A `20 Omega` resistor, 1.5 H inductor and `35 mu H` capacitor are connected in series with a 220 V, 50 ac supply. Calculate the impedance of the circuit and also find the current through the circuit.

Explanation

Given R = 20 W , L = 1.5 H,
` C = 35 xx 10^(-6) F ` ,
` V_(max ) = 220 V , f = 50 Hz `
w.k.t ` X_(L) = 2 pi fL = 2 xx 3.142 xx 50 xx 1.5 = 471 Omega ~~ 91 Omega `
` X_(C) = (1) /(2 pi f C) = (1)/(2xx 3.142 xx 50 xx 35 xx 10^(-6))
` (X_(L) - X_(C)) = 471.3 - 91 = 380.3 Omega `
w.k.t ` Z = sqrt(R^(2) + (X_(L) - X_(C))^(2)) = sqrt (400 + 144 704 )`
impenance , ` z ~~ 381 Omega `
` therefore I_(max) = (V_(max))/(Z) = (220) /(381) = 0.578 A ~~ 0.6 A ` .