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Here, I am solving this circuit where Capacitor acts as a voltage source of 10 V. Switch is closed at t=0,and Vc(0-)=10V,I2(0-)=0. When t≥0, what will be the time constant and frequency of the circuit?From my calculation,A=10 and A=B*omega. But how to calculate B? Is omega the angular frequency equal to resonant frequency?I have written the kvl equation I1(t)=dI2/dt Or, Ae^(-kt)cos(wt)=Be^(-kt)wcos(wt)+kBe^(-kt)sin(wt) When t=0, A=Bw and Vc(0-)=10V, hence Vc(t)=10e^(-kt)cos(wt)=I1(t) But what is k and w for parallel RLC circuit? I also got the equation, I1(t)+I2(t)= -dVc(t)/dt Or,Ae^(-kt)cos(wt)+Be^(-kt)sin(wt)=Ae^(-kt)wsin(wt)-Ake^(-kt)cos(wt) By solving it, k= (-1) and B=Aw Correct me if I am wrong.

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  • \$\begingroup\$ To get help with homework questions there is a minimum standard, basically you have to show your work so people can see where you're going wrong and help you. \$\endgroup\$
    – K H
    Feb 22, 2021 at 8:40
  • \$\begingroup\$ Hint \$v_L(t)=v_C(t)=v_R(t)\$ \$\endgroup\$ Feb 22, 2021 at 8:45
  • \$\begingroup\$ I have written the kvl equation I1(t)=dI2/dt Or, Ae^(-kt)cos(wt)=Be^(-kt)wcos(wt)+kBe^(-kt)sin(wt) \$\endgroup\$ Feb 22, 2021 at 9:34
  • \$\begingroup\$ @Camila'svoice Whatever you did you need to add to your question, not in the comments. But, if that's all you did, chances are nobody's going to answer. \$\endgroup\$ Feb 22, 2021 at 9:41
  • \$\begingroup\$ When t=0, A=Bw and as Vc(0-)=10V ,Vc(t)=10e^(-kt)cos(wt) but what's k and w? \$\endgroup\$ Feb 22, 2021 at 9:41

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