Shock Value B/Circuit Lab C Question Marathon

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wjnewhouse
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by wjnewhouse »

cellobix wrote:Since no one has posted in a while, I'll ask a question. Given the logic gate question on the previous page, show that its boolean algebraic representation Image is equivalent to the expression Image.
(Note that + signifies "OR", * signifies "AND", and (bar) signifies "NOT".)

Also, draw an equivalent logic gate circuit with one (4-input) gate and any necessary inverters.
[img]http://i.imgur.com/r6M4qwL.png[/img]
My partner is the one who posted the logic gate question on the previous page XD
Last edited by wjnewhouse on January 3rd, 2014, 9:14 pm, edited 1 time in total.
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by cellobix »

wjnewhouse is correct.
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by wjnewhouse »

Alright, let's try this one...

Image

At time t=0, the switch is closed. Prior to t=0 there was no charge on the capacitor C.

At what time does the capacitor have a charge equal to half the final charge? Express your answer in terms of R and C.

Disclaimer: My partner and I had taken a test where this question was marked wrong for us, but after reviewing the question we believe the graders marked it wrong on accident. We don't know for certain if what we have is the correct answer, but if you come up with the same answer that we got I will assume that you are correct.
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by lchs »

wjnewhouse wrote:Alright, let's try this one...

Image

At time t=0, the switch is closed. Prior to t=0 there was no charge on the capacitor C.

At what time does the capacitor have a charge equal to half the final charge? Express your answer in terms of R and C.

Disclaimer: My partner and I had taken a test where this question was marked wrong for us, but after reviewing the question we believe the graders marked it wrong on accident. We don't know for certain if what we have is the correct answer, but if you come up with the same answer that we got I will assume that you are correct.
Since the charge on a charging capacitor, Q, can be modeled by Q=CV[1-e^(-t/RC)], to solve the problem, you would have to set that equation equal to half the final charge.

The final charge is given by Q=CV so...

0.5CV = CV[1-e^(-t/RC)]      After canceling out the CV from both sides....

0.5 = 1-e^(-t/RC)                And after some algebra...

t = -RCln(0.5)                     Which I believe is the final answer!
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by wjnewhouse »

Since the charge on a charging capacitor, Q, can be modeled by Q=CV[1-e^(-t/RC)], to solve the problem, you would have to set that equation equal to half the final charge.

The final charge is given by Q=CV so...

0.5CV = CV[1-e^(-t/RC)]      After canceling out the CV from both sides....

0.5 = 1-e^(-t/RC)                And after some algebra...

t = -RCln(0.5)                     Which I believe is the final answer!
Correct! Your turn
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by tangentline »

A circuit I've made for a past challenge test for another school while still studying circuits last year.
Image

Out of the questions I've written, what people had most trouble on was finding the currents through certain resistors...
So: find the current through R8.
Just a little eye-opener question I'll be tossing out there as this question was missed by teammates on a team that made nationals.
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by Phys1cs »

3510/2999A

If you condense the 20 and 30 Ohm resistors, you get twelve ohms. [(1/20)+(1/30)]^-1
Then, you get the resistance of the "first" and "second" legs of the parallel circuit (going left)
and you have the leftmost leg as 22 ohms, and the middle leg as 130 ohms.
using the parallel circuit resistances, you add them up (1/22)+(1/130)+(1/15) to get that cumulative resistance as 2145/257
Take your voltage,and find the current going through the whole circuit, 300/((2145/257)+20+30), which equals, 15420/2999
Take that current, and multiply it by the resistance of the whole parallel circuit, (15420/2999)*(2145/257), which equals 128700/2999
Then take the first, leftmost leg of the parallel part of the circuit, (the one with R8) and multiply your current by 12(the parallel circuit containing R8), to get the total voltage drop of that parallel part. It ends up being 5850/2999.
Take that voltage drop, and divide from it the 20 Ohms of R8. (5850/2999)/20
and you end up with 3510/2999A
although I'm not positive, that is what I got
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by tangentline »

Yes, correct. I'm not sure if I found a way to do it the way AP Physics teaches you (or at least the way I was taught) by a system of equations method (which is always a waste of time for such circuit problems); it might not be possible for three branches.

~ Next question up for grabs
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by Phys1cs »

tangentline, what is the way AP physics teaches?

next question:

What quantities do these units represent?
Amps, Farads, Ohms, Henry, Volts, Coulomb, and Watt?
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Re: Shock Value B/Circuit Lab C Question Marathon

Post by tangentline »

Simple demonstration:
Take a 10 V battery and a 4 ohm resistor and a 1 ohm resistor in parallel connecting the terminals of this battery.

So you will use the parallel resistor formula to get: 4/5 ohms in total.

10V / .8 ohms = 12.5 A of current

Now we have to find some way to split this 12.5 amps of current between the two resistors.
Since more resistance means less current will go through (maybe they were hinting at some conductance conversion) the 1 ohm resistor will get a certain fraction of the 12.5 A and the 4 ohm resistor will get a lesser fraction. The 1 ohm resistor is 1/5 of the total resistance of the two resistors and the 4 ohm resistor is 4/5 of the total resistance
-> By this logic we can assume by the reciprocals of these fractions 5 and 5/4 (because apparently more will go to the one with the less resistance) will be the ratio of how much each amperage each resistor will get of the 12.5 Amps.
The fractions sum to 25/4 and 5 / (25/4) * 12.5 A will give you 10 A going through the 1 ohm resistor.

I mean, the teacher even showed a really good video tutor explaining this lengthy yet still a time waster method that the class understood well. Not sure if this method works with 3 resistors as of yet, but it's always fun to play around and try things and you may gain circuit techniques (AKA may save you from having to do a system of 3 node/mesh/whatever equations if you find a faster way) that actually make a problem easier in a different situation.
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