A 120Mu F capacitance is initially charged to 1050 V At t 0

A 120-Mu F capacitance is initially charged to 1050 V. At t = 0, it is connected to a 1 k ohm resistance. At what time t_2 has percent of the initial energy stored in the capacitance been dissipated in the resistance? Express your answer to four significant figures and include the appropriate units.

Solution

Initial energy is = (1/2)CV^2

0.5*120*10^-6*(1050)^2

66.15joules

50 percentage energy dispated means remains 50 having in capacitor

50% of energy is 0.5*66.15=33.075 joules

Voltage at 50 percentage energy is

33.075={0.5*C*v^2)

Then V=742.4671 volts

The discharge equation through resistance is

V=Voe^{-t/T}

T=RC= 100*120*10^-6=0.12 sec

742.4612=1050e^(-t/0.12)

Slove t = 41.5msec