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A 100 uF capacitor in series with a 40 Omegaresistance is connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit ? (b) What is the time lag between the current maximum and the voltage maximum ? |
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Answer» Solution :Here C = 100 `muF = 100 xx 10^(-6) F = 10^(-4) F, R = 40 Omega, V_(rms) = 110 V` and v= 60 Hz or `omega = 2pi v = 3.14 xx 60 xx 376.8 s^(-1)` `therefore` Impedance `Z = sqrt(R^(2) + X_(C)^(2)) = sqrt(R^(2) + (1/(Comega))^(2)) = sqrt((40)^(2) + (1/(10^(-4) xx 376.8))^(2)) = 48 Omega` `therefore` Maximum current `I_(m) = V_(m)/Z = (sqrt(2) V_(rms))/Z = (sqrt(2) xx 110)/48 = 3.24 `A (b) Phase difference between current and voltage. `PHI = tan^(-1)(X_(C))/R = tan^(-1) ((1//Comega)/R) = tan^(-1)(1/(10^(-4) xx 376.8 xx 40))` `=(33.6 xx pi)/180` rad `therefore` Time lag between current maximum and voltage maximum `t= phi/(2pi v) = (33.6 xx pi)/(180 xx 2pi xx 60) = 1.55 xx 10^(-3)`s. |
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