# Homework Help: 2 Problems

1. Mar 23, 2004

### NINHARDCOREFAN

https://hw.utexas.edu/tmp/Muddam1/1080079116Xuj.pdf [Broken]

for #5:
I know you have to use this equation:
$$Ve^\frac{-t}{RC}$$
R=3e6+2e6=5e6
So I did $$11e^\frac{-1.5}{5e6*1e-6}= 8.149000428$$
but it's wrong and I don't know why.

for #8:
q= 3.376e-6
I= 1.79839e-6
(both values are right)
C=1.1e-6
q=CV
V=q/C
3.06909=V
P=IV
5.5194224e-6=1.79839e-6*3.06909
because the answer should be in to the 1e-6 I didvided the answer by 1e-6, so my final answer is 5.5194224, which is also wrong.

Last edited by a moderator: May 1, 2017
2. Mar 23, 2004

### Staff: Mentor

First Problem

It looks like you found the voltage across the capacitor at t=1.5 sec, but the problem asks for the voltage across the 3M Ohm resistor. Hint: find the current, then use Ohm's law.

3. Mar 23, 2004

### NINHARDCOREFAN

So you want me to find current using this equation:
$$\frac{V}{R}e^\frac{-t}{RC}$$
I should just use 3e6 for R? Then use V=IR?

4. Mar 23, 2004

### Staff: Mentor

Second Problem

This problem asks for the power delivered by the battery, which is always P=IV, where V is the voltage of the battery.

5. Mar 23, 2004

### Staff: Mentor

Yes, where R is the total resistance.
To find the voltage across a resistor, use V=IR for that resistor.

6. Mar 23, 2004

### NINHARDCOREFAN

...

Thanks a LOT!