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A source of a.c. voltage V = V_(m) sin omega tis connected to a series combination of a capacitor C and a resistor R. Draw the phasor diagram and use it to obtain the expression for (i) impedance of the circuit, and (ii) phase angle. |
Answer» Solution :Consider an alternating voltage given by `V=V_(m) sin omega t`applied to a capacitor C and resistor R joined in SERIES as SHOWN in Fig. 7.28. As C and R are in series, same current I flows in the entire circuit. Let `vecV_(C), vecV_(R)` and `vecV`be the magnitudes of instantaneous values of voltage across the capacitance, resistance and the source. Then, as shown in phasor diagram:  `vecV_(C) = VECI.X_(C)` lagging behind in phaseby `pi/2` as compared to `vecl`, and `vecV_(R) = vecI R` in same phase as that of `vecl` These are being represented by phasors OB and OA, RESPECTIVELY. Then, total instantaneous voltage V is given by phasor OC such that: `V = sqrt(OA^(2) + OB^(2)) = sqrt(V_(R)^(2) + V_(C)^(2)) = Isqrt(R^(2) + X_(C)^(2))` `therefore` Impedance offered by the circuit, `Z = V/I = sqrt(R^(2) + X_(L)^(2)) = sqrt(R^(2) + (1/(Comega)^(2))` Moreover, the phasor diagram indicates that source voltage `vecV`lags behind the current I (or the circuit current leads the a.c. supply voltage) by a phase angle `phi`where, `tan phi = (OB)/(OA) = (IX_(C))/(IR) = X_(C)/R = (1//Comega)/R = 1/(R. Comega)` 
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