A capacitor of capacitance `1 muF` is charged to a potential of `1 V`, it is connected in parallel to an inductor of inductance `10^(-3)H`. The maximum current that will flow in the circuit has the value

Options

A. `sqrt(1000)mA`
B. `1mA`
C. `10mA`
D. `1000mA`

Answer

A

Explanation

Charge on the capacitor,
`q_0 = CV = 1 xx 10^(-6) xx 1 = 10^(-6) C = 1 muC`
Here , `q = q_0 sin omega t `
Maximum current , `I_0 = omega q_0`
Now `omega = (1)/(sqrt(LC)) = (1)/(sqrt(10^(-3) xx 10^(-6)) ) = (1)/(sqrt(10^(-9) )) = (10^9)^(1//2)`
`therefore I_0 = (10^9)^(1//2) xx (1 xx 10^(-6) ) = sqrt1000 mA`