A voltage `V_(PQ)= V_(0) cos omega t` (where `V_(0)` is a real amplitude) is applied between the point P and Q in the network shown in the figure. The values of capacitance and inductance are `C= (1)/(omega R sqrt3) and L= (R sqrt3)/(omega)`
Then, the total impedance between P and Q is

Options

A. 1.5R
B. 2R
C. 3R
D. 4R

Answer

C

Explanation

`because` Given, that, `C= (1)/(omega R sqrt3), L= (R sqrt3)/(omega)`

In the above figure, `Z_(1)= R+j omega L = R + j omega ((R sqrt3)/(omega))=R +jR sqrt3`
`Z_(2)= R- j (1)/(omega c)= R-j (1)/(omega ((1)/(omega R sqrt3)))=R-jR-sqrt3`
`because` Impedance, `Z_(1) and Z_(2)` are in parallel, So, `Z_(eq)= (Z_(1)Z_(2))/(Z_(1) +Z_(2))`
`= ((R+ jR sqrt3)(R-jR sqrt3))/(R+ jR sqrt3+R- jR sqrt3)`
`=(R^(2) + R^(2) (sqrt3)^(2))/(2R) ( because (a-b) (a+b) =a^(2)-b^(2))`
`Z_(eq)= (4R^(2))/(2R)=2R`
So, total impedance between P and Q is
`Z_(PQ)= R+Z_(eq)`
`Z_(PQ)= R + 2R= 3R`