In terms of q, the voltage equation for series L-C-R circuit is given by

Options

A. `L(dq)/(dt)=R(dq)/(dt)+q//C=V_(m)sinomegat`
B. `L(d^(2)q)/(dt^(2))+R(dq)/(dt)+q//C=V_(m)sinomegat`
C. `L(d^(2)q)/(dt)-R(dq)/(dt)+q//C=V_(m)sinomegat`
D. `L(d^(2)q)/(dt)-R(dq)/(dt)-q//C=V_(m)sin omegat`

Answer

B

Explanation

The voltage equation for the circuit is
`L(di)/(dt)+Ri+(q)/(C)=V=v_(m)sinomegat`
We know that, `i=dq//dt`. Therefore, `di//dt=d^(2)q//dt^(2)`. Thus,
in terms of q, the voltage equation becomes
`L(d^(2)q)/(dt^(2))+R(dq)/(dt)+q//C=V_(m)sin omega t`