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A 40 ohm resistor, 3 mH inductor and `2 mu F` capacitor are connected in series to a 110 V, 5000 Hz a.c. source. Calculate the value of current in the circuit. |
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Answer» Here, `R = 40 Omega, L = 3 mH = 3 xx 10^(-3) H`, `C = 2 mu F = 2 xx 10^(-6) F, E_(v) = 110 V, v = 5000 Hz` `X_(L) = omega L = 2 pi L = 2 xx 3.14 xx 5000 xx 3 xx 10^(-3)` `= 94.2 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (1)/(2 xx 3.14 xx 5000 xx 2 xx 10^(-6)) = 15.92 Omega` `Z = sqrt(R^(23) + (X_(L) - X_(C ))^(2))` `= sqrt(40^(2) + (94.2 - 15.92)^(2)) = 87.9 Omega` `I_(v) = (E_(v))/(Z) = (110)/(87.9) = 1.25 A` |
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