L-15(NKD)(ET) ((EE)NPTEL) - Module 4 Single-phase AC...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Module 4 Single-phase AC circuits Version 2 EE IIT, Kharagpur
Background image of page 1

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

View Full DocumentRight Arrow Icon
Lesson 15 Solution of Current in AC Series and Parallel Circuits Version 2 EE IIT, Kharagpur
Background image of page 2
In the last lesson, two points were described: 1. How to solve for the impedance, and current in an ac circuit, consisting of single element, R / L / C? 2. How to solve for the impedance, and current in an ac circuit, consisting of two elements, R and L / C, in series, and then draw complete phasor diagram? In this lesson, the solution of currents in simple circuits, consisting of resistance R, inductance L and/or capacitance C connected in series, fed from single phase ac supply, is presented. Then, the circuit with all above components in parallel is taken up. The process of drawing complete phasor diagram with current(s) and voltage drops in the different components is described. The computation of total power and also power consumed in the different components, along with power factor, is explained. One example of series circuit are presented in detail, while the example of parallel circuit will be taken up in the next lesson. Keywords: Series and 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 total reactance and impedance / admittance, of the series and 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? Solution of Current in R-L-C Series Circuit Series (R-L-C) circuit + - V R L C Fig. 15.1 (a) Circuit diagram O D E A I The voltage balance equation for the circuit with R, L and C in series (Fig. 15.1a), is t V dt i C dt di L i R v ω sin 2 1 = + + = Version 2 EE IIT, Kharagpur
Background image of page 3

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

View Full DocumentRight Arrow Icon
The current, i is of the form, ) ( sin 2 φω ± = t I i As described in the previous lesson (#14) on series (R-L) circuit, the current in steady state is sinusoidal in nature. The procedure given here, in brief, is followed to determine the form of current. If the expression for ) ( sin 2 = t I i is substituted in the voltage equation, the equation shown here is obtained, with the sides (LHS & RHS) interchanged. t V t I C t I L t I R ω ωφ sin 2 ) ( cos 2 ) / 1 ( ) ( cos 2 ) ( sin 2 = + or t V t I C L t I R ωω sin 2 ) ( cos 2 )] / 1 ( [ ) ( sin 2 = + The steps to be followed to find the magnitude and phase angle of the current I , are same as described there (#14). So, the phase angle is R C L / )] / 1 ( [ tan 1 φ = and the magnitude of the current is Z V I / = where the impedance of the series circuit is 2 2 )] / 1 ( [ C L R Z + = Alternatively, the steps to find the rms value of the current I, using complex form of impedance, are given here.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/03/2010 for the course ELECTRICAL EE212 taught by Professor Shetty during the Spring '10 term at International Institute of Information Technology.

Page1 / 18

L-15(NKD)(ET) ((EE)NPTEL) - Module 4 Single-phase AC...

This preview shows document pages 1 - 5. Sign up to view the full document.

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