A capasitor of capacitance `1 (mu)F` 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 valueA. (A) `sqrt(1000)mA`B. (B) `1mA`C. (C) `1(mu)A`D. (D) `1000 mA`

A capasitor of capacitance `1 (mu)F` is charged to a potential of 1 V,

1.	A capasitor of capacitance `1 (mu)F` 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 valueA. (A) `sqrt(1000)mA`B. (B) `1mA`C. (C) `1(mu)A`D. (D) `1000 mA`
Answer» Correct Answer - A Change on the capacitor, ltbRgt `(q_0)=CV = 1xx10^(-6) xx1 =10^(-6)C` or `(I_0) = omega (q_0)=`maximum current Now `omega =(1)/(sqrt(LC))=(1)/(sqrt(10^(-9)) =(10^(9))^(1//2)` `:. I_0=(10^(9))^(1//2) xx(1 xx 10^(-6)) = sqrt(10^(-3)) A=sqrt(1000)mA`.

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A capasitor of capacitance `1 (mu)F` 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 valueA. (A) `sqrt(1000)mA`B. (B) `1mA`C. (C) `1(mu)A`D. (D) `1000 mA`

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