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A 300 Omega resistor, a 0.250 H inductor...

A `300 Omega` resistor, a `0.250 H` inductor, and a `8.00muF` capacitor are in series with an Ac with voltage amplitude `120 V` and angular frequency `400 rad// B`.
(a) What is the current amplitude?
(b) Wheat is the phase angle of the source voltage with respect to the current? Does the source Lag, or lead the current?
(c) What are the voltage amplitudes across the resistor, inductor, and capacitor ?

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`X_L=omegaL=100Omega`
`X_C=1/(omegaC)=312.5Omega`
`Z=sqrt(R^2+(X_C-X_L)^2)`
`=sqrt((3000)^2+(312.5-100)^2)`
`=368Omega`
(a) `I_0=V_0/Z=120/368`
(b) Since `X_CgtX_L`, voltage lags the current by an angle given by
`phi=cos^-1(R/Z)=cos^-1(300/368)=35.3^@`
(c) `(V_0)_R=I_0R=(0.326)300=97.8V`
`(V_0)_L=I_0X_L=(0.326)(100)=32.6V`
`(V_0)_C=I_0X_c=(0.326)(312.5)=102V`.
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Knowledge Check

  • In the question number 44, the phase difference between the voltage across the source and current is

    A
    `80.2^(@)`
    B
    `31^(@)`
    C
    `50.2^(@)`
    D
    `38.2^(@)`

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