Archived Is There a More Efficient Way to Solve LC Circuit Problems?

[bishy]

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
http://tnaag.org/test3.jpg

 

[gneill]

A complete solution is offered.

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

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".