i = = 0.04678 A = 46.78 mA. 256.5 P2.3-4 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. Section 2-5 Resistors P2.5-1 i = is = 3 A and v = R i = 7 × 3 = 21 V v and i adhere to the passive convention ∴ P = v i = 21 × 3 = 63 W is the power absorbed by the resistor. P2.5-2 i = is = 3 m A a n d v = 2 4 V v 24 = = 8 0 00 = 8 k Ω i .003 P = (3×1 0 −3 )× 2 4 = 7 2×10 −3 = 7 2 m W R= P2.5-3 v = vs =10 V an d R = 5 Ω v 10 = =2 A R 5 v and i adhere to the passive convention ∴ p = v i = 2⋅10 = 2 0 W is the pow er absorbed by the resistor i= 2-4 P2.5-4 v = vs = 24 V and i = 2 A v 24 = = 12 Ω i 2 p = vi = 24⋅2 = 48 W R= P2.5-5 v1 = v 2 = vs = 150 V; R1 = 50 Ω; R2 = 25 Ω v 1 and i1 adhere to the passive convention so v 1 150 = =3 A R 1 50 v 150 v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = − = −6 A R2 25 i1 = The power absorbed by R1 is P = v1 i1 = 150 ⋅ 3 = 450 W 1 The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W P2.5-6 i1 = i 2 = is = 2 A ; R1 =4 Ω and R2 = 8 Ω v 1 and i 1 do not adhere to the passive convention so v 1 =− R 1 i 1 =−4⋅2= −8 V. The power absorbed by R 1 is P1 =−v 1i 1 =−(−8)(2) = 16 W. v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V . The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W. P2.5-7 Model the heater as a resistor, then (250) 2 v2 v2 ⇒R= = = 62.5 Ω 1000 R P v 2 (210) 2 = 705.6 W with a 210 V source: P = = R 62.5 with a 250 V source: P = 2-5 P2.5-8 The current required by the mine lights is: i = P 5000 125 A = = 3 v 120 Power loss in the wire is : i 2 R Thus the maximum resistance of the copper wire allowed is 0.05P 0.05×5000 = = 0.144 Ω (125/3) 2 i2 now since the length of the wire is L = 2×100 = 200 m = 20,000 cm R= thus R = ρ L / A with ρ = 1.7×10−6 Ω ⋅ cm from Table 2.5−1 A= ρL R = 1.7×10−6 ×20,000 = 0.236 cm 2 0.144 Section 2-6 Independent Sources P2.6-1 v s 15 2 = = 3 A a nd P = R i 2 = 5 ( 3 ) = 4 5 W R 5 (b) i and P do not depend on is . (a) i = The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A. P2.6-2 v 2 102 = 20 W (a) v = R i s = 5 ⋅ 2 = 10 V and P = = R 5 (b) v and P do not depend on v s . The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V 2-6 P2.6-3 Consider the current source: i s and v s do not adhere to the passive convention, so Pcs =i s v s =3⋅12 = 36 W is the power supplied by the current source. Consider the voltage source: i s and v s do adhere to the passive convention, so Pvs = i s vs =3 ⋅12 = 36 W is the power absorbed by the voltage source. ∴ The voltage source supplies −36 W. P2.6-4 Consider the current source: i s and vs adhere to the passive convention so Pcs = i s vs =3 ⋅12 = 36 W is the power absorbed by the current source. Current source supplies − 36 W. Consider the voltage source: i s and vs do not adhere to the passive convention so Pvs = i s vs = 3 ⋅12 =36 W is the power supplied by the voltage source. P2.6-5 (a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW 1 1 1 (b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t = 10 + 5 sin 2 mJ 2 4 0 1 1 2 2-7 Section 2-7 Voltmeters and Ammeters P2.7-1 (a) R = v 5 = = 10 Ω i 0.5 (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. P2.7-2 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 v ⇒ v = 20 V 2-8 Section 2-8 Dependent Sources P2.8-1 r= vb 8 = =4 Ω ia 2 P2.8-2 vb = 8 V ; g v b = i a = 2 A ; g = ia 2 A = = 0.25 vb 8 V i b = 8 A ; d i b = i a = 32 A ; d = i a 32 A = =4 ib 8 A va = 2 V ; b va = vb = 8 V ; b = vb 8 V = =4 va 2 V P2.8-3 P2.8-4 Section 2-9 Transducers P2.9-1 a= θ= θ 360 , θ= 360 vm Rp I (360)(23V) = 75.27° (100 kΩ)(1.1 mA ) P2.9-2 AD590 : k =1 µA , K v =20 V (voltage condition satisfied) ° 4 µ A < i < 13 µ A i T= k ⇒ 4 ° K < T < 1 3° K 2-9 Section 2-10 Switches P2.10-1 At t = 1 s the left switch is open and the right switch is closed so the voltage across the resistor is 10 V. i= v 10 = = 2 mA R 5×103 At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 15 V. i= v 15 = = 3 mA R 5×103 P2.10-2 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. Verification Problems VP2-1 vo =40 V and i s = − (−2) = 2 A. ( Notice that the ammeter measures − i s rather than i s .) So vo 40 V = = 20 2 A is Your lab partner is wrong. VP2-2 vs 12 = 0.48 A. The power absorbed by = R 25 this resistor will be P = i vs = (0.48) (12) = 5.76 W. We expect the resistor current to be i = A half watt resistor can't absorb this much power. You should not try another resistor. 2-10 Design Problems DP2-1 1.) 10 10 > 0.04 ⇒ R < = 250 Ω R 0.04 2.) 102 1 < ⇒ R > 200 Ω R 2 Therefore 200 < R < 250 Ω. For example, R = 225 Ω. DP2-2 1.) 2 R > 40 ⇒ R > 20 Ω 15 2.) 2 2 R < 15 ⇒ R < = 3.75 Ω 4 Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously. DP2-3 P = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W 1 2 2 P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W 2 2 P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W 2 2 2-11 ...
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