L-16(NKD)(ET) ((EE)NPTEL) - Module 4 Single-phase AC...

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Module 4 Single-phase AC circuits Version 2 EE IIT, Kharagpur
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Lesson 16 Solution of Current in AC Parallel and Series- parallel Circuits Version 2 EE IIT, Kharagpur
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In the last lesson, the following points were described: 1. How to compute the total impedance/admittance in series/parallel circuits? 2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac supply, and then draw complete phasor diagram? 3. How to find the power consumed in the circuit and also the different components, and the power factor (lag/lead)? In this lesson, the computation of impedance/admittance in parallel and series-parallel circuits, fed from single phase ac supply, is presented. Then, the currents, both in magnitude and phase, are calculated. The process of drawing complete phasor diagram is described. The computation of total power and also power consumed in the different components, along with power factor, is explained. Some examples, of both parallel and series-parallel circuits, are presented in detail. Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power factor. After going through this lesson, the students will be able to answer the following questions; 1. How to compute the impedance/admittance, of the parallel and series-parallel circuits, fed from single phase ac supply? 2. How to compute the different currents and also voltage drops in the components, both in magnitude and phase, of the circuit? 3. How to draw the complete phasor diagram, showing the currents and voltage drops? 4. How to compute the total power and also power consumed in the different components, along with power factor? This lesson starts with two examples of parallel circuits fed from single phase ac supply. The first example is presented in detail. The students are advised to study the two cases of parallel circuits given in the previous lesson. Example 16.1 The circuit, having two impedances of Ω + = ) 15 8 ( 1 j Z and Ω = ) 8 6 ( 2 j Z in parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10 A. Find each branch current, both in magnitude and phase, and also the supply voltage. Version 2 EE IIT, Kharagpur
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B I = 10A Fig. 16.1 (a) Circuit diagram A Z 2 = (6 – j8) Z 1 = (8 + j15) I 1 I 2 Solution Ω ° = + = 93 . 61 17 ) 15 8 ( 1 1 j Z φ Ω ° = = 13 . 53 10 ) 8 6 ( 2 2 j Z A j OC I ) 0 10 ( 0 10 ) ( 0 + = ° = ° The admittances, using impedances in rectangular form, are, 1 3 2 2 1 1 1 1 10 ) 9 . 51 68 . 27 ( 289 15 8 15 8 15 8 15 8 1 1 Ω = = + = + = = j j j j Z Y 1 3 2 2 2 2 2 2 10 ) 0 . 80 0 . 60 ( 100 8 6 8 6 8 6 8 6 1 1 Ω + = + = + + = = = j j j j Z Y Alternatively, using impedances in polar form, the admittances are, 1 3 1 1 1 1 10 ) 9 . 51 68 . 27 ( 93 . 61 05882 . 0 93 . 61 0 . 17 1 1 Ω = ° = ° = = j Z Y 1 3 2 2 2 2 10 ) 0 . 80 0 . 60 ( 13 . 53 1 . 0
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L-16(NKD)(ET) ((EE)NPTEL) - Module 4 Single-phase AC...

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