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An a.c. source of angular frequency omeg...

An a.c. 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 the source is changed to `omega//3` (but maintaining the same voltage), the current in the circuit is found to be halved. calculate the ratio of reactance to resistance at the original frequency `omega`.

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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 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 .

An a.c. source of angular frequency omega is fed across a resistor R and a capacitor C in series. The current registered is I. If you the frequency of source is changed to omega//3 , maintaining the same voltage, current in the circuits is found to be halved. Caculate the ratio of reactance to resistance at the original frequency omega .

Knowledge Check

  • An ac source of angular frequency o is fed across a resistor R and a capacitor C in series. The current registered is l. If now the frequency of source is changed to c/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 o will be

    A
    `sqrt(3/(5)`
    B
    `sqrt(5)/(3)`
    C
    3//5
    D
    `sqrtR`

  • An AC source of frequency omega when fed into a RC series circuit, current is recorded to be l. If now frequency is changed to (omega)/(4) (keeping voltage same), the current is found to 1/2. The ratio of reactance to resistance at original frequency omega is

    A
    2
    B
    `1/2`
    C
    `(1)/(sqrt(2))`
    D
    None of these

  • An AC voltage source V=V_0siomegat is connected across resistance R and capacitance C as shown in figureure. It is given that R=1/omegaC . The peak current is I_0 . If the angular frequency of the voltage source is changed to omega/sqrt3, then the new peak current in the circuit is .

    A
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    B
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    C
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    D
    `I_0/3`

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