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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» According to the given problems. `I=(V)/(Z)=(V)/([R^(2)+(1//Comega)^(2)]^(1//2))......(i)` `and (I)/(2)=(V)/([R^(2)+(3//Comega)^(2)]^(1//2))......(2)` Substituting the value of I from Equation (i) and (2). `4(R^(2)+(1)/(2C^(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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