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In an AC circuit, voltage V = V(0)sin om...

In an AC circuit, voltage V = `V_(0)sin omegat` and inductor L is connected across the circuit. Then the instantaneous power will be

A

`(V_(2)^(2))/(2omegaL)sin omegat`

B

`(-V_(0)^(2))/(2omegaL)sin omega t`

C

`(-V_(0)^(2))/(2omegaL)sin 2 omegat`

D

`(V_(0)^(2))/(omegaL)sin 2 omegat`

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(a) An AC source of voltage V = V_(0) sin omegat is connected across a series combination of an inductor a capacitor and a resistor . Use the phasor diagram to obtain the expression for the impedance of the circuit and phase angle between the voltage and the current. (b) A capacitor of unknown capacitance a resistor of 100 Omega and an inductor of self - inductance L = (4 //pi^(2)) henry are in series connected to an AC source of 200 V and 50 Hz . Calculate the value of the capacitance and the current that flows in the circuit when the current is in phase with the voltage .

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Knowledge Check

  • For an AC circuit V=15sin omegat and I=20cos omegat the average power consumed in this circuit is

    A
    `300` watt
    B
    `150` watt
    C
    `75` watt
    D
    zero

  • An ac source of voltage V=V_(m)sin omega t is connected across the resistance R as shown in figure. The phase relation between current and voltage for this circuit is

    A
    both are in phase
    B
    both are out of phase by `90^(@)`
    C
    both are out of phase by `120^(@)`
    D
    both are out of phase by `180^(@)`

  • In an AC circuit with voltage V and current I , the power dissipated is

    A
    `VI`
    B
    `1/2 VI`
    C
    `1/(sqrt(2)) VI`
    D
    Depened on the phase between `V` and `I`

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    The voltage and current in a series AC circuit are given by V = V_0 cos omegat and I = i_0 sin omegat . What is the power dissipated in the circuit?

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