Saved Bookmarks
| 1. |
An oscillating circuit consists of a capacitor with capacitance `C`, a coil of inductance `L` with negligible resistance, and aswitch. With the switch disconnected , the capacitor was charged to a voltage `V_(m)` and then at the moment `t=0` the switch was closed . Find `:` `(a)` the current `(I(t))` in the circuit as a function of time, `(b)` the emf of self- inductance in the coil at the moments when the electric energy of the capacitor is equal to that of the current in the coil. |
|
Answer» After the switch was closed, the circuit satisfies `-L(dI)/( dt)=(q)/(C)` or `(d^(2)q)/(dt^(2))+ omega_(0)^(2)q=0implies q=CV_(m) cos omega_(0)t` where we have used the fact that when the switch is closed we must have `V=(q)/(C)=V_(m), I=(dq)/( dt)=0` at `t=0` Thus `(a)` `I=(aq)/( dt)=-CV_(m)omega_(0) sin omega_(0)t` `=-V_(m) sqrt((C)/(L)) sin omega_(0)t` `(b)` The electrical energy of the capacitor is `(q^(2))/( 2 C) alpha cos^(2) omega_(0)t` and of the inductor is `(1)/(2) L I^(2) alphasin^(2) omega_(0)t.` The two are equal when ltbr` ` omega_(0)t=(pi)/( 4)` At that instant the emf of the self `-` inductance is `-L(di)/(dt)=V_(m) cos omega_(0)t =V_(m) //sqrt(2)` |
|