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A series LR circuit is connected to an ac source of frequency omega and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the old one is |
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Answer» `SQRT((2)/(3))` `= (R)/(sqrt(R^(2) + X_(L)^(2))) = (R)/(sqrt(R^(2) + (2R)^(2))) = (R)/(sqrt5R)` Power factor (new) ` = (R)/(sqrt(R^(2) + (X_(L) - X_(C))^(2))) = (R)/(sqrt(R^(2) + (2R - R)^(2))) = (R)/(sqrt 2R)` `therefore` (New power factor)/(Old power factor) = `((R)/(sqrt2R))/((R)/(sqrt5R)) = sqrt((5)/(2))` |
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