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For the circuit shown in the figure. Find the expressions for the impedance of the circuit and phase of current |
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Answer» Solution :Let ac SOURCE` V = V_(@) sin omegat` `I_(R)=(V_(@))/(R)SINOMEGAT,I_(L)=(V_(@))/(X_(L))sin(omegat-(pi)/(2))andI_(C)=(V_(@))/(X_(c))sin(omegat+(pi)/(2))` And as all these are in parallel, `I=I_(R)+I_(L)+I_(c)` `=[(V_(@))/(R)sinomegat+V_(@)(omegaC-(1)/(OMEGAL))cosomegat]` `(V_(@))/(R)=I_(0)sinphiV_(o)(omegaC-(1)/(omegaL)=I_(o)COsomegat` `I=I_(0)sin(omegat-phi)` `tanphi=(omegaC-(1)/(omegaL))//(1)(R)=R(omegaC-(1)/(omegaL))` `I_(0)=(V_(0))/(Z)V_(0)=[(10)/(R^(2))+(omegaC(1)/(omegaL))^(2)]^(t//2)` `z=(1)/(R^(2))+[(1)/(R^(2))+((1)/(X^(c))-(1)/(X_(L))^(2)]^(t//2)` |
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