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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

Respuesta :

Answer:

time = 1.94

so D is correct option

Explanation:

given data

capacitor C = 1 × [tex]10^{-6}[/tex] F

resistance R = 2 × [tex]10^{6}[/tex] ohms

to find out

how much time it take

solution

we know here charge that is

Q = Qo × [tex]e^{-t/RC}[/tex]    ...............1

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

so

E = Q²/ 2c

E = Qo²/ 2c × [tex]e^{-2t/RC}[/tex]

and for initial energy here

Eo = Qo²/ 2c

and for capacitor to drop to 1/7

1/ 7 =  [tex]e^{-2t/RC}[/tex]

take ln both side

ln 7 = 2t / RC

t =(  RC × ln7 )  / 2

time = ( 2 × [tex]10^{6}[/tex] × 1 × [tex]10^{-6}[/tex] × ln 7  ) / 2

time = 1.94

so D is correct option

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