Chap_3_Solut_Dorf

Chap_3_Solut_Dorf - Chapter 3 Resistive Circuits Exercises...

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Chapter 3 Resistive Circuits Exercises Ex 3.2-1 Apply KCL at node a to get 2 + 1 + i 3 = 0 i 3 = -3 A Apply KCL at node c to get 2 + 1 = i 4 i 4 = 3 A Apply KCL at node b to get i 3 + i 6 = 1 -3 + i 6 = 1 i 6 = 4 A Apply KVL to the loop consisting of elements A and B to get - v 2 – 3 = 0 v 2 = -3 V Apply KVL to the loop consisting of elements C , E , D , and A to get 3 + 6 + v 4 – 3 = 0 v 4 = -6 V Apply KVL to the loop consisting of elements E and F to get v 6 – 6 = 0 v 6 = 6 V Check: The sum of the power supplied by all branches is -(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0
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Ex 3.3-1 () 25 From voltage division = 8 2 V m 25+75 v ⇒= Ex 3.3-2 25 From voltage division = 8 2 V m 25+75 v ⇒− = Ex 3.4-1 3 3333 3 -3 1 11114 1 01 k 10 10 10 10 10 4 4 11 By current division, the current in each resistor (10 ) mA 44 R eq R eq = +++= == Ω Ex 3.4-2 10 From current division = 5 1 A 10+40 i m = Solution: Label the second circuit as follows: Corresponding elements are incident to corresponding nodes so the two drawings do indeed represent the same circuit. Choose node d to be the reference node. Apply KCL at nodes a , b and c to get:
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2 10 A i = ()() 26 6 2 52 0 7 7 1 0 3 ii i i +− +− + = = − = − =− A 33 10 0 10 A + =⇒ = Apply KVL to the meshes to get: ( ) 41 4 45 0 vv v v −+ −− − = = − 1 1 4 54 5 0 = () 55 60 6 −− − = = V Then 14 6 V and 1 6 1 7 V == =+ = + =
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Problems Section 3-2 Kirchhoff’s Laws P3.2-1 Apply KCL at node a to get 2 + 1 = i + 4 i = -1 A The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is power received by element B . The power supplied by element B is 12 W . Apply KVL to the loop consisting of elements D , F , E , and C to get 4 + v + (-5) – 12 = 0 v = 13 V The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W is the power supplied by element F . Check: The sum of the power supplied by all branches is -(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0 P3.2-2 Apply KCL at node a to get 2 = i 2 + 6 = 0 i 2 = 4 A
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Apply KCL at node b to get 3 = i 4 + 6 i 4 = -3 A Apply KVL to the loop consisting of elements A and B to get - v 2 – 6 = 0 v 2 = -6 V Apply KVL to the loop consisting of elements C , D , and A to get - v 3 – (-2) – 6 = 0 v 4 = -4 V Apply KVL to the loop consisting of elements E , F and D to get 4 – v 6 + (-2) = 0 v 6 = 2 V Check: The sum of the power supplied by all branches is (6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0 P3.2-3 2 22 1 1 1 KVL : 12 (3) 0 (outside loop) 12 12 3 or 3 12 KCL 3 0 (top node) 12 12 3 or 3 Rv v vR R i R iR Ri −+ = =+ = +− = =− = (a) () 12 3 3 21 V 12 31 A 6 v i = =− = (b) 21 2 1 0 1 2 ; 8 3 3 3 1.5 RR == Ω = = Ω (checked using LNAP 8/16/02) (c) 1 2 24 12 , because 12 and adhere to the passive convention. 12 2 A and 2.4 32 9 3 , because 3 and do not adhere to the passive convention 2 3 V and 3 3 ii vv ∴=− = = Ω + = = Ω The situations described in (b) and (c) cannot occur if R 1 and R 2 are required to be nonnegative.
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P3.2-4 2 Power absorbed by the 4 resistor = 4 = 100 W 2 2 Power absorbed by the 6 resistor = 6 = 24 W 1 2 Power absorbed by the 8 resistor = 8 = 72 W 4 i i i Ω⋅ 12 2A 1 6 20 5 A 2 4 3 2 32 3A 423 i i ii iii == =− =− =+= A
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Chap_3_Solut_Dorf - Chapter 3 Resistive Circuits Exercises...

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