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An AC source of angular frequency omega ...

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.

A

`sqrt((3)/(5))`

B

`sqrt((5)/(3))`

C

`3//5`

D

`5//3`

Text Solution

Verified by Experts

According to given problem,
`I=(V)/(Z)=V//[R^(2)+(1//Comega^(2)]^(1//2)`……`(1)`
and `(I)/(2)=(V)/([R^(2)+(3//Comega)^(2)]^(2))`……….`(2)`
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)/(5)R^(2)`
so that `(X)/(R)=((1//Comega))/(R )=([(3//5)R^(2)]^(1//2))/(R )=sqrt((3)/(5))`
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