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A resistor R and an inductor L are connected in series to a source V=V_m sin omegat. Find the (a) peak value of the voltage drops across R and across L, (b) phase difference betweenthe applied voltage and current. Which of them is ahead ? |
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Answer» Solution :Consider a circuit having an inductance land a resistance R, joined in series to an a.c. supply. Let voltage provided by a.c... supply be `V=V_(m) sin omegat`. (a) Let an instantaneous current l FLOWS through the coil. Then instantaneous VALUES of potential drops across inductance and resistance are verbs are given by `vecV_(L)=I X_(L) and vecV_(R)=IR," where "X_(L)=omega L` is the reactance due to the inductance. MOREVER phasor `vecV_(R)` is in phase with `vecI" but "vecV_(L)` leads in phase by `pi/2". Let "vecV_(R) and vecV_(L)` be represented by OA and OB in a phasor diagram. Then resultant voltage `vecV` will be given by the phasor OC. Hence. `V=OC =sqrt(OA^(2)+OB^(2))`  `=sqrt(V_(R)^(2)+V_(L)^(2))=I sqrt(R^2+X_(L)^(2))` `I=V/sqrt(R^(2)+X_(L)^(2)) and I_(m)=V_(m)/sqrt(R^(2)+X_(L)^(2))` Peak voltage drop across R, `(V_(m))_R=I_(m). R=(V_(m).R)/sqrt(R^(2)+X_(L)^(2))` and peak voltage drop across, `L (V_(m))_L=I_(m).X_(L)=(V_(m).X_(L))/sqrt(R^(2)+X_(L)^(2))`  (b) As per phasor diagram, the circuit voltage V is ahead as compared to circuit current I (or current I is lagging behind hte SOURCE voltage V) by a phase ANGLE `theta`, where `tan phi=(AC)/(OA)=(OB)/(OA)=(IX_(L))/(IR)=X_L/R=(L omega)/(R)` |
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