L-15(NKD)(ET) ((EE)NPTEL)

# L-15(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 15 Solution of Current in AC Series and Parallel Circuits Version 2 EE IIT, Kharagpur
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

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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.
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## 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.

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

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