Decay in Energy in an RC circuit?

In summary: Q and got Q = 1/sq(2)*Q_o. In summary, the problem at hand is finding the time T when a capacitor loses 50% of its energy. The given information includes the voltage, capacitance, and resistance. The equations involved are I=I_o(e^(-t/RC)), Q=Q_o(e^(-t/RC)), and Potential Energy = Q^2/2C. The approach to solving the problem is to use the equation for potential energy and relate it to the equations for current and charge in an RC circuit. By setting the potential energy at t=0 to be equal to half of the initial energy, and solving for the corresponding charge,
  • #1
oh.rry21
19
0

Homework Statement



They want us to find T, time, when the capacitor has lost 50% of its energy. They give us voltage, capacitance, and resistance.



Homework Equations



I=I_o(e^(-t/RC))
Q=Q_o(e^(-t/RC))

Potential Energy = Q^2/2C

The Attempt at a Solution



I have no idea how to relate charge/current decay to energy in an RC circuit. Does anyone have any ideas? :(
 
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  • #2
There's an additional formula for the energy stored in a capacitor in terms of the capacitance and the voltage.
 
  • #3
but even if i had that equation...how would i use it in relation to the RC equations? just because i have an energy equation in terms of capacitance and voltage doesn't mean i can find the time it takes for it to decay to 50%of its original energy.
 
  • #4
The charge on the capacitor is changing, in which the U = Q^2/2C. Since C is constant, the only variable you need to take account for is Q. If Q_o is the initial charge (and largest charge), it contributes to the highest potential energy, U_o of the capacitor. Half of U_o occors when about 1/sqr(2) of Q_o is left, in which sqr() is square root. Using Q=Q_o(e^(-t/RC)) for Q = (1/sqr(2)*Q_o, solve for t.
 
  • #5
wait how did you know that

U_o is 1/sq(2) of Q_o?

i know there's easy algebra involved x_x haha but i don't see how you got there
 
  • #6
He used E = Q^2/2C to find the energy at t=0 when the capacitor had a charge Q_o. that is simply E = Q_o^2/2C.
then used the same equation to get the charge of the capacitor when it has half that energy. (1/2)E = (1/2)(Q_o^2/2C) = Q^2/2C.
 

1. What is an RC circuit?

An RC circuit is a type of electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series or parallel. It is commonly used in electronic devices to control the flow of current and store electrical energy.

2. How does energy decay in an RC circuit?

Energy decay in an RC circuit occurs when the capacitor discharges its stored energy through the resistor. This process is governed by the time constant, which is the product of the resistance and capacitance in the circuit. As time passes, the energy in the circuit decreases until it reaches a steady state.

3. What factors affect the rate of energy decay in an RC circuit?

The rate of energy decay in an RC circuit depends on the values of the resistor and capacitor, as well as the voltage across the capacitor. A higher resistance or capacitance will result in a slower decay, while a higher voltage will result in a faster decay.

4. How can the energy decay in an RC circuit be calculated?

The energy decay in an RC circuit can be calculated using the formula E = 0.5CV2, where E is the energy, C is the capacitance, and V is the voltage across the capacitor. This formula assumes that the capacitor is initially fully charged.

5. What are some real-life applications of RC circuits?

RC circuits have many practical applications, such as in filters used in electronic devices to remove unwanted frequencies from signals. They are also used in timing circuits, such as in the flashing lights of a car turn signal. Additionally, RC circuits are commonly used in power supplies to smooth out fluctuations in voltage.

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