35042074-Solution-of-Network-Analysis-m-e-Van-Valkenburg-CH-1-2-3-4

35042074-Solution-of-Network-Analysis-m-e-Van-Valkenburg-CH-1-2-3-4

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Unformatted text preview: Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 1 Conventions for Describing Networks 2-1. For the controlled (monitored) source shown in the figure, prepare a plot similar to that given in Fig. 2-8(b). v 2 v 1 = V b V b v 1 = V a V a i 2 Fig. 2-8 (b) Solution: Open your book & see the figure (P/46) It is voltage controlled current source. i 2 +Ve axis v 2-Ve axis gv 1 i 2 gv 1 + v 2 current source- Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 2 2-2. Repeat Prob. 2-1 for the controlled source given in the accompanying figure. Solution: Open your book & see the figure (P/46) It is current controlled voltage source. v 2 ri 1 i 2 2-3. The network of the accompanying figure is a model for a battery of open-circuit terminal voltage V and internal resistance R b . For this network, plot i as a function v. Identify features of the plot such as slopes, intercepts, and so on. Solution: Open your book & see the figure (P/46) Terminal voltage v = V - iR b iR b = V - v i = (V - v )/R b When v = 0 i = (V - v )/R b i = (V - 0 )/R b i = V/R b amp When v = V i = (V - V )/R b i = (0 )/R b i = 0 amp v = 0 i = V/R v = V i = 0 i V/R b V v Slope: Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 3 y = mx + c (x 1 , y 1 ) = (0, V/R b ) (x 2 , y 2 ) = (V, 0) m = (y 2 y 1 )/(x 2 x 1 ) = (0 V/R b )/(V - 0) = (-V/R b )/V = (-V/Rb)(1/V) = -1/R b y-intercept = V/R b x-intercept = V Slope y-intercept x-intercept-1/R b V/R b V 2-4. The magnetic system shown in the figure has three windings marked 1-1, 2-2, and 3-3. Using three different forms of dots, establish polarity markings for these windings. Solution: Open your book & see the figure (P/46) Lets assume current in coil 1-1 has direction up at 1 (increasing). It produces flux (increasing) in that core in clockwise direction. 1 1 2 2 3 3 According to the Lenzs law current produced in coil 2-2 is in such a direction that it opposes the increasing flux . So direction of current in 2-2 is down at 2. Hence ends 1 & 2 are of same polarity at any instant. Hence are marked with . Similarly assuming the direction of current in coil 2-2, we can show at any instant 2 & 3 have same polarities and also 1 & 3 have same polarities. 2-5. Place three windings on the core shown for Prob. 2-4 with winding senses selected such that the following terminals have the same mark: (a) 1 and 2, 2 and 3, 3 and 1, (b) 1 and 2, 2 and 3, 3 and 1. Solution: Open your book & see the figure (P/47) Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 4 1 1 2 2 3 3 (a) 1 1 2 2 3 3 (b) 2-6. The figure shows four windings on a magnetic flux-conducting core. Using different shaped dots, establish polarity markings for the windings....
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35042074-Solution-of-Network-Analysis-m-e-Van-Valkenburg-CH-1-2-3-4

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