Solving the Capacitance Circuit: Time Constant & Charging Process

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    Ftl
In summary, the conversation discusses the time constant and charge on a capacitor in a circuit with resistors and capacitors in series and parallel. It also asks about the effect of changing the resistances and capacitances on the charging process. The charge constant is given by RC, where R is the total resistance and C is the total capacitance. The voltage across the parallel combination of capacitors is the same and the charge on C2 can be found using the charge distribution formula. Changing the resistances and capacitances will affect the time constant of the circuit, with a decrease in capacitance resulting in a faster charging process.
  • #1
yikes_physics
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capcitance again FTL :(

Homework Statement



All capacitors of the open circuit in Figure 3 are discharged at time t = 0, when the switch
S is turned on and capacitors starts to be charged. What is the time constant [tex]\tau[/tex] of the charging process? What is the charge on capacitor C2 at an arbitrary time t > 0? If the resistances R1 and R2 are increased by a factor of 2, while capacitances C1 and C2 are decreased by a factor of 2, would the charging process occur faster or slower?


Homework Equations


Resistors in sereis = R1 + R2 +...
capacitors in series = C1 + C2 + ...
capacitors in parallel = 1/C1 + 1/C2 + ...
Voltage across a capacitor = V(o)e^-(delta t/[tex]\tau[/tex])
capacitor discharging = I(o)e^-(t/[tex]\tau[/tex]) = Q(o)e^-(t/[tex]\tau[/tex])


The Attempt at a Solution


ok, bak with another one of these variable infested things, lol. First, i find R(eq) by adding the resistors in series and find Ceq by first combining the capacitors in series and then adding them in parallel. Now i have R(eq) and C(eq). Multiplying these together nets me [tex]\tau[/tex], correct?

if that's right, i then take [tex]\tau[/tex] and place it in the equation V(o)e^-(delta t/[tex]\tau[/tex]) where my voltage is half of C(eq). This is because both pieces of parallel capacitors get equal voltage.

As for the final part of the question, i would assume that the charge occurs faster due to the lower capacitance, but I am not sure by how much. since the resistors add in series, they simple become 2R(eq) while C(eq) goes up much more, or is this wrong to assume? any direction would be awesome.

ps. you guys are THE WIN...end of story
 
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  • #2
It would be great if you could post your Figure 3.
 
  • #3
doh! sry, i meant to put it as an attachment, but i guess my comp skills are down there with my physics ability, lol. here it is.
 

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  • #4
Charge constant is given by RC, where R is the total resistance in the circuit and C is the total capacitance in the circuit.
For the second part, at any instant voltage across the parallel combination of capacitors is the same. Using charge distribution formula you can find the charge on C2.
For third part, again find the total resistance and capacitor in the circuit and hence find the time constant.
 
  • #5
You switched the series/parallel capacitance equations, careful. I don't know if it was a typo or not, but I thought I would point that out.
 

Related to Solving the Capacitance Circuit: Time Constant & Charging Process

1. What is the time constant of a capacitance circuit and how is it calculated?

The time constant of a capacitance circuit is the amount of time it takes for the voltage across the capacitor to change by a factor of approximately 63% of its initial value. It is calculated by multiplying the resistance (R) in the circuit by the capacitance (C), or τ = RC.

2. How does the charging process of a capacitance circuit work?

When a voltage source is connected to a capacitance circuit, the capacitor begins to charge. Electrons flow from the voltage source to one plate of the capacitor, and an equal number of electrons are pushed from the other plate to the other side of the circuit. This creates an electric field between the plates, which stores energy in the form of an electric charge. As the capacitor charges, the voltage across it increases until it reaches the same voltage as the source.

3. What is the relationship between the time constant and the charging process in a capacitance circuit?

The time constant determines how quickly the capacitor charges. A smaller time constant means the capacitor will charge more quickly, while a larger time constant means it will take longer to charge. This relationship is described by the equation V(t) = V₀(1 - e^(-t/τ)), where V(t) is the voltage at a given time, V₀ is the initial voltage, and τ is the time constant.

4. How does the capacitance affect the time constant and charging process in a circuit?

The capacitance directly affects the time constant in a circuit. A larger capacitance value will result in a longer time constant, meaning it will take longer for the capacitor to charge. This is because a larger capacitance can hold more charge, so it takes more time for the capacitor to reach its maximum charge. However, the capacitance does not affect the rate at which the capacitor charges, only the time it takes to reach a certain voltage.

5. What is the practical application of understanding the time constant and charging process in a capacitance circuit?

A thorough understanding of the time constant and charging process in a capacitance circuit is essential for designing and analyzing electronic circuits. It allows engineers to accurately predict the behavior of the circuit and determine the appropriate components to use. This knowledge is also crucial for troubleshooting and repairing circuit failures.

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