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The series combination of R(Omega) and capacitor C(F) is connected to an A.C. source of V volts and angular frequency omega. If the angular frequency is reduced to omega/3, the current is found to be reduced to one-half without changing the value of the voltage. Determine the ratio of the capacitive reactance and the resistance. |
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Answer» Solution : First case : (Here, r.m.s. value of I and V as I and V for the SAKE of convenience) `I=V/sqrt(R^2 + 1/(omega^2C^2)) " " therefore I^2=V^2/(R^2+X_C^2)`...(i) SECOND case : `I/2=V/sqrt(R^2+9/(omega^2C^2)) "" therefore I^2/4 = V^2/(R^2+9X_C^2)`...(ii) Dividing EQU. (i) by (ii), we have `therefore 4=(R^2+9X_C^2)/(R^2+X_C^2)` `therefore 4R^2+4X_C^2=R^2+9X_C^2` `therefore 5X_C^2=3R^2` `therefore X_C/R=sqrt(3/5)` |
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