Initially, the capacitor in a series?

Question

Initially, the capacitor in a series LC circuit
is charged. A switch is closed, allowing the
capacitor to discharge, and 0.3 micros later the
energy stored in the capacitor is one-fourth
its initial value.
Find the inductance if the capacitance is
7 pF. Answer in units of H.

Answer 1

V = L Q'' + Q/C

The initial full charge can be calculated using

Q0 = CV

After you unhook the battery,
0 = L Q'' + Q/C

Q'' = (1/LC) Q

So Q = Q0 cos (omega t)

where omega = sqrt(1/LC)

The energy goes like Q^2.

So to go from zero to 1/4 energy in time t.

(sin (omega t)) ^2 = 1/4. Solve for omega. Then, because you know C, you can solve for L.

Answer 2

The energy in a cap is (1/2) x C x V ^2.

Calculate the voltage needed to have 1/4 the energy in it compared to a full charge. Set this voltage (it will be less than one) equal to a cosine. The arc cos will tell you what percentage of a cycle (how many degrees out of 360) the circuit went through. From there you know the frequency and you can calculate the inductance. The question is misleading through because the circuit could have gone through countless cycles before you measured it.

Answer 3

the present bypass ( and subsequently the volts around the resistor ) is I = I1 e ^ -t/T whilst the capacitor has reached seventy 9.5 % then the emf around the resistor has dropped to twenty.5% so e^ -t/T = 0.205 -t/T = ln(0.205) = - a million.fifty 8 ie t/T = a million.fifty 8 the place T is the time consistent and t is the time elapsed.