For the circuit shown in fig, the ammeter `A_(2)` reads 1.6A and ammeter `(A_3)` reads 0.4A. Then

Options

A. (A) `omega=(4)/(sqrt(LC))`
B. (B)`f=(2 pi )/(sqrt(LC))`
C. (C)The ammeter `(A_1)` reads 1.2 A
D. (D)the ammeter `(A_1)` reads 2A

Answer

C

Explanation

The current of 1.6 A algs emf in phase by `(pi)//(2)`. The current of 0.4 A leads emf in phase by `(pi)/(2)`. So, these two currents are `180^(@)` out of phase with each other.
`:.` Net current `(I_1)=(1.6-0.4)A=1.2 A`
`1.6 = (E_v)/(X_L) and 0.4 = (E_v)/(X_C)`
`implies (X_C)/(X_L)=4 implies (1)/("omega C omega L")=4`
`implies omega =(1).(2 sqrt(LC))`
`implies f=(omega)/(2 pi) =(1)/(4 pi sqrt(LC))`