RC Circuit discharge question -

AI Thread Summary
A charged 6μF capacitor discharges through a 2Ω resistor, and the question is how long it takes to lose half its stored energy. The energy stored in the capacitor is calculated using E = 1/2 * C * V0^2. To find the time for energy to reduce to E/2, the voltage at that point is determined to be V0/sqrt(2). The discussion highlights the challenge of solving for time without specific energy values, emphasizing the need for clarity in multiple-choice questions. The solution involves substituting the voltage into the discharge equation and solving for time.
jemitu
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RC Circuit discharge question - test prep for tomorrow

Homework Statement



A charged 6\muF capacitor is discharged across a 2\Omega resistor. How long does it take to lose half its stored energy?

Homework Equations


...so many:

Q = CV
V = IR
W = I2R
U = 1/2 QV
V = V0e-t/RC

The Attempt at a Solution



I'm not really sure where to start on this one. I'm trying to find a way to resolve Q and V where I only have one unknown, given C and R, but I'm having trouble. This is prep for an exam tomorrow, so any help is greatly appreciated.
 
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Hi jemitu, welcome to PF.
Energy stored in the capacitor is given by
E = 1/2*C*Vo^2.
In the expression C remains constant.
To change E to E/2, what should be the value of V in terms of Vo?
 
Thanks for responding so quickly, I'm getting a little overwhelmed here.

Given the energy equation you get Vo = sqrt(2E/C).
plugging into V(t), V= sqrt(2E/C) e^(-t/RC)

But when you do the math, you don't have the original E so you can't solve it. The question is multiple choice with values, not expressions, hence my severe confusion. I think I need a little more hand-holding.
 
V0 = sqrt(2E/C)
When E changes to E/2
V(t) = sqrt(2*E/2*1/C) = 1/Sqrt(2)*sqrt(2E/C) = V0/sqrt(2)
Substitute this in the equation and solve for t
 
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