An inductor `(X_(L) = 2 Omega)` a capacitor `(X_(C ) = 8 Omega)` and a resistance `(8 Omega)` are connected in series with an ac source. The voltage output of `A.C` source is given by `v = 10 cos 100 pi t`. The instantaneous p.d. between `A` and `B` when is half of the voltage output from source will be:

Options

A. `(24)/(7)` volt
B. `(24)/(5)` volt
C. `(7)/(24)` volts
D. `(5)/(24)` volts

Answer

B

Explanation

The current leads in phase by `( :.' X_(C ) gt X_(L))`
`phi = 37^(@)`
`:. I = (10 cos (100 pi t + 37^(@)))/(Z) = cos (100 pi + 37^(@))`
The instantaneous potential difference across `AB` is
`= I_(m) = (X_(C ) - X_(L)) cos (100 pi t + 37^(@) - 90^(@))`
The instantaneous potential difference across `AB` is half of the source voltage
`:. 6 cos (100 pi t - 53^(@)) = 5 cos 100 pi t`
Solving we get `cos 100 pi t = (1)/(sqrt(1 + (7//24)^(2))) = (24)/(5)`
`:.` Instantaneous potential difference
`= 5 xx (24)/(25) = (24)/(5)` volts