Maple TA Quiz 20-1

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Maple TA Quiz 20-1

Post  helping on Mon Mar 02, 2009 7:48 pm

Question 1: (Value in microF)*(Value in kiloOhm)*(10^-3)

Answer is negative.

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Re: Maple TA Quiz 20-1

Post  helping on Mon Mar 02, 2009 7:54 pm

Question 4: (Value in V)/(e^-((Value in ms*10^-3)/(Value in kiliOhm*Value in nF*10^-6)))

Answer is negative.

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Q2 and Q3

Post  sck105 on Tue Mar 03, 2009 11:41 am

Any ideas on q2 and q3?

sck105
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#2 and #3

Post  sjames on Tue Mar 03, 2009 6:53 pm

I cant figure out question 2 and 3 either can someone help plz?

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Re: Maple TA Quiz 20-1

Post  helping on Tue Mar 03, 2009 10:40 pm

Question 2: ((Value in kV*10^3)*(-((Value in s)/(Value in mF*Value in kOhm))))/(Value in kOhm*10^3)

Correct answer should be around 0.05.

Answer is negative.

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Re: Maple TA Quiz 20-1

Post  helping on Tue Mar 03, 2009 11:00 pm

Question 3: ln(1-(% provided/100))*(Value in microseconds)*10^-6

Answer is positive.

Sorry about the wait for these two. I was a bit tired the other day and simply had a bit of trouble with these (2,3) so I left them for today.

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question 2

Post  physics1 on Wed Mar 04, 2009 12:36 am

Sry helping, but #2 did not work for me.

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question 2

Post  guest786 on Wed Mar 04, 2009 12:49 am

I am having problems with this question, i used this equation provided above: ((Value in kV*10^3)*(-((Value in s)/(Value in mF*Value in kOhm))))/(Value in kOhm*10^3) and i am still not getting it right...

What is the discharging current in an RC circuit with a 0.2589 resistor and an initial voltage of 1.645 kV across a 50 mF capacitor after 27.49 s from the begin of the discharge (see sheet 8,8',9) ? Indicate with a negative (positive) sign whether the charging and discharging current have (do not have) the same functional dependence on time.

this is what i did:

((1.645*10^-3)*(-((27.49)/(50*.2589))))/(.2589*10^-3)

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q2

Post  sjames on Wed Mar 04, 2009 12:58 am

I cant figure out 2 either and I am following the method provided above anyone know how solve this question?

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Re: Maple TA Quiz 20-1

Post  helping on Wed Mar 04, 2009 1:06 am

Check your signs. You have to -3's where it should be positive.

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question 2

Post  guest786 on Wed Mar 04, 2009 1:12 am

i changed my signs and still no good

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#2 not working

Post  phystude on Wed Mar 04, 2009 1:43 am

hi helping. #2 is not working for me either. Could you take a shot at solving my problem for #2? I keep getting -20.96 as my answer when i do it the way you showed above but its incorrect when i put it in maple. i'd really appreciate it if you could help. thanks!

What is the discharging current in an RC circuit with a 0.1791 resistor and an initial voltage of 1.24 kV across a 50 mF capacitor after 27.11 s from the begin of the discharge (see sheet 8,8',9) ? Indicate with a negative (positive) sign whether the charging and discharging current have (do not have) the same functional dependence on time.

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Re: Maple TA Quiz 20-1

Post  helping on Wed Mar 04, 2009 1:51 am

My mistake. I believe I left out the e^ before the - sign once you open the parentheses.

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queston 2

Post  12345 on Wed Mar 04, 2009 2:02 am

followed the method above for #2 n i still don't get it >< thanks if anyone can help!

12345
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Question #2

Post  PHY000 on Wed Mar 04, 2009 11:46 am

The method above should work if done properly, however, your answer will not necessarily be around 0.05. Mine was actually 0.75.

((Value in kV*10^3)*(e^(-((Value in s)/(Value in mF*Value in kOhm)))/(Value in kOhm*10^3))

hope it helps!

PHY000
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question 2

Post  gamblerz on Wed Mar 04, 2009 3:00 pm

use the equation Vc=Vo*e^(-t/RC)
remember e is actually one of the functions on the graphing calculator such that ln(e^x) will give you x
once you solved for Vc then use the equation Vc=Ic*R
remember Vc is the voltage discharge which is what you are solving for and Vo is just the initial voltage and Ic is just the current discharged from Vc

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#3

Post  lost on Thu Mar 05, 2009 9:02 pm

i still couldnt get #3 with the equation provided, my answer came out to be negative. helppp

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20_1 #2

Post  this wor on Fri Mar 06, 2009 5:00 am

use equation 20.5'. This gives you voltage...then divide by the resistance they give you to get the current. FORMULA: V(given)*e^(-t/RC)--> then divide by R

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#3

Post  it works on Fri Mar 06, 2009 5:03 am

the power of e is negative. So when you do ln and the e comes off....you have -. Plus, ln(1-%/100) is negative. Multiply the 2 and you get positive.

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ques 4

Post  guest200 on Fri Mar 06, 2009 1:07 pm

anyone understand 4?

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Question 3

Post  123123 on Fri Mar 06, 2009 3:14 pm

Don't be bothered with the value of 0.05 or whatever.

I got 3.04 as my value and got it right.

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Re: Maple TA Quiz 20-1

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