InterviewSolution
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InterviewSolution
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Figure illustrates the simplest ripple filter. A voltage V=V_(0)(1+ cos omega t) is fed to the left input. Find : (a) the output voltabe V^(')(t), (b)the magnitude of the product is eta =7.0 times less than the direct voltage component, if omega=314 s^(-1). |
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Answer» Solution :Here, `V=IR+(int_(0)^(t)Idt)/(C)` or ` Rdot(I)+(1)/(C)I=dot(V)=- omega V_(0) sin omegat ` Ignoring transients , a solution has the form `I=I_(0)sin ( omegat- ALPHA)` ` omega RI_(0)COS ( omegat- alpha)+(I_(0))/(C) sin ( omegat- alpha) =- omegaV_(0) sin omegat` `=-omegaV_(0){sin ( omegat- alpha) cos alpha + cos ( omegat- alpha ) sin alpha}` s o` RI_(0)=- V_(0) sin alpha` `(I_(0))/( omegaC)=-V_(0) cos alpha` ` alpha=pi+TAN^(-1) ( omega RC)` `I_(0)=(V_(0))/( sqrt(R^(2)+((1)/( omegaC))^(2)))` `I=I_(0) sin ( omegat- tan ^(-1)omegarc- pi)=- I_(0) sin ( omegat - tan^(-1)omegaRC)` Then`Q=int_(0)^(t) I dt=Q_(0)+(I_(o))/( omega)cos ( omegat- tan ^(-1) omega RC)` It satisfies `V_(0)(1+cos omegat)=R(dQ)/(dt)+(Q)/(C)` if `V_(0) (1+ cos omegat)=- RI_(0) sin ( omegat- tan ^(-1) omega RC)` `+ (Q_(0))/( C)+(I_(0))/( omegaC) cos ( omega t - tan ^(-1)omegaRC)` and `{:((I_(0))/(omegaC)=V_(0)//sqrt(1+( omegaRC)^(2))),(RI_(0)=(V_(0) omega RC)/(sqrt(1+ omegaRC))):}}checks` Hence`V^(')=(Q)/(C)=V_(0)+(V_(0))/( sqrt(1+( omegaRC)^(2)))cos ( omegat-alpha)` `(b)``(V_(0))/( eta)=(V_(0))/( sqrt(1+( omegaRC)^(2)))` or `eta^(2)-1= omega^(2) ( RC)^(2)` or `RC=sqrt( eta^(2)-1 )//omega = 22 ms.` 
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