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Circuit: Time Constant, Net Charge in Capacitor

  • Thread starter ChelsM
  • Start date
  • #1
4
0

Homework Statement



In the circuit below, at time t = 0, the switch is closed, causing the capacitors to charge. The voltage across the battery is 12 V. **Image of circuit is attached.

A)Calculate the time constant for the RC circuit
B)Calculate the time required for the voltage across the capacitor to reach 6 V
C)What will be the net charge stored in the capacitor after 10 nanoseconds?


Homework Equations



Time constant (TC)= R*C
q= qo[1-e^(-t/TC)]

The Attempt at a Solution



A)
TC= [(C1+C2)*(R1*R2)]/(R1+R2)
TC= [(3μF+6μF)*(4kΩ*2kΩ)]/(4kΩ+2kΩ)
TC= 12 ms = 0.012 sec Correct/Incorrect???

B)
q= qo[1-e^(-t/TC)]
6= 12[1-e^(-t/12)]
t= 8.316 ms = 0.008316 sec Correct/Incorrect??? Also unsure about my set-up and math.

C) I cannot find an equation to solve for this and don't know how. Please Help!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
656
2
Dont got a calc. but time constant is something with 12 so I am going to trust your digits.

On b) the voltage drop across the capacitors at time 0 is zero. Remember when you connect this circuit no charge build up has occurred across the capacitors so there can be no pot. diff V, across the capacitors. So are you using the right exponential equation? There should be a voltage build up thru time that reaches some max. value at time = ∞; not a decrease in voltage...I see also you are using q instead of V.
And do you know the max value for voltage across the capacitors since you also have a voltage drop at the resistors?

c) And this is a conceptual question Im thinkin since it says net charge. If one + charge is placed on one plate, then one + charge leaves the other plate... yes? and the capacitor has built a small charge diff, thus a small voltage difference, this keeps happening thru time and you get a larger separation of charge, but net charge...? (This assumes conventional current as it is actually electrons acting as current)
 
Last edited:
  • #3
210
0
I get the same as you for the time constant and the time to reach 6V.
6V is half the final voltage and there is a short cut if t = T/2.
the 'half time' = 0.693xtime constant....t1/2 = Ln(2) x time constant
 
  • #4
656
2
emily has given you b basically

And you were using the right equation

And the reason you use 12V as V initial is that at t=∞ the current will approach zero (which they might have you know and uses a slightly diff exp. equation) so the resistors have no current flowing through them (or approach this) and the battery and capacitor must have the same diff in potential, only opposite in direction.

And c...
 
  • #5
4
0
Thanks for your responses. I think I might have part C. What I did was use this equation again: q= q0[1-e^(-t/TC)].

q0= CeqV
q0= (9μF)*(12V)
q0= 108 μC

q= q0[1-e^(-t/TC)]
q= 108[1-e^(-0.00001/12)] 10 nanoseconds= 0.00001 milliseconds
q= 9E-5 μC

Does this seem reasonable?
 
  • #6
656
2
I think that you are trying to calculate the amount of charge that moved after a short period of time.

Its says net charge. So if one plate has a certain negative charge, then the opposing plate must have the same + charge. So there is a voltage across the two plates, but net charge?

So I am not sure which one they want. If it said how much charge moved or the net charge on one plate, same thing, it would be clearer.
 

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