Answer:
Explanation:
Given that,
The capacitance of the capacitor is
C = 4 µF
Discharging through a resistor of resistance
R = 2 MΩ
Time the energy stored in the capacitor be one-third of its initial energy
i.e. u(E)_ final = ⅓u(E)_initial
U / Uo = ⅓
Energy stored in a capacitor (discharging) can be determined using
U = Uo•RC•exp(-t/RC)
U / Uo = -RC exp(-t/RC)
U / Uo = ⅓
RC = 2 × 10^6 × 4 × 10^-6 = 8s
⅓ = 8 exp( -t / 8)
Divide both side by -8
1/24 = exp(-t/8)
Take In of both sides
In(1/24) = In•exp(-t/8)
-3.1781 = -t / 8
t = -3.1781 × -8
t = 25.42 seconds
It will take 25.42 seconds for he capacitor to be ⅓ of it's initial energy
