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When an AC source of emf E=E(0)sin(100t)...

When an AC source of emf `E=E_(0)sin(100t)` is connected across a circuit, the phase difference between the emf E and the current I is observed be `pi/4`, as shown in the figure. If the circuit consists possibly only of R-C or R-L in series, which of the following combinations is possible?

A

`R=1 kOmega, C=10 (mu)F`

B

`R=1 kOmega, C=1(mu)F`

C

`R=1 kOmega,L=10H`

D

`R=1 kOmega, L=H`

Text Solution

Verified by Experts

The correct Answer is:
A
Since currents leads emf, therefore, this is an R-C circuit.
`tan phi=(X_(C )-X_(L))/(R ) here phi=45^(@)`
`:. X_(C )=R` [`X_(L)=0` as there is no inductor]
`(1)/(omega C)=R`
L.H.S `=(1)/(100 xx 10 xx 10^(-6)) = R = (10^3) Omega`.
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Knowledge Check

  • When an ac source of emf e=E_(0) sin (100 t) is connected across a circuit, the phase difference between emf e and currnet I in the circuit is observed to be (pi)//(4) as shown in fig. If the circuit consists possibly only of R-C or R-C of L-R series, find the relationship find the relationship between the two elements.

    A
    `R=1 k Omega, C=10 mu F`
    B
    `R = 1 k Omega, C = 1 mu F`
    C
    `R=1 k Omega, L = 10 mH`
    D
    `R=10 k Omega, L = 10 mH`

  • In an electrical circuit R,L,C and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is pi//3 . If instead, C is removed from the circuit, difference the phase difference is again pi//3 . The power factor of the circuit is

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    `(1)/(2)`
    B
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