An alternative voltage `V = 10 sin omega t` (in volts) is applied across a parallel arrangement as shown current (in `A`) through the source is best described by

Options

A. `i = 0.05 sin omega t`
B. `i = 0.7 sin (omega t + (pi)/(6))`
C. `i = 0.7 sin (omega t + (pi)/(4))`
D. `i = 0.05 sin (omega t + (pi)/(2))`

Answer

C

Explanation

Phasor diagram
`|vec(i)|^(2) = |vec(i_(R ))|^(2) + |vec(i_(C ) - i_(L))|^(2)`
`i_(0) = ((V_(0))/(R ))^(2) + V_(0)^(2) ((1)/(X_(C )) - (1)/(X_(L)))^(2)`
`i_(0) = V_(0) [(1)/(R^(2)) + ((1)/(X_(C )) - (1)/(X_(L)))^(2) ]^(1//2) = (1)/(sqrt(2)) A`
`tan Q = ((1)/(X_(C )) - (1)/(X_(L)))/(1//R) = 1 [Q = (pi)/(4)]`