Solving RC Circuit Problem: Find Capacitance from 40V Voltage

AI Thread Summary
In a series RC circuit with an 80V battery, two equal capacitors each reach 40V after a certain time, leading to questions about calculating capacitance. The discussion highlights the use of the equation V = Vmax(1-e^(-t/RC)) and the approximation t = 5RC to determine the time constant for charging. However, it is noted that without a specific time value, calculating individual capacitor values is challenging. The voltage across the capacitors approaches but never fully reaches 80V, complicating accurate capacitance measurements. The conversation emphasizes the importance of considering potential leakage currents and suggests measuring lower voltages for better accuracy in future calculations.
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I have a series rc circuit.

80 v battery, capacitor1, resistor1: 10M ohm, capacitor 2, then back to the negative end of the battery.

Capacitor 1 and 2 are equal, after a certain time, the voltage on each capacitor is 40.0v. What is the capacitance of each capacitor?

Thanks!
 
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My whole physics class could not colve this, we used the V = Vmax(1-e^(-t/RC)) equation, and Q= CV, but the Q could not be found to find C. Is there a trick to this?
 
Probably you made a mistake with the time constant. Not sure but it could be.
 
We used t = 5RC for time, that's the only thing we had.
 
54stickers said:
We used t = 5RC for time, that's the only thing we had.

I asked about the time constant. It is 5C here.

Lets forget about the resistor for a moment and consider only the capacitors and battery. What can be the maximum potential difference across both the capacitors?
 
Would that be 2q/C ? From Ctot = C/2 and q = CV
 
54stickers said:
Would that be 2q/C ? From Ctot = C/2 and q = CV

It is but what's the value of it? :smile:

You have a 80 V battery.
 
I don't know how to find q in that. Is it 40v?
 
  • #10
54stickers said:
I have a series rc circuit.

80 v battery, capacitor1, resistor1: 10M ohm, capacitor 2, then back to the negative end of the battery.

Capacitor 1 and 2 are equal, after a certain time, the voltage on each capacitor is 40.0v. What is the capacitance of each capacitor?

Thanks!

Is this the full statement of the problem? Is there any information missing (such as the "certain time" having a specified value)?
 
  • #11
That's everything, our teacher said that we should use t = 5RC when the problems say 'after sufficient time' to charge the resistors fully
 
  • #12
54stickers said:
That's everything, our teacher said that we should use t = 5RC when the problems say 'after sufficient time' to charge the resistors fully

Then there is insufficient information to calculate specific capacitor values. Any two equal valued capacitors in the circuit will eventually end with a 40V potential across each of them. The only difference will be the time constant RC that applies, that is, what the value of t = 5RC will be when the circuit has essentially reached steady state.
 
  • #13
I think you need more help ...

54stickers said:
We used t = 5RC for time, that's the only thing we had.

The equation t=5RC is an approximation. It means that after t (=5RC) the voltage will be 99% of the way towards 80V. In reality it never quite gets to 80V.

http://en.wikipedia.org/wiki/File:Series_RC_capacitor_voltage.png

So if you measure or are given t you can solve t=5RC to give C.

Then you need to remember that the calculated value for C is is the total capacitance. eg for two capacitors in series. The question asks you for the individual values so you have to work those out.

That should be sufficiently accurate for this problem. However for the future...

This method is pretty inaccurate because the voltage curve is very flat as it approaches the supply voltage 80V. Any error in measuring the voltage changes t a lot and hence C. If doing this for real you would do better to measure the time to charge to say 8V or 16V (but still using the 80V supply) and then use the more general equation for the voltage on a charging capacitor...

Vc = VCC (1-e-t/RC)

There is another potential problem with the method as well. Many large capacitors have a high leakage current. You may find that this leakage current through the 10MOhm resistor produces a noticable voltage drop...so it never gets to even 99% of 80V.
 
  • #14
Thanks everyone! I'll let the class know.
 
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