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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 i in the circuit, the phase difference between the emf e and the current i in the circuit is observed to be `(pi//4)`, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements

A

`R=1k Omega, C=10 mu F`

B

`R=1k Omega, C=1 mu F`

C

`R=1k Omega, L=10 mu F`

D

`R=1k Omega, L=1 mu F`

Text Solution

Verified by Experts

The correct Answer is:
A
Since,current leads emf (as seen from the graph), 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 implies RC(omega)=1`
`:. RC=(1)/(100)(s^(-1))`.
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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.

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