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 freqency of source is chaged to `(omega)//3` (but maintainging the same voltage), the current in the circuit is found to be halved. Calculate the ration of hte reactance to resistance at the original frequency `omega`.

Answer 1

Correct Answer - `sqrt((3)/(5))`
According to the given problem,
`I=(V)/(Z)=(V)/([R^(2)+(1//C omega)^(2)]^(1//2))` …(i)
and `(I)/(2) =(V)/([R^(2)+(3//C omega)^(2)]^(1//2))` …(ii)
From eqs. (i) in (ii) , `(1)/(C^(2)omega^(2)) =(3)/(5)R^(2)`
So that ,` (X)/(R ) =(1//C omega)/(R ) = ((3/5 R^(2))^(1//2))/(R ) =sqrt((3)/(5))`.