Unraveling the Enigma: Understanding the Equation

  • Thread starter Thread starter hidemi
  • Start date Start date
  • Tags Tags
    Mystery
Click For Summary
The discussion focuses on calculating the charge on a capacitor after five complete cycles using a specific equation from the textbook. To find the charge, one must first calculate the angular frequency (ω') using the values for resistance (R), inductance (L), and capacitance (C). Initial conditions, including charge (q(0) = 25μC) and current (I(0) = 0), are used to determine q0 and φ. The duration of one cycle (T') is derived from ω', and then the charge at five cycles (q') is calculated by substituting t = 5T' into the initial equation for q(t). Finally, the energy stored in the capacitor is computed using the formula E = (1/2)(q'^2/C).
hidemi
Messages
206
Reaction score
36
Homework Statement
An RLC circuit has a resistance of 200 Ω, an inductance of 15 mH, and a capacitance of 34nF. At time t = 0, the charge on the capacitor is 25
μC and there is no current flowing. After five complete cycles the energy stored in the capacitor is:

A) 0.64 μJ
B) 77 μJ
C) 0.49 mJ
D) 1.8 mJ
E) 9.2 mJ

The answer is A.
Relevant Equations
(see better expression below)
The equation is the only equation discussed in the textbook.
Is there a hint of how I could start this?
 

Attachments

  • 3.jpg
    3.jpg
    20.5 KB · Views: 241
Physics news on Phys.org
Yes well, you should find the charge on the capacitor after five complete cycles. To find it you will use the first equation you give in the attached picture.

You can calculate ##\omega'## from the given values for R L C, and also ##q_0## and ##\phi## from the initial conditions of ##q(0)=25\mu C,I(0)=0##.

Also once you find ##\omega'## find ##T'=\frac{2\pi}{\omega'}## which is the duration of one cycle. Finally put ##t=5T'## in the initial equation for ##q(t)## and find ##q'=q(5T')##. The energy stored will be $$E=\frac{1}{2}\frac{q'^2}{C}$$.
 
Thanks for the hint. I got it.
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 1 ·
Replies
1
Views
779
  • · Replies 5 ·
Replies
5
Views
2K
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
817
  • · Replies 61 ·
3
Replies
61
Views
5K
  • · Replies 19 ·
Replies
19
Views
1K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K