# Homework Help: Capacitors and the time constant

1. Mar 27, 2006

### dan greig

I have a question that has confused me a little,

A circuit is set up with a 10v battery,a 100 micro farad capacitor and a variable resistor. The capacitor is left to charge with the switch closed.

After the switch is opened the variable resistor is adjusted to maintain a current of 100 micro amps. This current is sustaind for 10s. After 10s no more current flows.

Calculate the charge in the capacitor the moment the switch is opened.

Calculate the capacitance of the capacitor.

some help would be much appreciated, thanks, Dan

2. Mar 27, 2006

### Euclid

Isn't the capacitance of the capacitor given?

3. Mar 27, 2006

### dan greig

it seems so, this is one of the things that has confused me!

I was thinking the question has something to do with the time constant??

The previous questions in the same area have been about the same circuit but with the switch closed, the time constant just seems like the next step in the question.

4. Mar 27, 2006

### Staff: Mentor

Well the problem is stated poorly, but whatever. The capacitance is obviously 100uF. The charge on the cap at the moment the switch is opened is related to the capacitance and that initial voltage of 10V. Do you know the equation relating the charge on a capacitor to the capaciatnce and voltage?

5. Mar 27, 2006

### dan greig

so you guys think it's as straight forward as,

Q = CV

and

C = Q/V

6. Mar 27, 2006

### Staff: Mentor

You can cross-check your answer using the 100uA current for 10s. Since I=dQ/dt, how much charge flows off the capacitor in the 10 second interval?

7. Mar 27, 2006

### nrqed

The question does not make sense as it stands...What is the circuit?? If it's only one loop and there is only one capacitor in series with a battery and a resistor, this does not make sense. I have the gut feeling there is more than one loop and more than one capacitor involved. Can you describe the circuit??

8. Mar 27, 2006

### dan greig

there is a battery, switch, variable resistor, micro ammeter in in series in a loop with the capacitor across the loop with the switch and battery to the left of it and the resistor and ammeter to the right of it.

9. Mar 27, 2006

### nrqed

So there is a single loop then? But if the switch is opened, there can't be any current at all in the circuit. So I am confused:surprised

10. Mar 27, 2006

### Staff: Mentor

I think that the cap and resistor are in parallel. The switch feeds both, so the battery charges up the cap and also provides some current to the resistor during the chargeup time. After the switch is opened, the cap discharges through the "variable resistor".

11. Mar 27, 2006

### nrqed

Ah ok. Now I see... (I had not understood "across the loop").Thanks

I stil don't understand the question asking the capacitance when it is already given. I don't understand why the current would suddenly stop after 10 seconds. And I wonder if the question is meant to imply a resistance which would be a function of time (otherwise, how to keep the current constant?).
SO I am a bit baffled.

12. Mar 27, 2006

### rahuldandekar

The question is very confusing...

I guess the 10 seconds is related to the time constant... after 5 times the time constant, very little current (0.05%) flows in the circuit. It could be that.

Since the capacitance is given, and we have the time constant (from what I assumed above), all we have yet to find is the resistance.

I am just guessing, though.

13. Mar 28, 2006

### Staff: Mentor

No, there is nothing going on with an RC time constant. The (poorly worded) problem is using a "variable resistor" to hold the discharge current constant. The variable resistance tracks the capacitor voltage to hold the 100uA constant. Check out the two ways of calculating the initial charge Q that I alluded to -- the constant output current for 10 seconds is useful for that.

14. Mar 28, 2006

### rahuldandekar

Oh, got it. Thanks. :)