Calculating Voltage Across a Capacitor in a Series Circuit

[roam]

Homework Statement

A capacitor of 4.50 × 10³ μF is connected in series with a resistor of 820. kΩ, a perfect switch, and a 19.0 V battery. Initially there is no charge on the capacitor, and the switch is open.
What will the voltage across the capacitor be 4.62 × 10³ s after the switch is closed?

The Attempt at a Solution

First I used the equation for current as a function of time for a capacitor being charged:

I(t)=\frac{\epsilon}{R}e^{-t/RC}

RC= 820000 \times (45 \times 10^{-6}) = 36.9

t/RC= (4.62 \times 10^3)/36.9=125.2, so

\frac{19}{820 \times 10^3} e^{-125.2} = 9.8 \times 10^{-6}

Now I use V=IR to find the voltage:

(9.8 \times 10^{-6}) \times (820000) =8 \times 10^{-54}

But this is the wrong, the correct answer must be 13.6 V. So what's wrong? I appreciate it if anyone could show me how to do this problem...

 

[tiny-tim]

hi roam!

your RC has the wrong power of 10

 

[Dadface]

In addition to the above it seems that you are attempting to calculate the voltage across the resistor(Vr) and not the capacitor(Vc).That's okay if you remember and use the fact that Vc+Vr=19V

 

[collinsmark]

First I used the equation for current as a function of time for a capacitor being charged:

I(t)=\frac{\epsilon}{R}e^{-t/RC}

RC= 820000 \times (45 \times 10^{-6}) = 36.9

For the capacitance, don't you mean 4.5 X 10^-3? (There is already a 10^-6 factor built into the "μ" part of μF)

t/RC= (4.62 \times 10^3)/36.9=125.2, so

\frac{19}{820 \times 10^3} e^{-125.2} = 9.8 \times 10^{-6}

Now I use V=IR to find the voltage:

(9.8 \times 10^{-6}) \times (820000) =8 \times 10^{-54}

Your time constant is off by a couple orders of magnitude, so you'll have to do the above operation again.

But using V = IR gives you the voltage drop across the resistor. In order to find the voltage across the capacitor, you'll need to subtract that from the 19 V supply.

[Edit: Oh my, looks like tiny-tim and Dadface beat me to the answer. I must learn to work/type quicker. ]

 

[vk6kro]

And that V=IR calculation is out by a factor of 10^54. Calculator error.

 

[roam]

It worked. Thanks a lot guys.