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.