How Do You Calculate the Oscillation Period and Energy Gain in an LC Circuit?

AI Thread Summary
To calculate the oscillation period of an LC circuit with a current amplitude of 7.20 mA, a potential amplitude of 250 mV, and a capacitance of 240 nF, the relevant equations involve the energy stored in the inductor and capacitor. The maximum rate at which the inductor gains energy can be derived by differentiating the energy stored in the inductor with respect to time. The participant initially miscalculated the charge (Q) and the energy equations, leading to confusion about the maximum energy gain. After clarification, they recognized the need to correctly apply the formulas for energy and charge in the context of the circuit. The discussion highlights the importance of accurately interpreting equations to solve problems related to oscillating LC circuits.
Oijl
Messages
102
Reaction score
0

Homework Statement


An oscillating LC circuit has a current amplitude of 7.20 mA, a potential amplitude of 250 mV, and a capacitance of 240 nF.

(a) What is the period of oscillation?

(e) What is the maximum rate at which the inductor gains energy?

Homework Equations


Possibly: Energy stored in an inductor = 1/2 * L * I^2
i = -wQsin(wt+ø)

The Attempt at a Solution


Energy per unit of time is Watts.

If I find the equation for energy stored in an inductor, I can differenceate that with respect to time and find its maximum value.

I thought that this would be:
dE/dt = P = -L(w^3)(Q^2)sin(wt)cos(wt)

And isn't the maximum value for this just

(1/2)L(w^3)(Q^2) ?

And isn't that just

0.9 mW ?

I've done this problem wrong several times so far, so is this one right?
 
Last edited:
Physics news on Phys.org
U = ((Q^2)/2C) cos(^2)(wt+ø) is the equation for the energy contained in the capacitor,
not for the potential across it.
 
How are you determining Q? I get Q = VC = .25 *2.4e(-7) = 6e(-8) C.

Q = VC = \int_0^{\pi/2} dQ = \int_0^{\pi/2} idt = \int_0^{\pi/2} i_{max}\sin{\omega t}dt

so:

Q = VC = i_{max}/\omega



AM
 
Oh, I see. I misunderstood the equation for U I had written, and therefore got a bad Q. Thanks to both.
 
But now I have another question for this problem, and I've edited the first post for it.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top