चित्र [V.39 देखें और `V_("out")//V_(s)` निकालें जब `omega`ज्यादा हो।

Answer

1

Explanation

क्योंकि `V_("out")= sqrt(V_(R)^(2) + V_(L)^(2)) = I sqrt(R^(2) + (omega L)^(2))`
` = V_(s)/Z sqrt(R^(2)+ omega^(2)L^(2)) `, इसलिए
`V_("out")/V_(s) = sqrt(R^(2)+ omega^(2)L^(2))/sqrt(R^(2)+ (omega L - 1//omega C)^(2))`
जब `omega` ज्यादा हो , तो `V_("out")/V_(s) approx sqrt((omega L)^(2))/sqrt((omegaL)^(2)) = 1`