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In an oscillating circuit shown in figure, the coil inductance is equal to L=2.5 mu F. The capacitor have capacitances C_(1)=2.0 mu F and C_(2)=3.0 mu F. The capacitors were charged to a voltage V=180 V, and then the switch S w was closed. Find : (a) the naturaloscillation frequency, (b) the peak value of the currentflowing through the coil. |
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Answer» Solution :The equations of the `L-C` circuit are `L(d)/(dt)(I_(1)+I)=(C_(1) V-int I_(1)dt )/( C_(1))=(C_(2)V-int I_(2) dt)/( C_(2))` Differentiating again `L(I_(1)+I_(2))=-(1)/(C_(2))I_(1)=-(1)/(C_(2))I_(2)` Then ` I_(1)=(C_(1))/(C_(1)-C_(2))I, I_(2)=(C_(2))/(C_(1)+C_(2))I`, `I=I_(1)+I_(2)` so `L(C_(1)+C_(2))I+I=0` or `I=I_(0) sin ( omega_(0)t + ALPHA)` where `omega_(0)^(2)=(1)/( L(C_(1)+ C_(2))) (` PART `a)` `(` Hence `T=(2PI)/( omega_(0))=0.7 ms )` At ` t=0 , I=0` so `alpha=0` `I=I_(0)sin omega_(0)t` The peak value of the CURRENT if `I_(0)` and it is RELATED to the voltage `V` by the first equation `LI=V-intI dt//(C_(1)+C_(2))` or `+ L omega_(0) I_(0) cos omega_(0)t = V-(1)/( C_(1)+ C_(2))int_(0)^(t)I_(0) sin omega_(0) t dt ` `(` The `P.D.` across the inductance is `V` at `t=0)` `=V+(1)/(C_(1)+C_(2)). (I_(0))/(omega_(0))( cos omega_(0)t -1)` Hence `I_(0)=(C_(1)+C_(2))omega_(0) V=Vsqrt((C_(1)+ C_(2))/( L))=8.05A.` 
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