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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. |
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Answer» Solution :`R = 20 Omega, L = 1.5 H, C = 35 xx 10^(-6)F, V = 220 V, v = 50 Hz, Z = ?, I = ?` We have, impedance , `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2))` `X_(L) = omega L = 2PI VL = 2 xx 3.14 xx 50 xx 1.5` `X_(L) = 471 Omega` `X_(C) = (1)/(omega C) = (1)/(2pi v C) = (1)/(2 xx 3.14 xx 50 xx 1.5)` `X_(C) = 90.99 Omega` `Z = sqrt((20)^(2) + (471 - 90.99)^(2))` `Z = 380.27 Omega` `I = (V)/(Z) = (220)/(380.27) = 0.578 A` |
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