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

time = 1.94

so D is correct option

Explanation:

given data

capacitor C = 1 × F

resistance R = 2 × ohms

to find out

how much time it take

solution

we know here charge that is

Q = Qo ×    ...............1

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

so

E = Q²/ 2c

E = Qo²/ 2c ×

and for initial energy here

Eo = Qo²/ 2c

and for capacitor to drop to 1/7

1/ 7 =  

take ln both side

ln 7 = 2t / RC

t =(  RC × ln7 )  / 2

time = ( 2 × × 1 × × ln 7  ) / 2

time = 1.94

so D is correct option

i think the answer is tomato juice? not entirely sure though

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