Find capacitance from time, power, and voltage

[ikjadoon]

Homework Statement

A flashlight uses 6.0J in 1.8ms (0.0018 seconds). First question: find the power. Got it: 3333W (3.3E3 with two sig figs, which is the accepted answer in the online system).

The real problem: find the capacitance if the flashlight uses 200V.

Homework Equations

Power = Energy used / time
Capacitance = charge / voltage
Power = Current * voltage
Current = delta charge / delta time

The Attempt at a Solution

OK, P = 3333W and V = 200V. Divide to get current, which is 16.67 amps.

16.67 amps and it took 0.0018 seconds to do so. Thus, the change in charge is .03 coulombs via the last equation listed.

Now we have both charge (.03) and voltage (200V).

The capacitance is just the quotient, charge / voltage. I get 1.5 * 10^-4 farads (1.5E-4 F). I have done it at least four times now. No dice. I tried rounding at different times, but I tried 1.5E-4 F and 1.4E-4 F and neither worked. The online system wants 2 significant figures and the appropriate units. I've got two and capacitance is definitely in farads!

What unfortunate mistake am I making?

~Ibrahim~

 

[gneill]

Have you considered directly equating the energy delivered to that stored on a capacitor?

 

[ikjadoon]

Have you considered directly equating the energy delivered to that stored on a capacitor?

Energy delivered to that stored on the capacitor...the U = .5QV formula!

No: I just tried it and it worked!

This time, instead, I got .06 coulombs as the charge, twice the charge compared to my previous attempt.

I completely forgot about that one. But, out of curiosity, why did my initial approach not work? It's bugging me because I can't find the mistake I made or the assumption I took that wasn't valid?

Thank you so much for your help, either way! :D

~Ibrahim~

EDIT: Oh, wait! Is it because the initial charge I calculated was just the charge on one plate?! And the total charge is actually the charge of BOTH plates?

 

[gneill]

I think the problem lies in assuming that the voltage and current were constant over the discharge. This will never happen in a real capacitive situation; everything follows exponential decay curves. So the only thing you could really count on from the given data is the total energy delivered and the initial voltage.

 

[ikjadoon]

I think the problem lies in assuming that the voltage and current were constant over the discharge. This will never happen in a real capacitive situation; everything follows exponential decay curves. So the only thing you could really count on from the given data is the total energy delivered and the initial voltage.

Ohhhhh, OK, right, right, right. We just went over those exponential curves today. That makes sense: it wasn't 16.67 amps at the very beginning.

Whew...all right; thank you!

~Ibrahim~