InterviewSolution
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InterviewSolution
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Find the value of phase lag/lead between the current and voltage in the given CLR circuit, fig. Without making any other change, find the value of additional capacitor, such that when joined switably to `C = 2 mu F`, would make the power factor of this circuit unity. |
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Answer» (a) ` V=V_(0) sin (1000 t + phi)` ` V=V_(0) sin ( omegat+ phi) ` `omega=1000` ` X_(L) = omegaL` ` X_(L) = 1000 xx 100 xx10^(-3) =10^(2) = 1000 Omega` `X_(C) = 1/(omegaC) = 1/(1000 xx 2xx 10^(-6)) =1/2 xx 10^(3) = 500 Omega` Phase difference ` tan phi= ( X_(L) -X_(C))/R` ` tan phi = (X_(C)-X_(L))/R ,X_(C) gt X_(L)` ` tan phi = (500 -100)/400 = 400/400 =1` ` tan phi =1` ` tan phi tan 45^(@)` `phi = 45` Now ` sqrt(R^(2) +(X_(L) -X_(C))^(2))` ` = sqrt(400^(@)(100-500)^(2)) = sqrt(160000 + 160000) = sqrt(320000) = 565.68` power factor = ` cos phi R/Z= 400/565.68 = 0.707` (b) ` X_(L)=X_(C) . " Let new , capatior " = C_(1)` ` omegaL= 1/(omegaC)` `C= = 1/(omega^(2)L) =1/(2pif)^(2) X_(L)) = 1/(4pi^(2)f^(2)L)` ` 2xx 1-^(-6) +C_(1) = 1/( 4xx (3.14)^(2) xx (50)^(2) xx 100 xx 10^(-3)) = 1/( 4xx 9.8596 xx 2500 xx 10^(-1))` ` 1/(4xx 9.8596 xx250) = 1/(9859.6)` ` C_(1) =1/(9859.6) -2 xx 10^(-6) = 1/(9859.6) -2/10^(6) = 0.000101424-0.000002= 0.00009924=99.42 muF` |
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