A LC circuit has zero resistance, no battery, a capacitor C=66microFarads, and a

Question

A LC circuit has zero resistance, no battery, a capacitor C=66microFarads, and an inductor L=0.7H connected in a simple loop. The capacitor and inductor store a compined total energy of U=24microJoules.
(a) If all the circuits energy is stored in the capacitor at time t = zero, when is the soonest time that all the energy will be stored in the inductor?
ms
(b) How much charge is stored on one side of the capacitor at t = 0?
microCoulombs.
(c) How much current flows through the inductor at the time you found in (a)?
mA
(d) If the inductor is a cylindrical solenoid of length l=11centimeters and radius r=3centimeters, find the number of coils in the solenoid.
coils
(e) What is the maximum magnetic field inside the inductor?
mT
(f) If the maximum electric field in the capacitor is E=1.3x103N/C and the space between the plates if filled with a dielectric of constant 100,000, what is the separation between the plates?
mm
(g) What is the area of one of the plates?
cm2

Explanation / Answer

a)

T = 2/ = 2*(LC) = 2*3.14*(0.7*66e-6) = 42.686 ms

the time that all the energy will be stored in the inductor = T/2 = 21.3 ms

b)

U = (1/2) Q2/C

>>>> Q = (2CU) = (2*66e-6*24e-6) = 56.3 C

c)

U = (1/2) L i2

>>>> i = (2U/L) = (2*24e-6/0.7) = 8.28 mA

d)

L = N2A/l

>>> N = (l L/(A))

= (0.11*0.7/(4*3.14e-7*(3.14*0.03*0.03)))

= 4658 turns

e)

B = Ni/L = (4*3.14e-7*4658*0.00828079/0.11) = 0.440 mT

f)

U = (1/2)CV2

>>>> V = (2U/C) = (2*24e-6/66e-6) = 0.8528 V

>>>> d = V/E = 0.8528/1.3e3 = 0.656 mm

g)

C = k0A/d

>>> A = dC/(k0) = 0.656e-3*66e-6/(100000*8.85e-12) = 489 cm2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.