Could someone please tell me what equations i am suppose to use to work out these problems.
A capacitor in RC circuit has two overlapping plates of width 3.40 m and length 8.90 m separated by 2.60 cm. A dielectric of dielectric constant being 73.0 fills the space between the plates. A capacitor of 400. ìF is connected in series with a resistor of 13.0 kiloOhms, a perfect switch, and a 28.0 V battery. Initially there is no charge on the capacitor, and the switch is open.
a) What will the voltage across the capacitor be 9.00 s after the switch is closed?
2.The circuit is then left for a long time with the switch closed so that the capacitor fully charges to 28.0 V. The switch is now carefully opened so that no charge leaves the capacitor and the battery is also replaced by a resistor of resistance 11.0 kiloOhms
a) What will be the voltage across the capacitor 20.0 s after the switch is closed again?
Is thgis one problem or two?
In other words is the first capacitor in series with the second and with the resistor?
For the first capacitor:
C = e A/d = K eo A/d
here k = 73
eo = 8.85 * 10^-12
A = 3.4 * 8.9
d = .026
If the second problem is the second capacitor in series with the resistor then at the instant the switch is closed the charge on the capacitor is zero and the whole 28 volts appears across the resistor.
A current flows at that initial instant with value
Io = E/R = 28/R
that current fills the capacitor up with charge and as time goes on more voltage appears over the capacitor and less across the resistor so the current goes down. By the time the current is zero the entire 28 volts is across the capacitor.
I assume he derivation of the process is in your book and gives
i = Io e^-t/(RC)
From that compute i at t = 9
then the voltage on the resistor is
Vresistor = i R
and
Vcapacitor = 28 - Vresistor
Hey thsnks for your help, regarding the first question i found the capacitance to be 7.522e-7 F. Do i then use the formula Q=CV to work out voltage, if so how do i get Q? and how do i use the time of 9 seconds given to work out my final answer which in the textbook was 13.3V
To solve both parts of this problem, we need to use the equations for charging and discharging of a capacitor in an RC circuit:
1. Charging of a capacitor:
- Q = Q₀ * (1 - e^(-t / RC))
- Vc = Q / C
2. Discharging of a capacitor:
- Q = Q₀ * e^(-t / RC)
- Vc = Q / C
Where:
- Q is the charge on the capacitor at time t.
- Q₀ is the initial charge on the capacitor (usually 0 for a fully discharged capacitor).
- Vc is the voltage across the capacitor at time t.
- C is the capacitance of the capacitor.
- R is the resistance in the circuit.
- t is the time in seconds.
a) Let's solve part 1 of the problem:
1. First, let's calculate the time constant (τ) for the circuit:
- τ = RC = (13.0 kiloOhms) * (400. μF)
2. Now, we can calculate the charge on the capacitor at t = 9.00 s after the switch is closed:
- Q = Q₀ * (1 - e^(-t / τ))
3. Using the charge, we can determine the voltage across the capacitor at t = 9.00 s:
- Vc = Q / C
b) Now, let's solve part 2 of the problem:
1. The circuit is fully charged, so we can use the discharging equation to find the voltage across the capacitor at t = 20.0 s after the switch is closed again.
- Q = Q₀ * e^(-t / τ)
- Vc = Q / C
2. However, this time the battery is replaced by a resistor. We need to consider the new time constant (τ') for the circuit, which is determined by the new resistance:
- τ' = R' * C = (11.0 kiloOhms) * (400. μF)
3. Using the new time constant, we can calculate the charge on the capacitor at t = 20.0 s:
- Q = Q₀ * e^(-t / τ')
4. Finally, we find the voltage across the capacitor at t = 20.0 s:
- Vc = Q / C
By substituting the appropriate values into these equations, you can find the answers to both parts (a) and (b) of the problem.