A capacitor of capacitance C=10-6 F is discharged through the resistor of resistance
R=10 Ohm. (Switch is closed at t=0) When the energy stored in the capacitor will be half
of its initial value?
2007-10-10 07:16:51 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
Energy in a capacitor is 1/2 * C * V^2
So, 1/2 the energy represents 1/sqrt(2) of the intital voltage
The cap discharges at a rate of e^(-t/RC), so at 1/2 the energy:
0.7071 = e^(-t/RC)
.
2007-10-10 08:16:26 · answer #1 · answered by tlbs101 7 · 1⤊ 1⤋
Your time constant tao is Rth*C and that is the amount of time it requires to decay to roughly 36.8% of initial charge.
2007-10-10 07:45:42 · answer #2 · answered by ? 1 · 2⤊ 1⤋
dude if you cant figure that one out... quit electronics engineering... for your own sake.
2007-10-10 08:26:01 · answer #3 · answered by Progen P 2 · 1⤊ 1⤋