Calculating Voltage Across a Capacitor in a Series Circuit

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Homework Help Overview

The problem involves calculating the voltage across a capacitor in a series circuit with a resistor, switch, and battery. The capacitor has a specified capacitance, and the scenario begins with no charge on the capacitor before the switch is closed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the current equation for a charging capacitor and question the accuracy of the calculated time constant. There are attempts to clarify the correct values for capacitance and resistance, as well as the interpretation of voltage across the capacitor versus the resistor.

Discussion Status

Some participants have pointed out potential errors in the original poster's calculations and assumptions, particularly regarding the powers of ten and the distinction between voltage across the capacitor and the resistor. Guidance has been offered on how to approach the problem, but no consensus has been reached on a final solution.

Contextual Notes

There appears to be confusion regarding the units and values used in the calculations, particularly with the capacitance and the resulting time constant. The original poster also expresses uncertainty about the correctness of their approach and results.

roam
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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:

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

[tex]RC= 820000 \times (45 \times 10^{-6}) = 36.9[/tex]

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

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

Now I use V=IR to find the voltage:

[tex](9.8 \times 10^{-6}) \times (820000) =8 \times 10^{-54}[/tex]

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...
 
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hi roam! :smile:

your RC has the wrong power of 10 :cry:
 
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
 
roam said:
First I used the equation for current as a function of time for a capacitor being charged:

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

[tex]RC= 820000 \times (45 \times 10^{-6}) = 36.9[/tex]
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)
[tex]t/RC= (4.62 \times 10^3)/36.9=125.2[/tex], so

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

Now I use V=IR to find the voltage:

[tex](9.8 \times 10^{-6}) \times (820000) =8 \times 10^{-54}[/tex]
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. :-p]
 
And that V=IR calculation is out by a factor of 10^54. Calculator error.
 
It worked. Thanks a lot guys.
 

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