- #1
ikjadoon
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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~