Question

If an 87 -uF capacitor is charged to 120 volts, How much power would be produced if the capacitor is discharged in 50 micro-seconds. They gave me a hint but I don't understand it (T=50*10^-6s)

Answer 1

The hint is that they're asking you to use the Joules per Second definition of power.

Figure out how many Joules are stored in an 87uF cap at 120V, and plug that into the Joules per Second definition of Watts.

Answer 2

0.6264 J ?

Answer 3

Sounds about right

Answer 4

626400 W ?

Answer 5

50 micro seconds, not 1...

Answer 6

I'm a little confused now...

Answer 7

12528 W ?

Answer 8

The hint is to convert 50 micro-seconds to seconds

Answer 9

Sounds better. Seat of the pants cross-check - Watts = Volts * Amps. You're starting at 120V. Is 100 Amps a believable figure for the discharge?

Answer 10

I don't understand your Q

Answer 11

Congrats @tjjavi It seems like a lot of power but it didn't last long lol

Answer 12

It's always nice to be able to sanity check your answer, using an independent route. In this case, there is another formula for electrical power (Watts) that uses Volts and Amps. You know your initial Volts, and using your hypothesized 12528 Watts, come up with a number for Amps that would need to be true, if the answer was true.

Doing this, you get roughly 100Amps, and the question is, does that sound like a reasonable number? If the number seems unreasonably high, or unreasonably low, then you probably want to check to see if you've made a math error somewhere. You know the number needs to be subjectively "big", because it's discharging the capacitor in a very brief time, but it's lightning-bolt-like big. 100 fits the bill for big-but-not-too-big, so just as a seat of the pants sanity check, that estimation says that your 12528W number sounds reasonable.

This kind of sanity-checking is far from exact, but it can give you some confidence in your answer without going through the work of an actual full alternative solution.

Answer 13

@tjjavid , The FAA long range radar ARSR-1D, has a peak power of 4 million watts, but it was contained in a pulse that was only 2 microseconds wide and repeated every 2,000 microseconds, so the average power didn't sound so impressive.

Answer 14

@wilray well said

Answer 15

aargh - typing error -- make that "_not_ lightning-bolt-big"...

Answer 16

There is a formula for energy stored by a capacitor, but I don't know if it was in tjjavid study text, and that is joules is equal to:\[Q ^{2} \over2C\] Where Q is charge, and C is capacitance in Farad, Q in Coulombs.

Answer 17

Yeah - given their hint, it seemed unlikely that they wanted the students to calculate the stored charge and then integrate the discharge curve. I've had a few profs who might be that cruel, but it doesn't look like many of those machiavellian questions end up here.

Answer 18

I would like to emphasize the importance of the "sanity check" you have mentioned in an earlier post. Look for reasonableness in your solution, for example if you are solving for the equivalent total resistance of a group of parallel resistors, the total should be less than any of them, or those problems where jack can mow a lawn in 1 day, and jill takes 2 days then both mowing (the same yard) should be less than 1 day (i hope).

Answer 19

Thanks a lot for the extra eplanations @radar and @willray

Answer 20

You're welcome and wishing you good luck in your studies.

Answer 21

No problem at all. Something to keep in mind - it's the little things, like dividing by 1, rather than 50, that'll trip you up, more frequently than the big things like finding the right formulas.

Math and Physics are a game. They're a game of fiddly little details, but they're a game. Treat them like one, and revel in the victories when you catch a fiddly little detail that someone else missed, and you'll have fun.

Good luck!