Question

An `AC` source of angular frequency `omega` is fed across a resistor `R` and a capacitor `C` in series. The current registered is `I`. If now the frequency of source is changed to `omega//3` (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency `omega` will be.

Answer 1

According to given problem `I=V/Z=(V)/([R^(2)-(1//Comega)^(2)]^(1//2))`...(i)
and `1/2=(V)/([R^(2)+(3//Comega)^(2)]^(1//2))`...(ii)
Substituting the value of `I` from Equation (1) in (2) `4(R^(2)+1/(C^(2)omega^(2)))=R^(2)9/(C^(2)omega^(2))i.e.,1/(C^(2)omega^(2))=3/5R^(2)`
So that,`X/R=(1//Comega)/R=((3/5R^(2)))/R=sqrt(3/5)`