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Consider the RLC circuit shown below connected to an AC source of constant peak voltage V_(0) and variable frequency omega_(0).The value of L is 20 mH.For a certain value omega_(0) = omega_(1), rms voltage across L,C, R are shown in the diagram. At omega_(0) = omega_(2), it is found that rms voltage across resistance is 50 V. Then the value of omega_(2) is |
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Answer» `sqrt((3)/(5))omega_(1)` At resonance, `X_(L) = X_(C)` `omega_(2)L = (1)/(omega_(2)C) , omega_(2) = (1)/(sqrtLC)` Also, At `omega = omega_(1)` `I = (100)/(X_(L) = (60)/(X_(C)), (100)/(omega_(1)L) = (60)/(1/omega_(1)C)` `C = (100)/(omega_(1)^(2)L xx 60) = (5)/(3omega_(1)^(2)L)` `omega_(2) = (1)/(sqrt(L xx (5)/(3omega_(1)^(2)L))) = sqrt((3)/(5))omega_(1)` |
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