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.

Options

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`

Answer

A

Explanation

Figure given in the question shows that current U leads the voltage V by a phase angle `pi//4`. Therefore, the circuit can be RC circuit alone.
`tan phi=(X_(C ))/(R )=(1)/(omega CR) " "(because X_(C )=(1)/(omega C))`
`"tan"(pi)/(4)=(1)/(omega CR) , 1=(1)/(omega CR)` ....(i)
From `V=V_(0)` sin 100t, we get, `omega = 100"rad "s^(-1)`
`therefore CR = (1)/(omega)=(1)/(100)` (Using (i))
When `R=1l Omega = 10^(3)Omega, C=(1)/(10^(5))=10^(-5)F=10 mu F`