Question

A 7.50-nF capacitor is charged up to 12.0 V. then disconnected from the power supply and connected in series through a coil. The period of oscillation of the circuit is then measured to be 8.60 X 10^-5 s. Calculate:
(a) the inductance of the coil:
(b) the maximum charge on the capacitor:
(c) the total energy of the circuit;
(d) the maximum current in the circuit.

Answer 1

Answer:

a

Explanation:

Answer 2

Answer: a) 25 *10^-3 H ; b) 90 nC; c) 540 nJ ; d) 6.6 mA

Explanation: In order to solve this problem we have to consider the expression for a L-C electric circuit. Also we should take into account the charge of the capacitor.

The oscillation frequency of the L-C circuit is given by:

$\omega= \sqrt{1/L*C}$

Then we have that 2*π/ω= T ( period)

Combining both expression we have that:

L=1/(ω^2*C)=T^2/((2*π)^2*C)=(8.6*10^-5)^2/(2*π^2*7.5*10^-9)=25 *10^-3 H

The charge of teh capacitor is given by:

Qo=C¨*V=7.5*10^-9*12= 90 *10^-9 C

The total energy can be obtained at t=0, the current is equal to zero then the total energy is stored at the capacitor:

Uc=1/2*C*V^2=540 nJ

At any tim, this the total energy is divided in one part stored at the coil and other part stored in the capacitor.

Finally, the maximun current is given;

as Q(t)= Qo*cos (ω*t) then I(t)=-ω*Qo sin (ω*t)

I(t) maximun;  Imax=ω*Qo= 6.6 *10^-3A