What is the capacitance of the circuit problem.

AI Thread Summary
An LC circuit with a 1mH inductor has a capacitor that is initially fully charged, and after 2.1 ms, its energy is reduced to half. To find the capacitance, the relationship between the period of oscillation and the inductance is used, leading to the formula C = (T^2) / (4 * π^2 * L). Given that the capacitor is half discharged at 2.1 ms, the full discharge occurs at 4.2 ms, which is the full period. By substituting the values into the formula, the capacitance is calculated to be 4.4 microfarads. The charge on the capacitor varies over time according to a cosine function, with the angular frequency being derived from the energy condition at 2.1 ms.
tag16
Messages
95
Reaction score
0

Homework Statement


An LC circuit consists of a 1mH inductor. The capacitor is fully charged. After 2.1 ms,
the energy in the capacitor is ½ of its original value. What is the capacitance of the
circuit?


Homework Equations


(Q/C)+ L(d^2Q/dt^2)=0

d^2Q/dt^2= -(1/LC)Q


The Attempt at a Solution



I found the above equations in my physics book but I don't know what Q would be (it says that the capacitor is fully charged but it doesn't give an actual number), or how to get into a C= ? format in order to find the capacitance. Also I'm not even sure these are the formulas I need for this problem and if they're not then I have no idea what to do.
 
Physics news on Phys.org


you can try to solve the task using Thomson's formula T=2*П * root of L*C
as i understood there is given that after 2,1 ms the capacitor is half discharged and it means that it will be fully dischardged after 4,2 ms, that is 4,2*10^-3 s - it is full period of time.
transforming the formula C= (T^2) / (2^2 * П^2 * L) and the result is 4,4 micro farads.
 


The charge varies with time according to a cosine function - Q=QoCos(wt)
w is the angular frequency.
From the fact that is 1/2 from the maximum at t=2.1 ms you can find w.
 
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