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When a 60 mH inductor and a resistor are connected in series with an AC voltage source, the voltage leads the current by 60^(@). If the inductor is replaced by a 0.5 mu F capacitor, the voltage lags behind the current by 30^(@). What is the frequency of the AC supply ? |
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Answer» `(1)/(2pi)xx10^(4)Hz`  Now,`tan 60^(@)=(X_(L))/(R )=(omega L)/(R ) ""`….(i) Again, `tan 30^(@)=(X_(C ))/(R )=(1)/(omega CR)` Now, from EQN. (i) and (ii), we get `(tan 60^(@))/(tan 30^(@))=(omega L)/(R )xx (1)/((1)/(omega CR))=omega^(2)LC` `RARR (sqrt(3))/(1//sqrt(3))=omega^(2)xx 60xx10^(-3)xx0.5xx10^(-6)rArr omega = 10^(4)` As `omega = 2pi upsilon` or `2pi upsilon=10^(4)rArr upsilon = (10^(4))/(2pi)Hz` |
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