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 mu H`

Answer

A

Explanation

Fig. shows that current `i` leads the emf `e` by a phase angle `pi//4`. Therefore, the circuit can be `R - C` circuit alone.
`tan phi = (X_(C ))/(R ) = (1)/(omega CR) = "tan" (pi)/(4) = 1` or `CR = (1)/(omega)`
From `e = E_(0) sin 100 t, omega = 100 rad//s`
`:. CR = (1)/(omega) = (1)/(100)`
when `R = 1 K Omega = 10^(3) Omega`
`C = (1)/(omega R) = (1)/(10^(5)) = 10^(-5) F = 10 mu F`