A box `P` and a coil `Q` are connected in series with an ac source of variable freguency The emf of the source is constant at `28V` The frequency is so adjusted that the maximum current flows in `P` and `Q` Find
(a) impedance of `P` and `Q` at this frequency
(b) voltage across `P` and `Q`
.

Explanation

Current is amximum i.e. Case of series resonance,
`omega_(0)= 1/(sqrt(LC)` = `1/(sqrt(1.2 xx 10^(-3) xx 25/3 xx 10^(-6)) = 10^(4)rads^(-1)`
At resonance, `X_(C) = X_(L)`
`X_(C) = 1/(omega_(0)C) = 1/(10^(4)xx25/3xx10^(-6)) = 12 Omega=X_(L)`
Resistance `R_(1)` and `R_(2)` are in series, so `R_(net)= R_(1) + R_(2)`
`therefore I= V/(R_(1)+R_(2)) = 28/(16+12) = 28/28 = 1A`
a) `Z_(P) = sqrt((16)^(2) + (X_(C))^(2)) = sqrt(16^(2)+(12)^(2))=20Omega`
`Z_(Q) = sqrt(12^(2)_X_(L)^(2)) = sqrt((12)^(2)+(12)^(2))` = `20Omega`
b) `V_(P) = IZ_(P)= 1 xx 20 = 20V`
`V_(Q) = IZ_(Q) = 1 xx 12sqrt(2) = 12sqrt(2)V`