An L-C-R series circuit with `R=100Omega` is connected to a `200V`, `50Hz` a.c. source .When only the capacitance is removed, the voltage leads the current by `60^(@)` and when only the inductance is removed, the current leads the voltage by `60^(@)`. The current in the circuit is

Options

A. 1A
B. 2A
C. 1.732 A
D. 0.866 A

Answer

B

Explanation

( b) When capacitor is removed then
`tanPhi= (omegaL) /( R)`
`implies omegaL = RtanPhi`
`implies omegaL = 100xxtan 60^(@) = 100sqrt(3) Omega`
Similarly when inductor is removed then
`tan Phi= (1//omegaC)/( R)`
`implies (1)/(omegaC)= R tanPhi = 100 tan 60^(@)`
`implies (1)/(omegaC) = 100sqrt(3)Omega`
Hence when all three components are connected then net impedance can be written as follows :
`Z= sqrt(R^(2)+(omegaL-(1)/(omegaC))^(2))`
`implies Z= 100Omega`
Current `I = V//Z= 200//100=2 A.`