# Camara Sample Exam - Morning #16



## NVRSTOP (Feb 24, 2008)

16. A voltage of v(t) = (339.5V)sin2512t is applied to a 30 microF capacitor that was initially uncharged. (t is measured in seconds.) What is the approximate current in the capacitor at 0.625s?

Solution indicates:

i(t) = C * dv(t)/dt

= C (d((339.5V)sin 2512t)/dt)

= C (339.5V)(d(sin 2512t)/dt)

= C (339.5V)((2512s^-1)cos 2512t)

= (30 x 10^-6 F)(339.5 V)(2512 s^-1) x cos((2512s^-1)(0.625 x 10^-3 s))

= 0.020 A

I am getting 1.17 (I'm converting 2512 to degrees (mult by (180/pi) = 143,927.

I'm sure it's a conversion that I am missing here, I just do not see it right now.

NVRSTOP


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## Flyer_PE (Feb 24, 2008)

I'm away from my calculator right now so I'm struggling a little bit with the math.

Are you converting 2512 radians to degrees or are you converting 2512*0.625*10-3?

If you don't get a better answer, I'll take another shot at it when I get home.

Jim


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## NVRSTOP (Feb 24, 2008)

IFR_Pilot said:


> I'm away from my calculator right now so I'm struggling a little bit with the math.
> Are you converting 2512 radians to degrees or are you converting 2512*0.625*10-3?
> 
> If you don't get a better answer, I'll take another shot at it when I get home.
> ...



Converting 2512 radians to degrees - 2512(180/pi) = 143,927

Thanks


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## Flyer_PE (Feb 24, 2008)

NVRSTOP said:


> 16. A voltage of v(t) = (339.5V)sin2512t is applied to a 30 microF capacitor that was initially uncharged. (t is measured in seconds.) What is the approximate current in the capacitor at 0.625s?
> Solution indicates:
> 
> i(t) = C * dv(t)/dt
> ...


You need to convert the entire expression "(2512s-1)(0.625*10^10-3)" to degrees.

2512*0.625*10-3=1.57radians

1.57radians*180/pi= 89.95 deg.

I'm teaching myself how to do trig functions on excel while doing this so double-check the math.

Jim


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## NVRSTOP (Feb 24, 2008)

IFR_Pilot said:


> You need to convert the entire expression "(2512s-1)(0.625*10^10-3)" to degrees.
> 2512*0.625*10-3=1.57radians
> 
> 1.57radians*180/pi= 89.95 deg.
> ...



Nice work!

I worked it this way, but I converted the 2512 in front of the cos function to degrees and it gave me an answer of 1.27 A. Not thinking as clearly as usual. A little sleep deprived this weekend. Been hitting the books pretty hard the past two(2) weeks.

Thanks again Jim


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## nmjwolf (Feb 26, 2008)

Could someone point me to the correct spot in the EERM or the practice problems to review a few poblems like this...I can't seem to find them?

Thanks,

Jason


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## mudpuppy (Feb 27, 2008)

Chapters 30 &amp; 31 in the EERM (6th ed.) cover problems similar to this.


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