chapter 03 - Chapter 3 Resistive Circuits Exercises Ex...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 3 – Resistive Circuits Exercises Ex 3.3-1 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 3-1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Ex 3.3-2 Apply KCL at node a to determine the current in the horizontal resistor as shown. Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4( i ) - 24 = 0 i = 4 A Ex 3.3-3 2 18 0 12 0 30 V and 3 9 A 5 aa m am vv i vi −+−−= ⇒ = = + ⇒ = Ex 3.3-4 18 10 4 8 0 6 V and 4 24 V 3 a m a v v v −− + −= ⇒ = = = = Ex 3.4-1 3 3 From voltage division 3 12 = 3V 39 = = 1A 3 then v v i    = + The power absorbed by the resistors is: ( )( ) ( )( ) ( )( ) 222 16 13 131 2 W ++= The power supplied by the source is (12)(1) = 12 W. 3-2
Background image of page 2
Ex 3.4-2 1 1 01 6 W and 6 6 2 = 1 or =1 A 6 =(1) (6)=6V PR P ii R vi R = =Ω == = from KVL: (2 4 6 2) 0 14 14 V s s ++ + += ⇒= = Ex 3.4-3 () 25 From voltage division = 8 2 V m 25+75 v ⇒= Ex 3.4-4 25 From voltage division = 8 2 V m 25+75 v ⇒− = Ex. 3.5-1 3 3333 3 -3 1 11114 1 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 3.5-2 10 From current division = 5 1 A 10+40 i m = 3-3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problems Section 3-3 Kirchoff’s Laws P3.3-1 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 3-4
Background image of page 4
P3.3-2 Apply KCL at node a to get 2 = i 2 + 6 = 0 i 2 = -4 A 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 3-5
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
P3.3-3 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 33 3 1 . 5 RR == = =
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 28

chapter 03 - Chapter 3 Resistive Circuits Exercises Ex...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online