Inductance (L), capacitance (C) and resistance (R) are constained in a box. When 250 V DC is applied to the terminals of the box, a current of 1.0A floes in the circuit. When an AC source of `250V_(rms)` at `2250 rad sec^(-1)` is connected, a current of `1.25 A_(rms)` flows. It is observed that the current rises with frequency and becomes maximum at `4500 rad sec^(-1)`. find the values of L,C and R. draw the circuit diagram.

Explanation

When DC sources is applied, current = 1A
In DC circuit, inductor offers zero resistance and capacitor offer infinite resistance, R and C cannot be in series.
`I=V_(rms)/I_(rms) = (250)/(1.25)=200Omega`
As `ZltR`, R L and C cannot be in series.
R cannot be in series with C as explained earlier.
As current is maximum at resonance and current increase with frequency, R cannot be in series with L. In fact, L and C must be in series.

`I_(1) = 250/250 = 1A`, in phase with voltage impedance of lower branch.
`rArr Z^(')=|(omegaL-1/(omegaC))|`
The phase difference between `I_(2)` and the voltage is `pi/2`
The angle between `I_(1)` and `I_(2)` is `90^(@)`.

`I_(1)^(2)+I_(2)^(2)=I^(2)rArr (1)^(2)+I_(2)^(2)=(1.25)^(2)`
`rArr I^(2) = (1.25)^(2)-(1)^(2) = (2.25)(0.25)`
`rArr I_(2)=1.5 xx 0.5 = 0.75`A
`rArr I_(2)=V/(Z^('))`
`therefore Z^(')=V/I_(2)`
`rArr |omegaL - 1/(omegaC)|=0.25/0.75 rArr |(omega^(2)LC-1)/(omegaC)|=1000/3`
`rArr |(2250)^(2)LC-1)/(2250C)|1000/3`..........(i)
At resosance, `I_(2)` is maximum, `Z^(')` is minimum
`rArr Z^(')=omegaL-1/(omegaC)=0`
`rArr 1/(LC)=omega^(2)=omega_(r)^(2)=(4500)^(2)`............(ii)