LC circuit and peak current

In summary: If we accept the 59.9 ms as the correct time, then the first peak of the current will occur at the instant when the capacitor is at its maximum voltage. (This is true because the initial conditions are such that all the energy was originally stored in the capacitor, so the first peak of the current will occur exactly one quarter period later.) So we can check that answer by solving for the time when the capacitor is at its maximum voltage and seeing if that corresponds to t = 59.9 ms. To do this we use the equation for the natural frequency from (1) and the equation for the capacitor voltage when the current is at its zero crossing:##ω_o t_{Vmax} = \frac
  • #1
bishy
13
0

Homework Statement



An LC circuit starts at t=0 with its 2000 microF capacitor at its peak voltage of
14V. At t=35 ms the voltage has dropped to 8.5 V.
a) What will be the peak current?
b) At what time will the peak current occur?
c) What must be the value of the inductor in the circuit?

The Attempt at a Solution



the answers I found are:
a) 1.73*10^-1 A
b) 7.05 s
c) 6.474 H

I've attached the work in two pictures, should be able to zoom into it if you are having problems reading it. I can do it easy doing windows picture media Is this right? Also, it seems like a did a whole lot of work when something probably really obvious that I am not seeing could have been used. Is there a shorter more efficient way to solving this problem?

edit - well attaching isn't going to work, the files are here than:

http://tnaag.org/test.jpg [Broken]
http://tnaag.org/test3.jpg [Broken]
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
A complete solution is offered.

The OP's images are no longer available, but we can surmise that the circuit appears similar to this:

upload_2016-2-5_19-14-39.png


When the switch S closes the circuit will oscillate. In particular, starting at t = 0 the capacitor potential will follow a cosine function while the current follows a sine function. The natural frequency of an LC circuit being given by

##ω_o = \frac{1}{\sqrt{L C}}~~~~~~~~~~~~~~...(1)##

It is convenient to start with part (c) of the question, determining the value of the inductor from the given information.

Part (c) The Value of the Inductor
Writing the capacitor voltage as:

##V_c(t) = V_o cos(ω_o t)##

We can solve for ω_o:

## ω_o = \frac{1}{t} cos^{-1} \left( \frac{V_c}{V_o} \right) ##

We are given that ##V_c = 8.5~V~~## at time ##t = 35~ms##. So that gives us:

## ω_o = \frac{1}{35~ms} cos^{-1} \left( \frac{8.5}{14} \right) ##

##ω_o = 26.238~rad/sec##

Now, using the equation for the natural frequency from (1) and solving for L we have:

##L = \frac{1}{ω_o^2 C} = 726.28~mH##

Part (a) The peak Current
By conservation of energy the maximum current occurs when all the energy originally available in the capacitor is momentarily stored in the inductor during an oscillation cycle. So we write:

##\frac{1}{2}C V_o^2 = \frac{1}{2} L I_{max}^2##

Solving for ##I_{max}## we find:

##I_{max} = V_o \sqrt{\frac{C}{L}} = 735 mA##

Part (b) The Time of Peak Current
As already mentioned the current will follow a sine function, starting at zero and rising to a peak at the first quarter period of the cycle (and repeating every half period after that, corresponding to the positive and negative peaks of the sine curve).

So for the time of the first peak after t = 0 we find the period of the oscillation cycle and divide by four.

## T = \frac{2 \pi}{ω_o} = 239.5 ms##
## t_{Imax} = 59.9 ms##

That's close enough to 60 ms to suspect that the circuit part values were chosen "nicely".
 

1. What is an LC circuit?

An LC circuit is an electronic circuit that consists of an inductor (L) and a capacitor (C) connected in series or parallel. It is also known as a resonant circuit because it can store energy at a specific resonant frequency.

2. What is the purpose of an LC circuit?

The main purpose of an LC circuit is to generate and amplify an alternating current (AC) signal at a specific frequency. It is commonly used in radio and television circuits, as well as in other electronic devices such as filters and amplifiers.

3. How does an LC circuit work?

When an alternating current is applied to an LC circuit, the capacitor stores electrical energy while the inductor stores magnetic energy. As the current alternates, the energy is transferred back and forth between the inductor and capacitor, resulting in a resonant frequency that depends on the values of the inductor and capacitor.

4. What is peak current in an LC circuit?

Peak current is the maximum current that flows through the circuit at the resonant frequency. It occurs when the capacitor is fully charged and the inductor is storing maximum energy. The peak current value can be calculated using the formula I=V/R, where V is the voltage across the circuit and R is the total resistance.

5. How can peak current be controlled in an LC circuit?

Peak current can be controlled by adjusting the values of the inductor and capacitor in the circuit. Increasing the inductance or capacitance will result in a lower resonant frequency and therefore a lower peak current. Additionally, adding a resistor in series with the circuit can also limit the peak current by increasing the overall resistance.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
5K
  • Introductory Physics Homework Help
Replies
28
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
537
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Electromagnetism
Replies
6
Views
961
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top