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A box contains L,C and R When 250V dc is applied to the terminals of the box a current of 0.1 A flows in the circuit When an ac source of 250V rms at 2250 rad//sec is connected a current of 1.25 A rms flows it is observed that the current rises with frequency and becomes maximum at 4500rad/sec Find the value of L,Cand R Draw the circuit diagram . |
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Answer» Solution :When dc source is applied current `=1A` In dc circuit inductor offers ZERO resistance and capacitor offers infinite resistance `R` and `C` cannot be in series `I = (V_(dc))/ (R ) implies 1 = (250)/R` ` R = 250 Omega` Force source IMPEDANCE `Z = (V_(rms))/(i_(rms)) = (250)/(1.25) = 200 Omega` As `Z lt R, R L C` cannot be in series `R` cannot be in series with `C` as explained earlier As current is maximum at resonance and current increases with frequency `R` cannot be in series with `L` In fact `L` and `C` must be in series `i_(1) = (250)/(250) =1` A in phase with voltage Impedance of lower branch `Z' = | (omegaL -(1)/(omegaC))|` `i_(2) = (250)/(Z)` The phase difference between `i_(2)` and voltage is `pi//2` The angle between `i_(1)` and `i_(2)`i is `90^(@)` `i_(1)^(2) + i_(2)^(2) = i^(2)` `(1)^(2) + i_(2)^(2) = (1.25)^(2)` `i_(2)^(2) = (1.25)^(2) -(1)^(2) = (2.25) (0.25)` `i_(2) = 1.5 xx 0.5 =0.75 A` `i_(2) = (V)/(Z')` `Z' = (V)/(i_(2))` `|omegaL - (1)/(omegaC)| = (250)/(0.75)` `|(omega^(2) LC -1)/(omegaC)|= (1000)/(3) ` `|((2250)^(2) LC -1)/(2250C)|= (1000)/(3)` At resonance `i_(2)` is maximum `Z` is minimum `Z' = omegaL - (1)/(omegaC) =0` `(1)/(LC) = omega^(2) = omega_(r)^(2) = (4500)^(2)` Solving (i) and (ii) we get `|(((2250)/(4500))^(2) -1)/(2250 C)|= (1000)/(3)` `C = (3//4)/(2250)X (3)/(1000) = 10^(-6) F = 1mu F` `LC = (1)/((4500)^(2)` `L = (1)/((4500)^(2) xx 10^(-6)) = (100)/(45 xx 45) =(4)/(81)H` `R = 250 Omega, L = 4//81 H ,C =1 MUF`   .
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