Question

While discharging a fully charged capacitor C=1x10^-6 F through a resistance R=2x10^6 ohms, how much time (in seconds) will it take for the energy of the capacitor to drop to 1/7 of the initial energy? A) 2.00

B) 10.0
C) None of these
D) 1.90
E) 0.50

Answer 1

time = 1.94

so D is correct option

Step-by-step explanation:

given data

capacitor C = 1 ×
`10^(-6)` F

resistance R = 2 ×
`10^(6)` ohms

to find out

how much time it take

solution

we know here charge that is

Q = Qo ×
`e^(-t/RC)` ...............1

here RC is time constant and q is charge and t is time

so

E = Q²/ 2c

E = Qo²/ 2c ×
`e^(-2t/RC)`

and for initial energy here

Eo = Qo²/ 2c

and for capacitor to drop to 1/7

1/ 7 =
`e^(-2t/RC)`

take ln both side

ln 7 = 2t / RC

t =( RC × ln7 ) / 2

time = ( 2 ×
`10^(6)` × 1 ×
`10^(-6)` × ln 7 ) / 2

time = 1.94

so D is correct option