Decay in Energy in an RC circuit?

  • Thread starter oh.rry21
  • Start date
  • #1
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? :(
 

Answers and Replies

  • #2
537
2
There's an additional formula for the energy stored in a capacitor in terms of the capacitance and the voltage.
 
  • #3
19
0
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 doesnt mean i can find the time it takes for it to decay to 50%of its original energy.
 
  • #4
1,179
5
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
19
0
wait how did you know that

U_o is 1/sq(2) of Q_o?

i know theres easy algebra involved x_x haha but i dont see how you got there
 
  • #6
454
0
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.
 

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