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<strong>www</strong>.<strong>GOALias</strong>.<strong>blogspot</strong>.<strong>com</strong>7.6.1 Phasor-diagram solutionFrom the circuit shown in Fig. 7.12, we see that the resistor, inductorand capacitor are in series. Therefore, the ac current in each element isthe same at any time, having the same amplitude and phase. Let it bei = i msin(ωt+φ ) (7.21)where φ is the phase difference between the voltage across the source andthe current in the circuit. On the basis of what we have learnt in the previoussections, we shall construct a phasor diagram for the present case.Let I be the phasor representing the current in the circuit as given byEq. (7.21). Further, let V L, V R, V C, and V represent the voltage across theinductor, resistor, capacitor and the source, respectively. From previoussection, we know that V Ris parallel to I, V Cis π/2behind I and V Lis π/2 ahead of I. V L, V R, V Cand Iare shown in Fig. 7.13(a) with apppropriate phaserelations.The length of these phasors or the amplitudeof V R, V Cand V Lare:v Rm= i mR, v Cm= i mX C, v = i X (7.22)Lm m LThe voltage Equation (7.20) for the circuit canbe written asv L+ v R+ v C= v (7.23)The phasor relation whose vertical <strong>com</strong>ponentgives the above equation isV L+ V R+ V C= V (7.24)This relation is represented in Fig. 7.13(b). SinceV Cand V Lare always along the same line and inopposite directions, they can be <strong>com</strong>bined into a single phasor (V C+ V L)which has a magnitude ⏐v Cm– v Lm⏐. Since V is represented as thehypotenuse of a right-traingle whose sides are V Rand (V C+ V L), thepythagorean theorem gives:( ) 2v = v + v −v2 2m Rm Cm LmSubstituting the values of v Rm, v Cm, and v Lmfrom Eq. (7.22) into the aboveequation, we havev = ( i R) + ( i X −i X )2 2 2m m m C m LAlternating CurrentFIGURE 7.13 (a) Relation between thephasors V L, V R, V C, and I, (b) Relationbetween the phasors V L, V R, and (V L+ V C)for the circuit in Fig. 7.11.2 2 2= i ⎡m ⎣R + ( XC − XL)⎤⎦vmor, im=2 2R + ( X − X )CL[7.25(a)]By analogy to the resistance in a circuit, we introduce the impedance Zin an ac circuit:vmim= [7.25(b)]ZwhereZ = R + ( X − X )(7.26)2 2C L245

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