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A source of sinusoidal emf with constant voltage is connected in series with an oscillating circuit with quality factor `Q=100`. At a certain frequency of the external voltage the heat power generated in the circuit reached the maximum value. How much `(` in per cent`)` should this frequency be shifted to decrease the power generated `n=2.0` times ? |
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Answer» `P=(V^(2)R)/(R^(2)+(X_(L)-X_(C))^(2))` At resonance `X_(L)=X_(C)implies omega_(0)=(1)/( sqrt(LC))` Power generated will decrease `n` times when `(X_(L)-X_(C))-( omegaL-(1)/( omegaC))^(2)=(n-1)R^(2)` or ` omega-(omega_(0)^(2))/(omega)=+-sqrt(n-1)(R)/(L)=+-sqrt(n-1) 2 beta` Thus `omegabar(+)2 sqrt(n-1)beta omega - omega_(0)^(2)=0` `( omegabar+sqrt(n-1)beta)^(2)=omega_(0)^(2)+(n-1) beta^(2)` or `(omega)/(omega_(0)=sqrt(1+(n-1)beta^(2)//omega_(0)^(2)))+-sqrt(n-1)beta//omega_(0)` `(` taking only the positive sign in the first term to ensure positive valur for `(omega)/( omega_(0)).)` Now `Q=(omega)/( 2 beta)=(1)/(2) sqrt(((omega_(0))/( beta))^(2) -1)` `(omega_(0))/( beta)=sqrt(1+ 4 Q^(2))` Thus `(omega)/( omega_(0))=sqrt(1+(n-1)/( ( 1+4Q)))+-sqrt(n-1)//sqrt(1+ 4 Q)` For large `Q` `|( omega-omega_(0))/( omega_(0))|=(sqrt( en -1))/( 2Q)-(sqrt(en -1))/(2Q)xx100%=0.5%` |
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