Discharging a capacitor

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In summary: Voltage is the measure of the energy in a circuit. It is measured in volts and is usually represented by the symbol V. The unit of voltage is the joule per coulomb. When you have a lot of charge moving through a small area, the voltage can be high. When you have a lot of charge moving through a large area, the voltage can be low.
  • #1
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The capacitor C = 3.00 microF at t = 0 has a stored energy U0 = 1.67 x 10^-7 J. At t = 0
the switch S is closed and the capacitor discharges through resistor R = 8700 Ohms. What's the initial charge of the capacitor?

I'm doing q0 = iRC. I'm stuck on finding what the current through the circuit is.
 
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  • #2
I=Cdu/dt. You have to find the voltage on the capacitor.Since the energy before and after the commutation is equal W(0-)=W(0+). You have a relation about energy W=C*U^2/2. You find the intial value for voltage.The final value of the capacitor`s voltage is zero,because it has been discharged.So u(t)=U(0+)*e^-t/T. T=R*C-time constant for discharging the capacitor. and Qo=C*U(0-). Think that is enough
 
  • #3
What do the symbols 0- and 0+ mean?
 
  • #4
0- means before the commutation and 0+ after commutation. In capacitor the voltage before and after are equal U(0-)=U(0+).Or there would be a infinite change in voltage du/dt.
 
  • #5
There are 3 versions of the equation for the energy stored on a charged capacitor :
0.5 x Q x V; 0.5 x C x V^2 ; 0.5 Q^2 /C
You could use the third one directly to find the charge, Q, on the capacitor.
You can now also calculate the voltage across the capacitor, this is the voltage at t = 0.
When S is closed the voltage is connected across the 8700Ω resistor so you should be able to calculate the initial current.
The current (and voltage) decrease exponentially ( I x e^-t/T ; and V x e^-t/T )
T is the time constant of the C R combination
 
  • #6
K, thanks for the equations. I was trying to relate the voltage and the potential energy somehow, and I thought maybe U = qV. But that's only for particles, right?
 
  • #7
U = qV is certainly the way to calculate the energy of particles with charge 'q' moving through a potential difference (voltage) of V.
You have probably met eV (electron volt) as a unit of energy when the charge is that of an electron (e = 1.6 x 10^-19 C)
A really useful thing to know is that Voltage means Joule per Coulomb.
1 Volt means 1 Joule of energy given to 1Coulomb of charge when it passes through 1Volt
It will help you in lots of electricity problems if you translate 'Volt' to be 'Joule per coulomb'
 
  • #8
Thanks. Units do indeed go a long way.
 

1. How do you discharge a capacitor?

To discharge a capacitor, you can simply connect a resistor or wire between the two terminals of the capacitor. This will allow the stored energy in the capacitor to dissipate.

2. Why is it important to discharge a capacitor?

It is important to discharge a capacitor before handling it or working on a circuit that contains a capacitor. This is because capacitors can store a significant amount of electrical energy, and if not discharged properly, can cause electric shocks or damage to the circuit.

3. How long does it take for a capacitor to discharge?

The time it takes for a capacitor to discharge depends on the capacitance and resistance of the circuit. A larger capacitor with a higher capacitance will take longer to discharge, while a smaller capacitor with a lower capacitance will discharge more quickly.

4. What happens if a capacitor is not discharged?

If a capacitor is not discharged, it will still hold a charge and can potentially cause harm to anyone who comes in contact with it. It can also cause damage to the circuit or other components if it is not properly discharged before handling.

5. Can you discharge a capacitor without a resistor?

Yes, you can discharge a capacitor without a resistor, but it is not recommended. Without a resistor, the discharge can happen very quickly and can cause sparks or damage to the circuit. It is best to use a resistor to control the discharge and ensure safety.

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