An AC voltage of rms value 2V is applied to a parallel combination of L and C in which L=2mH and C=3.2`muF`. The current through each branch at resonance is

Options

A. 80 mA, 80 mA
B. 80 mA, 60 mA
C. 60 mA, 80 mA
D. 40 mA, 40 mA

Answer

A

Explanation

`f=(1)/(2pisqrt(LC))`
`=(1)/(2xx3.14sqrt(2xx10^(-3)xx3.2xx10^(-6)))=1989Hz`.
`I_(L)=(e_(rms))/(X_(L))=(e_(rms))/(2pifL)=(2)/(2xx3.14xx1989xx2xx10^(-3))`
`~~80mA`
At resonance, current through inductor and condenser is same. Thus `I_(C)=80mA`