# Maple TA Quiz 20-1

## Maple TA Quiz 20-1

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

Answer is negative.

Answer is negative.

**helping**- Guest

## Re: Maple TA Quiz 20-1

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

Answer is negative.

Answer is negative.

**helping**- Guest

## Re: Maple TA Quiz 20-1

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.

Correct answer should be around 0.05.

Answer is negative.

**helping**- Guest

## Re: Maple TA Quiz 20-1

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.

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.

**helping**- Guest

## question 2

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)

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)

**guest786**- Guest

## q2

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

**sjames**- Guest

## #2 not working

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.

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.

**phystude**- Guest

## Re: Maple TA Quiz 20-1

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

**helping**- Guest

## queston 2

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

**12345**- Guest

## Question #2

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!

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

hope it helps!

**PHY000**- Guest

## question 2

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

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

**gamblerz**- Guest

## #3

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

**lost**- Guest

## 20_1 #2

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

**this wor**- Guest

## #3

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.

**it works**- Guest

## Question 3

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

I got 3.04 as my value and got it right.

I got 3.04 as my value and got it right.

**123123**- Guest

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