Question

\[F= 1/2\Pi \sqrt{LC}\]
for C

Answer 1

Just some algebraic manipulation going on here.\[F= 1/2\Pi \sqrt{LC}\]\[{2F \over \Pi}= \sqrt{LC}\]\[\left({2F \over \Pi}\right)^2= LC\]\[{\left({2F \over \Pi}\right)^2 \over L}= C\]\[{4F^2 \over L\Pi ^2 }= C\]

Answer 2

supposed to be a fraction 1 over 2 pi sq rt LC